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Say I have a set of integers ranging between 1-100. I will only have 5 of these integers drawn out of a hat. I want to then take those 5 integers and place them into 5 buckets guaranteed unique (without having to deduplicate or anything using something like quadratic probing). Wondering how to do that.
For example, say I have these numbers (random from 1-100):
1 5 20 50 100
I then want to take those numbers and place them into these 5 buckets:
a b c d e
Using some hash function to accomplish it. For example, perhaps like this:
hash(1) -> b
hash(5) -> a
hash(20) -> e
hash(50) -> d
hash(100) -> c
Wondering how to write the hash function so that it takes a number x from a domain of numbers D and a set of numbers D(X) from that domain, and outputs 1 bucket b from the set of buckets B.
H : D(X) -> B
Next time around I might have 6 numbers between 1 and 1,000, going into 6 buckets. So then I would need a new hash function that works using those constraints (6 numbers, 6 buckets, range 1-1,000).
The goal is as few steps as possible.
Note: The hash function for this example won't take integers in a domain larger than 10,000 lets say, as well as the size of the set of integers limited to some small number too like 1,000.
Update
Basically I am trying to get this to happen:
// var domain = [1, 2, ..., 100]
// var set = [1, 5, 20, 50, 100]
// var buckets = [1, 2, 3, 4, 5]
hash(1) // 2
hash(5) // 1
hash(20) // 5
hash(50) // 4
hash(100) // 3
function hash(integer) {
if (integer == 1) return 2
if (integer == 5) return 1
if (integer == 20) return 5
if (integer == 50) return 4
if (integer == 100) return 3
}
But I don't know how to construct that hash function dynamically.
One solution (in JavaScript) would be to just create a map like this:
var map = {
1: 2,
5: 1,
20: 5,
50: 4,
100: 3
}
But that's sort of cheating because the object in JavaScript is implemented as a hashtable underneath (or something like that). So I am looking for how to do this at a low level, just using basically what assembly gives you.
Pretty much, I want to do this:
1
5 |
| | 20
| | 50 |
| | 100 | |
[ slot1, slot2, slot3, slot4, slot5 ]
Where 1 is somehow "hashed" to go into that slot2 in an array of size 5 (that slot is arbitrary for this example), etc.
Suppose the domain of your integer values is the range from 0 to n-1, and you want the set of values [x0, x1, ..., xk-1] to map to values from 0 to k-1.
Create an array of n values containing the numbers from 0 to k-1 in roughly equal amounts, for example [a0 = 0, a1 = 1, ..., ak = 0, ..., an = n%k].
Then for each of the k values in the initial set (xi, where i = 0 .. k-1), change the k-th element of this array to i, either by direct assignment or by swapping with a value from elsewhere (taking care not to clobber a value set for a previous element of the initial set).
Then to hash a value y, just fetch the y-th value from this array.
DEMO
Here's a Javascript demo that basically implements the above algorithm, except that instead of pre-filling the array with values from 0 to k-1, it first inserts the hash values for the selected items, then fills the remaining items with the repeating sequence of numbers from 0 to k-1. You will probably get better collision resistance by using a random sequence instead of incrementing values, but I hope you get the picture.
var hash_array;
function generate_hash() {
var i, j, k;
var v = document.getElementById;
var n = document.getElementById("n").value;
// Create a new hash lookup table
hash_array = Array(n);
// Initialize every value to -1
for (i=0; i<n; i++) hash_array[i] = -1;
// Map the given values to the first k hash buckets
var initial_values = document.getElementById("init").value.split(/ +/);
k = initial_values.length;
for (i=0; i<k; i++) {
hash_array[initial_values[i]] = i;
}
// Fill the remaining buckets with values from 0 to k-1
// This could be done by selecting values randomly, but
// here we're just cycling through the values from 0 to k-1
for (i=j=0; i<hash_array.length; i++) {
if (hash_array[i] == -1) {
hash_array[i] = j;
j = (j + 1) % k;
}
}
document.getElementById("gen").innerHTML = "Hash lookup table:<br>" + hash_array.join(", ");
}
<h2>Demo</h2>
<p>Creating a hash function that works on integer values less than <i>n</i>. What is the value of <i>n</i>?<br>
<input type="number" id="n" min="6" max="100" value="20"/></p>
<p>Enter a few different values separated by spaces. These will hash to the first buckets<br/>
<input type="text" size="40" id="init" value="2 3 5 6 9"/></p>
<p id="gen"><button onclick="generate_hash(); return false">Generate hash table</button></p>
Something like this should work:
Create a set of bucket IDs and populate it ahead of hashing (assumption here is that set guarantees uniqueness). This means that you have to know in advance how many buckets you want.
For each element from the input set calculate hash(element) modulo bucketIds.size to find index of the next ID to use.
Remove the resulting bucket ID from the set of bucked IDs
Repeat (until you are done or the set of IDs is exhausted)
Feel free to inspect the noddy implementation in JS using arrays (Node8).
If you'd like a function that's not a straight map, you could also experiment with Polynomial Regression.
Here's a JavaScript example using some free code under the GNU license.
/***************************************************************************
* Copyright (C) 2018 by Paul Lutus *
* lutusp#arachnoid.com *
* *
* This program is free software; you can redistribute it and/or modify *
* it under the terms of the GNU General Public License as published by *
* the Free Software Foundation; either version 2 of the License, or *
* (at your option) any later version. *
* *
* This program is distributed in the hope that it will be useful, *
* but WITHOUT ANY WARRANTY; without even the implied warranty of *
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the *
* GNU General Public License for more details. *
* *
* You should have received a copy of the GNU General Public License *
* along with this program; if not, write to the *
* Free Software Foundation, Inc., *
* 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. *
***************************************************************************/
// classic Gauss-Jordan matrix manipulation functions
var gj = gj || {}
gj.divide = function(A, i, j, m) {
for (var q = j + 1; q < m; q++) {
A[i][q] /= A[i][j];
}
A[i][j] = 1;
}
gj.eliminate = function(A, i, j, n, m) {
for (var k = 0; k < n; k++) {
if (k != i && A[k][j] != 0) {
for (var q = j + 1; q < m; q++) {
A[k][q] -= A[k][j] * A[i][q];
}
A[k][j] = 0;
}
}
}
gj.echelonize = function(A) {
var n = A.length;
var m = A[0].length;
var i = 0;
var j = 0;
var k;
var swap;
while (i < n && j < m) {
//look for non-zero entries in col j at or below row i
k = i;
while (k < n && A[k][j] == 0) {
k++;
}
// if an entry is found at row k
if (k < n) {
// if k is not i, then swap row i with row k
if (k != i) {
swap = A[i];
A[i] = A[k];
A[k] = swap;
}
// if A[i][j] is != 1, divide row i by A[i][j]
if (A[i][j] != 1) {
gj.divide(A, i, j, m);
}
// eliminate all other non-zero entries
gj.eliminate(A, i, j, n, m);
i++;
}
j++;
}
}
// a simple data class
function Pair(x,y) {
this.x = x;
this.y = y;
};
Pair.prototype.toString = function() {return x + ',' + y};
// matrix functions
var matf = matf || {}
// a weak substitue for printf()
matf.number_format = function(n,p,w) {
s = n.toExponential(p);
while(s.length < w) {
s = ' ' + s;
}
return s;
}
// produce a single y result for a given x
matf.regress = function(x, terms) {
var y = 0;
var m = 1;
for (var i = 0; i < terms.length;i++) {
y += terms[i] * m;
m *= x;
}
return y;
}
// compute correlation coefficient
matf.corr_coeff = function(data, terms) {
var r = 0;
var n = data.length;
var sx = 0;
var sx2 = 0, sy = 0, sy2 = 0, sxy = 0;
var x, y;
for (var i = 0;i < data.length;i++) {
pr = data[i];
var x = matf.regress(pr.x, terms);
var y = pr.y;
sx += x;
sy += y;
sxy += x * y;
sx2 += x * x;
sy2 += y * y;
}
var div = Math.sqrt((sx2 - (sx * sx) / n) * (sy2 - (sy * sy) / n));
if (div != 0) {
r = Math.pow((sxy - (sx * sy) / n) / div, 2);
}
return r;
}
// compute standard error
matf.std_error = function(data, terms) {
var r = 0;
var n = data.length;
if (n > 2) {
var a = 0;
for (var i = 0;i < data.length;i++) {
pr = data[i];
a += Math.pow((matf.regress(pr.x, terms) - pr.y), 2);
}
r = Math.sqrt(a / (n - 2));
}
return r;
}
// create regression coefficients
// for provided data set
// data = pair array
// p = polynomial degree
matf.compute_coefficients = function(data, p) {
p += 1;
var n = data.length;
var r, c;
var rs = 2 * p - 1;
//
// by request: read each datum only once
// not the most efficient processing method
// but required if the data set is huge
//
// create square matrix with added RH column
m = Array();
for (var i = 0; i < p; i++) {
mm = Array();
for (var j = 0; j <= p; j++) {
mm[j] = 0;
}
m[i] = mm;
}
//double[][] m = new double[p][p + 1];
// create array of precalculated matrix data
mpc = Array();
for(var i = 0;i < rs;i++) {
mpc[i] = 0;
}
mpc[0] = n;
for (var i = 0;i < data.length;i++) {
pr = data[i];
// process precalculation array
for (r = 1; r < rs; r++) {
mpc[r] += Math.pow(pr.x, r);
}
// process RH column cells
m[0][p] += pr.y;
for (r = 1; r < p; r++) {
m[r][p] += Math.pow(pr.x, r) * pr.y;
}
}
// populate square matrix section
for (r = 0; r < p; r++) {
for (c = 0; c < p; c++) {
m[r][c] = mpc[r + c];
}
}
// reduce matrix
gj.echelonize(m);
// extract result column
terms = Array();
for (var i = 0;i < m.length;i++) {
mc = m[i];
terms[i] = mc[p];
}
return terms;
}
// test the system using known data
matf.test = function() {
var xd = [-1,0,1,2,3,5,7,9];
var yd = [-1,3,2.5,5,4,2,5,4];
data = Array();
for(var i = 0;i < xd.length;i++) {
data[i] = new Pair(xd[i],yd[i]);
}
terms = compute_coefficients(data,6);
var prec = 16;
var width = 24;
for(var i = 0;i < terms.length;i++) {
print(number_format(terms[i],prec,width) + ' * x^' + i);
}
cc = corr_coeff(data,terms);
print ('cc = ' + number_format(cc,prec,width));
se = std_error(data,terms);
print('se = ' + number_format(se,prec,width));
}
//test();
// "data" is an array of Pair(x,y) data
// p = polynomial degree
matf.process_data = function(data,p) {
var terms = matf.compute_coefficients(data,p);
var cc = matf.corr_coeff(data,terms);
var se = matf.std_error(data,terms);
return [terms,cc,se];
}
/**** END Paul Lutus' code ****/
function f(cs, x){
let n = cs.length - 1;
let result = 0;
for (let i=0; i<cs.length; i++)
result += cs[i] * Math.pow(x, i);
return result;
}
var data = [[1,1], [5,2], [20,3], [50,4], [100,5]];
var xy_data = []
for (let i of data)
xy_data.push(new Pair(i[0], i[1]));
var result = matf.process_data(xy_data, xy_data.length - 1);
for (let i=0; i<data.length; i++)
console.log(data[i][0], f(result[0], data[i][0]));
My solution scored only 40% correctness on Codility.
What am I doing wrong?
Here is the test result (https://codility.com/demo/results/trainingU7KSSG-YNX/)
Problem:
A zero-indexed array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.
Your goal is to find that missing element.
Solution:
function solution(A) {
var output = 1;
var arrayLength = A.length;
if(!arrayLength){
return output;
}
if(arrayLength == 1) {
return A[0] + 1;
}
var sorted = A.sort(sortingFn);
for(var i = 0; i < A.length - 1; i++) {
if(A[i+1] - A[i] > 1) {
output = A[i] + 1;
break;
}
}
return output;
}
function sortingFn(a, b) {
return a - b;
}
Result
Your algorithm find the missing element by comparing neighboring elements in the array. This means it is incapable of handling cases where the first or last element is missing, as these only have a single neighbor.
Consider as an example [1, 2, 3]. The missing element would be 4. But since 4 has precisely one neighbor (3), it can't be found by the algorithm, and 1 will be returned.
In addition your algorithm is rather inefficient, as sorting takes O(n lg n), while the problem is solveable in O(n):
find_missing(arr):
s = sum(arr)
s' = (len(arr) + 1) * (len(arr) + 2) / 2
return s' - s
This code works by summing up all elements in the array and comparing it to expected sum, if all elements were present. The advantage of this approach is that it only requires linear operations and will find the missing element with relative simplicity.
Try this in c#:
using System;
using System.Linq;
private static int PermMissingElem(int[] A)
{
if (!A.Any() || !A.Any(x => x == 1)) { return 1; }
var size = A.Length;
var numberTwoList = Enumerable.Range(1, size);
var failNumber = numberTwoList.Except(A);
if (!failNumber.Any()) { return A.Max() + 1; }
return failNumber.FirstOrDefault();
}
Well, when the last element is missing, you obviously return 1, since your if statement's condition is always false. Same for first element.
Take as example this input:
1 2 3 4 5
the difference will be always 1, but element 6 is missing.
The reason for this incapability of your algorithm to catch these cases, is that it examines neighboring elements (A[i + 1] and A[i]).
JS solution #1
function solution(A) {
if (A.length === 1) {
return A[0] > 1 ? 1 : 2;
}
const map = {};
let max = 0;
for (let i = 0, len = A.length; i < len; i++) {
map[A[i]] = A[i];
if (A[i] > max) {
max = A[i]
}
}
for (let i = 0, len = A.length; i < len; i++) {
if (!map[i + 1]) {
return i + 1;
}
}
return max + 1
}
JS solution #2
function solution(A) {
const s = A.reduce((a, b) => {return a + b}, 0);
const s2 = (A.length + 1) * (A.length + 2) / 2
return s2 - s;
}
try this arrays (like the end test):
[1,2,3] -> must return 4;
[1] -> must return 2;
[2] -> must return 1;
[2,3] -> must return 1;
[1, 3] -> 2
But for #2 solution [4] returns -1 and for [123] returns -120. The test will show 100 points. But actually, it doesn't work as expected on my opinion.
Both solutions work with the same performance.
Try this javascript function:
function solution(A) {
let givenSum = 0;
let expectedSum = 0;
let size = A.length;
for(let i = 1; i <= size +1; i++){
expectedSum = expectedSum + i;
}
for(let i = 0; i < size; i++){
givenSum += A[i];
}
return expectedSum - givenSum;
}
here is my solution:
https://app.codility.com/demo/results/trainingMZWVVT-55Y/
function solution(A) {
A = A.sort((a,b)=>a-b)
if(A[0]!==1) return 1
for(let i = 0; i < A.length; i++)
{
if(A[i+1]-A[i]!==1) return A[i] + 1
}
return A[A.length] + 1
}
Tested on Codility with 100% score see here
The solution i implemented is using set difference. since the question guaranties exactly one element is missing.
def solution(A):
# write your code in Python 3.6
N = len(A)
difference = set(range(1, N+2)) - set(A)
return difference.pop()
Given a sorted list of numbers, I would like to find the longest subsequence where the differences between successive elements are geometrically increasing. So if the list is
1, 2, 3, 4, 7, 15, 27, 30, 31, 81
then the subsequence is 1, 3, 7, 15, 31. Alternatively consider 1, 2, 5, 6, 11, 15, 23, 41, 47 which has subsequence 5, 11, 23, 47 with a = 3 and k = 2.
Can this be solved in O(n2) time? Where n is the length of the list.
I am interested both in the general case where the progression of differences is ak, ak2, ak3, etc., where both a and k are integers, and in the special case where a = 1, so the progression of difference is k, k2, k3, etc.
Update
I have made an improvement of the algorithm that it takes an average of O(M + N^2) and memory needs of O(M+N). Mainly is the same that the protocol described below, but to calculate the possible factors A,K for ech diference D, I preload a table. This table takes less than a second to be constructed for M=10^7.
I have made a C implementation that takes less than 10minutes to solve N=10^5 diferent random integer elements.
Here is the source code in C: To execute just do: gcc -O3 -o findgeo findgeo.c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <memory.h>
#include <time.h>
struct Factor {
int a;
int k;
struct Factor *next;
};
struct Factor *factors = 0;
int factorsL=0;
void ConstructFactors(int R) {
int a,k,C;
int R2;
struct Factor *f;
float seconds;
clock_t end;
clock_t start = clock();
if (factors) free(factors);
factors = malloc (sizeof(struct Factor) *((R>>1) + 1));
R2 = R>>1 ;
for (a=0;a<=R2;a++) {
factors[a].a= a;
factors[a].k=1;
factors[a].next=NULL;
}
factorsL=R2+1;
R2 = floor(sqrt(R));
for (k=2; k<=R2; k++) {
a=1;
C=a*k*(k+1);
while (C<R) {
C >>= 1;
f=malloc(sizeof(struct Factor));
*f=factors[C];
factors[C].a=a;
factors[C].k=k;
factors[C].next=f;
a++;
C=a*k*(k+1);
}
}
end = clock();
seconds = (float)(end - start) / CLOCKS_PER_SEC;
printf("Construct Table: %f\n",seconds);
}
void DestructFactors() {
int i;
struct Factor *f;
for (i=0;i<factorsL;i++) {
while (factors[i].next) {
f=factors[i].next->next;
free(factors[i].next);
factors[i].next=f;
}
}
free(factors);
factors=NULL;
factorsL=0;
}
int ipow(int base, int exp)
{
int result = 1;
while (exp)
{
if (exp & 1)
result *= base;
exp >>= 1;
base *= base;
}
return result;
}
void findGeo(int **bestSolution, int *bestSolutionL,int *Arr, int L) {
int i,j,D;
int mustExistToBeBetter;
int R=Arr[L-1]-Arr[0];
int *possibleSolution;
int possibleSolutionL=0;
int exp;
int NextVal;
int idx;
int kMax,aMax;
float seconds;
clock_t end;
clock_t start = clock();
kMax = floor(sqrt(R));
aMax = floor(R/2);
ConstructFactors(R);
*bestSolutionL=2;
*bestSolution=malloc(0);
possibleSolution = malloc(sizeof(int)*(R+1));
struct Factor *f;
int *H=malloc(sizeof(int)*(R+1));
memset(H,0, sizeof(int)*(R+1));
for (i=0;i<L;i++) {
H[ Arr[i]-Arr[0] ]=1;
}
for (i=0; i<L-2;i++) {
for (j=i+2; j<L; j++) {
D=Arr[j]-Arr[i];
if (D & 1) continue;
f = factors + (D >>1);
while (f) {
idx=Arr[i] + f->a * f->k - Arr[0];
if ((f->k <= kMax)&& (f->a<aMax)&&(idx<=R)&&H[idx]) {
if (f->k ==1) {
mustExistToBeBetter = Arr[i] + f->a * (*bestSolutionL);
} else {
mustExistToBeBetter = Arr[i] + f->a * f->k * (ipow(f->k,*bestSolutionL) - 1)/(f->k-1);
}
if (mustExistToBeBetter< Arr[L-1]+1) {
idx= floor(mustExistToBeBetter - Arr[0]);
} else {
idx = R+1;
}
if ((idx<=R)&&H[idx]) {
possibleSolution[0]=Arr[i];
possibleSolution[1]=Arr[i] + f->a*f->k;
possibleSolution[2]=Arr[j];
possibleSolutionL=3;
exp = f->k * f->k * f->k;
NextVal = Arr[j] + f->a * exp;
idx=NextVal - Arr[0];
while ( (idx<=R) && H[idx]) {
possibleSolution[possibleSolutionL]=NextVal;
possibleSolutionL++;
exp = exp * f->k;
NextVal = NextVal + f->a * exp;
idx=NextVal - Arr[0];
}
if (possibleSolutionL > *bestSolutionL) {
free(*bestSolution);
*bestSolution = possibleSolution;
possibleSolution = malloc(sizeof(int)*(R+1));
*bestSolutionL=possibleSolutionL;
kMax= floor( pow (R, 1/ (*bestSolutionL) ));
aMax= floor(R / (*bestSolutionL));
}
}
}
f=f->next;
}
}
}
if (*bestSolutionL == 2) {
free(*bestSolution);
possibleSolutionL=0;
for (i=0; (i<2)&&(i<L); i++ ) {
possibleSolution[possibleSolutionL]=Arr[i];
possibleSolutionL++;
}
*bestSolution = possibleSolution;
*bestSolutionL=possibleSolutionL;
} else {
free(possibleSolution);
}
DestructFactors();
free(H);
end = clock();
seconds = (float)(end - start) / CLOCKS_PER_SEC;
printf("findGeo: %f\n",seconds);
}
int compareInt (const void * a, const void * b)
{
return *(int *)a - *(int *)b;
}
int main(void) {
int N=100000;
int R=10000000;
int *A = malloc(sizeof(int)*N);
int *Sol;
int SolL;
int i;
int *S=malloc(sizeof(int)*R);
for (i=0;i<R;i++) S[i]=i+1;
for (i=0;i<N;i++) {
int r = rand() % (R-i);
A[i]=S[r];
S[r]=S[R-i-1];
}
free(S);
qsort(A,N,sizeof(int),compareInt);
/*
int step = floor(R/N);
A[0]=1;
for (i=1;i<N;i++) {
A[i]=A[i-1]+step;
}
*/
findGeo(&Sol,&SolL,A,N);
printf("[");
for (i=0;i<SolL;i++) {
if (i>0) printf(",");
printf("%d",Sol[i]);
}
printf("]\n");
printf("Size: %d\n",SolL);
free(Sol);
free(A);
return EXIT_SUCCESS;
}
Demostration
I will try to demonstrate that the algorithm that I proposed is in average for an equally distributed random sequence. I’m not a mathematician and I am not used to do this kind of demonstrations, so please fill free to correct me any error that you can see.
There are 4 indented loops, the two firsts are the N^2 factor. The M is for the calculation of the possible factors table).
The third loop is executed only once in average for each pair. You can see this checking the size of the pre-calculated factors table. It’s size is M when N->inf. So the average steps for each pair is M/M=1.
So the proof happens to check that the forth loop. (The one that traverses the good made sequences is executed less that or equal O(N^2) for all the pairs.
To demonstrate that, I will consider two cases: one where M>>N and other where M ~= N. Where M is the maximum difference of the initial array: M= S(n)-S(1).
For the first case, (M>>N) the probability to find a coincidence is p=N/M. To start a sequence, it must coincide the second and the b+1 element where b is the length of the best sequence until now. So the loop will enter times. And the average length of this series (supposing an infinite series) is . So the total number of times that the loop will be executed is . And this is close to 0 when M>>N. The problem here is when M~=N.
Now lets consider this case where M~=N. Lets consider that b is the best sequence length until now. For the case A=k=1, then the sequence must start before N-b, so the number of sequences will be N-b, and the times that will go for the loop will be a maximum of (N-b)*b.
For A>1 and k=1 we can extrapolate to where d is M/N (the average distance between numbers). If we add for all A’s from 1 to dN/b then we see a top limit of:
For the cases where k>=2, we see that the sequence must start before , So the loop will enter an average of and adding for all As from 1 to dN/k^b, it gives a limit of
Here, the worst case is when b is minimum. Because we are considering minimum series, lets consider a very worst case of b= 2 so the number of passes for the 4th loop for a given k will be less than
.
And if we add all k’s from 2 to infinite will be:
So adding all the passes for k=1 and k>=2, we have a maximum of:
Note that d=M/N=1/p.
So we have two limits, One that goes to infinite when d=1/p=M/N goes to 1 and other that goes to infinite when d goes to infinite. So our limit is the minimum of both, and the worst case is when both equetions cross. So if we solve the equation:
we see that the maximum is when d=1.353
So it is demonstrated that the forth loops will be processed less than 1.55N^2 times in total.
Of course, this is for the average case. For the worst case I am not able to find a way to generate series whose forth loop are higher than O(N^2), and I strongly believe that they does not exist, but I am not a mathematician to prove it.
Old Answer
Here is a solution in average of O((n^2)*cube_root(M)) where M is the difference between the first and last element of the array. And memory requirements of O(M+N).
1.- Construct an array H of length M so that M[i - S[0]]=true if i exists in the initial array and false if it does not exist.
2.- For each pair in the array S[j], S[i] do:
2.1 Check if it can be the first and third elements of a possible solution. To do so, calculate all possible A,K pairs that meet the equation S(i) = S(j) + AK + AK^2. Check this SO question to see how to solve this problem. And check that exist the second element: S[i]+ A*K
2.2 Check also that exist the element one position further that the best solution that we have. For example, if the best solution that we have until now is 4 elements long then check that exist the element A[j] + AK + AK^2 + AK^3 + AK^4
2.3 If 2.1 and 2.2 are true, then iterate how long is this series and set as the bestSolution until now is is longer that the last.
Here is the code in javascript:
function getAKs(A) {
if (A / 2 != Math.floor(A / 2)) return [];
var solution = [];
var i;
var SR3 = Math.pow(A, 1 / 3);
for (i = 1; i <= SR3; i++) {
var B, C;
C = i;
B = A / (C * (C + 1));
if (B == Math.floor(B)) {
solution.push([B, C]);
}
B = i;
C = (-1 + Math.sqrt(1 + 4 * A / B)) / 2;
if (C == Math.floor(C)) {
solution.push([B, C]);
}
}
return solution;
}
function getBestGeometricSequence(S) {
var i, j, k;
var bestSolution = [];
var H = Array(S[S.length-1]-S[0]);
for (i = 0; i < S.length; i++) H[S[i] - S[0]] = true;
for (i = 0; i < S.length; i++) {
for (j = 0; j < i; j++) {
var PossibleAKs = getAKs(S[i] - S[j]);
for (k = 0; k < PossibleAKs.length; k++) {
var A = PossibleAKs[k][0];
var K = PossibleAKs[k][17];
var mustExistToBeBetter;
if (K==1) {
mustExistToBeBetter = S[j] + A * bestSolution.length;
} else {
mustExistToBeBetter = S[j] + A * K * (Math.pow(K,bestSolution.length) - 1)/(K-1);
}
if ((H[S[j] + A * K - S[0]]) && (H[mustExistToBeBetter - S[0]])) {
var possibleSolution=[S[j],S[j] + A * K,S[i]];
exp = K * K * K;
var NextVal = S[i] + A * exp;
while (H[NextVal - S[0]] === true) {
possibleSolution.push(NextVal);
exp = exp * K;
NextVal = NextVal + A * exp;
}
if (possibleSolution.length > bestSolution.length) {
bestSolution = possibleSolution;
}
}
}
}
}
return bestSolution;
}
//var A= [ 1, 2, 3,5,7, 15, 27, 30,31, 81];
var A=[];
for (i=1;i<=3000;i++) {
A.push(i);
}
var sol=getBestGeometricSequence(A);
$("#result").html(JSON.stringify(sol));
You can check the code here: http://jsfiddle.net/6yHyR/1/
I maintain the other solution because I believe that it is still better when M is very big compared to N.
Just to start with something, here is a simple solution in JavaScript:
var input = [0.7, 1, 2, 3, 4, 7, 15, 27, 30, 31, 81],
output = [], indexes, values, i, index, value, i_max_length,
i1, i2, i3, j1, j2, j3, difference12a, difference23a, difference12b, difference23b,
scale_factor, common_ratio_a, common_ratio_b, common_ratio_c,
error, EPSILON = 1e-9, common_ratio_is_integer,
resultDiv = $("#result");
for (i1 = 0; i1 < input.length - 2; ++i1) {
for (i2 = i1 + 1; i2 < input.length - 1; ++i2) {
scale_factor = difference12a = input[i2] - input[i1];
for (i3 = i2 + 1; i3 < input.length; ++i3) {
difference23a = input[i3] - input[i2];
common_ratio_1a = difference23a / difference12a;
common_ratio_2a = Math.round(common_ratio_1a);
error = Math.abs((common_ratio_2a - common_ratio_1a) / common_ratio_1a);
common_ratio_is_integer = error < EPSILON;
if (common_ratio_2a > 1 && common_ratio_is_integer) {
indexes = [i1, i2, i3];
j1 = i2;
j2 = i3
difference12b = difference23a;
for (j3 = j2 + 1; j3 < input.length; ++j3) {
difference23b = input[j3] - input[j2];
common_ratio_1b = difference23b / difference12b;
common_ratio_2b = Math.round(common_ratio_1b);
error = Math.abs((common_ratio_2b - common_ratio_1b) / common_ratio_1b);
common_ratio_is_integer = error < EPSILON;
if (common_ratio_is_integer && common_ratio_2a === common_ratio_2b) {
indexes.push(j3);
j1 = j2;
j2 = j3
difference12b = difference23b;
}
}
values = [];
for (i = 0; i < indexes.length; ++i) {
index = indexes[i];
value = input[index];
values.push(value);
}
output.push(values);
}
}
}
}
if (output !== []) {
i_max_length = 0;
for (i = 1; i < output.length; ++i) {
if (output[i_max_length].length < output[i].length)
i_max_length = i;
}
for (i = 0; i < output.length; ++i) {
if (output[i_max_length].length == output[i].length)
resultDiv.append("<p>[" + output[i] + "]</p>");
}
}
Output:
[1, 3, 7, 15, 31]
I find the first three items of every subsequence candidate, calculate the scale factor and the common ratio from them, and if the common ratio is integer, then I iterate over the remaining elements after the third one, and add those to the subsequence, which fit into the geometric progression defined by the first three items. As a last step, I select the sebsequence/s which has/have the largest length.
In fact it is exactly the same question as Longest equally-spaced subsequence, you just have to consider the logarithm of your data. If the sequence is a, ak, ak^2, ak^3, the logarithmique value is ln(a), ln(a) + ln(k), ln(a)+2ln(k), ln(a)+3ln(k), so it is equally spaced. The opposite is of course true. There is a lot of different code in the question above.
I don't think the special case a=1 can be resolved more efficiently than an adaptation from an algorithm above.
Here is my solution in Javascript. It should be close to O(n^2) except may be in some pathological cases.
function bsearch(Arr,Val, left,right) {
if (left == right) return left;
var m=Math.floor((left + right) /2);
if (Val <= Arr[m]) {
return bsearch(Arr,Val,left,m);
} else {
return bsearch(Arr,Val,m+1,right);
}
}
function findLongestGeometricSequence(S) {
var bestSolution=[];
var i,j,k;
var H={};
for (i=0;i<S.length;i++) H[S[i]]=true;
for (i=0;i<S.length;i++) {
for (j=0;j<i;j++) {
for (k=j+1;k<i;) {
var possibleSolution=[S[j],S[k],S[i]];
var K = (S[i] - S[k]) / (S[k] - S[j]);
var A = (S[k] - S[j]) * (S[k] - S[j]) / (S[i] - S[k]);
if ((Math.floor(K) == K) && (Math.floor(A)==A)) {
exp= K*K*K;
var NextVal= S[i] + A * exp;
while (H[NextVal] === true) {
possibleSolution.push(NextVal);
exp = exp * K;
NextVal= NextVal + A * exp;
}
if (possibleSolution.length > bestSolution.length)
bestSolution=possibleSolution;
K--;
} else {
K=Math.floor(K);
}
if (K>0) {
var NextPossibleMidValue= (S[i] + K*S[j]) / (K +1);
k++;
if (S[k]<NextPossibleMidValue) {
k=bsearch(S,NextPossibleMidValue, k+1, i);
}
} else {
k=i;
}
}
}
}
return bestSolution;
}
function Run() {
var MyS= [0.7, 1, 2, 3, 4, 5,6,7, 15, 27, 30,31, 81];
var sol = findLongestGeometricSequence(MyS);
alert(JSON.stringify(sol));
}
Small Explanation
If we take 3 numbers of the array S(j) < S(k) < S(i) then you can calculate a and k so that: S(k) = S(j) + a*k and S(i) = S(k) + a*k^2 (2 equations and 2 incognits). With that in mind, you can check if exist a number in the array that is S(next) = S(i) + a*k^3. If that is the case, then continue checknng for S(next2) = S(next) + a*k^4 and so on.
This would be a O(n^3) solution, but you can hava advantage that k must be integer in order to limit the S(k) points selected.
In case that a is known, then you can calculate a(k) and you need to check only one number in the third loop, so this case will be clearly a O(n^2).
I think this task is related with not so long ago posted Longest equally-spaced subsequence. I've just modified my algorithm in Python a little bit:
from math import sqrt
def add_precalc(precalc, end, (a, k), count, res, N):
if end + a * k ** res[1]["count"] > N: return
x = end + a * k ** count
if x > N or x < 0: return
if precalc[x] is None: return
if (a, k) not in precalc[x]:
precalc[x][(a, k)] = count
return
def factors(n):
res = []
for x in range(1, int(sqrt(n)) + 1):
if n % x == 0:
y = n / x
res.append((x, y))
res.append((y, x))
return res
def work(input):
precalc = [None] * (max(input) + 1)
for x in input: precalc[x] = {}
N = max(input)
res = ((0, 0), {"end":0, "count":0})
for i, x in enumerate(input):
for y in input[i::-1]:
for a, k in factors(x - y):
if (a, k) in precalc[x]: continue
add_precalc(precalc, x, (a, k), 2, res, N)
for step, count in precalc[x].iteritems():
count += 1
if count > res[1]["count"]: res = (step, {"end":x, "count":count})
add_precalc(precalc, x, step, count, res, N)
precalc[x] = None
d = [res[1]["end"]]
for x in range(res[1]["count"] - 1, 0, -1):
d.append(d[-1] - res[0][0] * res[0][1] ** x)
d.reverse()
return d
explanation
Traversing the array
For each previous element of the array calculate factors of the difference between current and taken previous element and then precalculate next possible element of the sequence and saving it to precalc array
So when arriving at element i there're already all possible sequences with element i in the precalc array, so we have to calculate next possible element and save it to precalc.
Currently there's one place in algorithm that could be slow - factorization of each previous number. I think it could be made faster with two optimizations:
more effective factorization algorithm
find a way not to see at each element of array, using the fact that array is sorted and there's already a precalculated sequences
Python:
def subseq(a):
seq = []
aset = set(a)
for i, x in enumerate(a):
# elements after x
for j, x2 in enumerate(a[i+1:]):
j += i + 1 # enumerate starts j at 0, we want a[j] = x2
bk = x2 - x # b*k (assuming k and k's exponent start at 1)
# given b*k, bruteforce values of k
for k in range(1, bk + 1):
items = [x, x2] # our subsequence so far
nextdist = bk * k # what x3 - x2 should look like
while items[-1] + nextdist in aset:
items.append(items[-1] + nextdist)
nextdist *= k
if len(items) > len(seq):
seq = items
return seq
Running time is O(dn^3), where d is the (average?) distance between two elements,
and n is of course len(a).
I was asked this question in a job interview, and I'd like to know how others would solve it. I'm most comfortable with Java, but solutions in other languages are welcome.
Given an array of numbers, nums, return an array of numbers products, where products[i] is the product of all nums[j], j != i.
Input : [1, 2, 3, 4, 5]
Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)]
= [120, 60, 40, 30, 24]
You must do this in O(N) without using division.
An explanation of polygenelubricants method is:
The trick is to construct the arrays (in the case for 4 elements):
{ 1, a[0], a[0]*a[1], a[0]*a[1]*a[2], }
{ a[1]*a[2]*a[3], a[2]*a[3], a[3], 1, }
Both of which can be done in O(n) by starting at the left and right edges respectively.
Then, multiplying the two arrays element-by-element gives the required result.
My code would look something like this:
int a[N] // This is the input
int products_below[N];
int p = 1;
for (int i = 0; i < N; ++i) {
products_below[i] = p;
p *= a[i];
}
int products_above[N];
p = 1;
for (int i = N - 1; i >= 0; --i) {
products_above[i] = p;
p *= a[i];
}
int products[N]; // This is the result
for (int i = 0; i < N; ++i) {
products[i] = products_below[i] * products_above[i];
}
If you need the solution be O(1) in space as well, you can do this (which is less clear in my opinion):
int a[N] // This is the input
int products[N];
// Get the products below the current index
int p = 1;
for (int i = 0; i < N; ++i) {
products[i] = p;
p *= a[i];
}
// Get the products above the current index
p = 1;
for (int i = N - 1; i >= 0; --i) {
products[i] *= p;
p *= a[i];
}
Here is a small recursive function (in C++) to do the modification in-place. It requires O(n) extra space (on stack) though. Assuming the array is in a and N holds the array length, we have:
int multiply(int *a, int fwdProduct, int indx) {
int revProduct = 1;
if (indx < N) {
revProduct = multiply(a, fwdProduct*a[indx], indx+1);
int cur = a[indx];
a[indx] = fwdProduct * revProduct;
revProduct *= cur;
}
return revProduct;
}
Here's my attempt to solve it in Java. Apologies for the non-standard formatting, but the code has a lot of duplication, and this is the best I can do to make it readable.
import java.util.Arrays;
public class Products {
static int[] products(int... nums) {
final int N = nums.length;
int[] prods = new int[N];
Arrays.fill(prods, 1);
for (int
i = 0, pi = 1 , j = N-1, pj = 1 ;
(i < N) && (j >= 0) ;
pi *= nums[i++] , pj *= nums[j--] )
{
prods[i] *= pi ; prods[j] *= pj ;
}
return prods;
}
public static void main(String[] args) {
System.out.println(
Arrays.toString(products(1, 2, 3, 4, 5))
); // prints "[120, 60, 40, 30, 24]"
}
}
The loop invariants are pi = nums[0] * nums[1] *.. nums[i-1] and pj = nums[N-1] * nums[N-2] *.. nums[j+1]. The i part on the left is the "prefix" logic, and the j part on the right is the "suffix" logic.
Recursive one-liner
Jasmeet gave a (beautiful!) recursive solution; I've turned it into this (hideous!) Java one-liner. It does in-place modification, with O(N) temporary space in the stack.
static int multiply(int[] nums, int p, int n) {
return (n == nums.length) ? 1
: nums[n] * (p = multiply(nums, nums[n] * (nums[n] = p), n + 1))
+ 0*(nums[n] *= p);
}
int[] arr = {1,2,3,4,5};
multiply(arr, 1, 0);
System.out.println(Arrays.toString(arr));
// prints "[120, 60, 40, 30, 24]"
Translating Michael Anderson's solution into Haskell:
otherProducts xs = zipWith (*) below above
where below = scanl (*) 1 $ init xs
above = tail $ scanr (*) 1 xs
Sneakily circumventing the "no divisions" rule:
sum = 0.0
for i in range(a):
sum += log(a[i])
for i in range(a):
output[i] = exp(sum - log(a[i]))
Here you go, simple and clean solution with O(N) complexity:
int[] a = {1,2,3,4,5};
int[] r = new int[a.length];
int x = 1;
r[0] = 1;
for (int i=1;i<a.length;i++){
r[i]=r[i-1]*a[i-1];
}
for (int i=a.length-1;i>0;i--){
x=x*a[i];
r[i-1]=x*r[i-1];
}
for (int i=0;i<r.length;i++){
System.out.println(r[i]);
}
Travel Left->Right and keep saving product. Call it Past. -> O(n)
Travel Right -> left keep the product. Call it Future. -> O(n)
Result[i] = Past[i-1] * future[i+1] -> O(n)
Past[-1] = 1; and Future[n+1]=1;
O(n)
C++, O(n):
long long prod = accumulate(in.begin(), in.end(), 1LL, multiplies<int>());
transform(in.begin(), in.end(), back_inserter(res),
bind1st(divides<long long>(), prod));
Here is my solution in modern C++. It makes use of std::transform and is pretty easy to remember.
Online code (wandbox).
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
vector<int>& multiply_up(vector<int>& v){
v.insert(v.begin(),1);
transform(v.begin()+1, v.end()
,v.begin()
,v.begin()+1
,[](auto const& a, auto const& b) { return b*a; }
);
v.pop_back();
return v;
}
int main() {
vector<int> v = {1,2,3,4,5};
auto vr = v;
reverse(vr.begin(),vr.end());
multiply_up(v);
multiply_up(vr);
reverse(vr.begin(),vr.end());
transform(v.begin(),v.end()
,vr.begin()
,v.begin()
,[](auto const& a, auto const& b) { return b*a; }
);
for(auto& i: v) cout << i << " ";
}
Precalculate the product of the numbers to the left and to the right of each element.
For every element the desired value is the product of it's neigbors's products.
#include <stdio.h>
unsigned array[5] = { 1,2,3,4,5};
int main(void)
{
unsigned idx;
unsigned left[5]
, right[5];
left[0] = 1;
right[4] = 1;
/* calculate products of numbers to the left of [idx] */
for (idx=1; idx < 5; idx++) {
left[idx] = left[idx-1] * array[idx-1];
}
/* calculate products of numbers to the right of [idx] */
for (idx=4; idx-- > 0; ) {
right[idx] = right[idx+1] * array[idx+1];
}
for (idx=0; idx <5 ; idx++) {
printf("[%u] Product(%u*%u) = %u\n"
, idx, left[idx] , right[idx] , left[idx] * right[idx] );
}
return 0;
}
Result:
$ ./a.out
[0] Product(1*120) = 120
[1] Product(1*60) = 60
[2] Product(2*20) = 40
[3] Product(6*5) = 30
[4] Product(24*1) = 24
(UPDATE: now I look closer, this uses the same method as Michael Anderson, Daniel Migowski and polygenelubricants above)
Tricky:
Use the following:
public int[] calc(int[] params) {
int[] left = new int[n-1]
in[] right = new int[n-1]
int fac1 = 1;
int fac2 = 1;
for( int i=0; i<n; i++ ) {
fac1 = fac1 * params[i];
fac2 = fac2 * params[n-i];
left[i] = fac1;
right[i] = fac2;
}
fac = 1;
int[] results = new int[n];
for( int i=0; i<n; i++ ) {
results[i] = left[i] * right[i];
}
Yes, I am sure i missed some i-1 instead of i, but thats the way to solve it.
This is O(n^2) but f# is soooo beautiful:
List.fold (fun seed i -> List.mapi (fun j x -> if i=j+1 then x else x*i) seed)
[1;1;1;1;1]
[1..5]
There also is a O(N^(3/2)) non-optimal solution. It is quite interesting, though.
First preprocess each partial multiplications of size N^0.5(this is done in O(N) time complexity). Then, calculation for each number's other-values'-multiple can be done in 2*O(N^0.5) time(why? because you only need to multiple the last elements of other ((N^0.5) - 1) numbers, and multiply the result with ((N^0.5) - 1) numbers that belong to the group of the current number). Doing this for each number, one can get O(N^(3/2)) time.
Example:
4 6 7 2 3 1 9 5 8
partial results:
4*6*7 = 168
2*3*1 = 6
9*5*8 = 360
To calculate the value of 3, one needs to multiply the other groups' values 168*360, and then with 2*1.
public static void main(String[] args) {
int[] arr = { 1, 2, 3, 4, 5 };
int[] result = { 1, 1, 1, 1, 1 };
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < i; j++) {
result[i] *= arr[j];
}
for (int k = arr.length - 1; k > i; k--) {
result[i] *= arr[k];
}
}
for (int i : result) {
System.out.println(i);
}
}
This solution i came up with and i found it so clear what do you think!?
Based on Billz answer--sorry I can't comment, but here is a scala version that correctly handles duplicate items in the list, and is probably O(n):
val list1 = List(1, 7, 3, 3, 4, 4)
val view = list1.view.zipWithIndex map { x => list1.view.patch(x._2, Nil, 1).reduceLeft(_*_)}
view.force
returns:
List(1008, 144, 336, 336, 252, 252)
Adding my javascript solution here as I didn't find anyone suggesting this.
What is to divide, except to count the number of times you can extract a number from another number? I went through calculating the product of the whole array, and then iterate over each element, and substracting the current element until zero:
//No division operation allowed
// keep substracting divisor from dividend, until dividend is zero or less than divisor
function calculateProducsExceptCurrent_NoDivision(input){
var res = [];
var totalProduct = 1;
//calculate the total product
for(var i = 0; i < input.length; i++){
totalProduct = totalProduct * input[i];
}
//populate the result array by "dividing" each value
for(var i = 0; i < input.length; i++){
var timesSubstracted = 0;
var divisor = input[i];
var dividend = totalProduct;
while(divisor <= dividend){
dividend = dividend - divisor;
timesSubstracted++;
}
res.push(timesSubstracted);
}
return res;
}
Just 2 passes up and down. Job done in O(N)
private static int[] multiply(int[] numbers) {
int[] multiplied = new int[numbers.length];
int total = 1;
multiplied[0] = 1;
for (int i = 1; i < numbers.length; i++) {
multiplied[i] = numbers[i - 1] * multiplied[i - 1];
}
for (int j = numbers.length - 2; j >= 0; j--) {
total *= numbers[j + 1];
multiplied[j] = total * multiplied[j];
}
return multiplied;
}
def productify(arr, prod, i):
if i < len(arr):
prod.append(arr[i - 1] * prod[i - 1]) if i > 0 else prod.append(1)
retval = productify(arr, prod, i + 1)
prod[i] *= retval
return retval * arr[i]
return 1
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5]
prod = []
productify(arr, prod, 0)
print(prod)
Well,this solution can be considered that of C/C++.
Lets say we have an array "a" containing n elements
like a[n],then the pseudo code would be as below.
for(j=0;j<n;j++)
{
prod[j]=1;
for (i=0;i<n;i++)
{
if(i==j)
continue;
else
prod[j]=prod[j]*a[i];
}
One more solution, Using division. with twice traversal.
Multiply all the elements and then start dividing it by each element.
{-
Recursive solution using sqrt(n) subsets. Runs in O(n).
Recursively computes the solution on sqrt(n) subsets of size sqrt(n).
Then recurses on the product sum of each subset.
Then for each element in each subset, it computes the product with
the product sum of all other products.
Then flattens all subsets.
Recurrence on the run time is T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
Suppose that T(n) ≤ cn in O(n).
T(n) = sqrt(n)*T(sqrt(n)) + T(sqrt(n)) + n
≤ sqrt(n)*c*sqrt(n) + c*sqrt(n) + n
≤ c*n + c*sqrt(n) + n
≤ (2c+1)*n
∈ O(n)
Note that ceiling(sqrt(n)) can be computed using a binary search
and O(logn) iterations, if the sqrt instruction is not permitted.
-}
otherProducts [] = []
otherProducts [x] = [1]
otherProducts [x,y] = [y,x]
otherProducts a = foldl' (++) [] $ zipWith (\s p -> map (*p) s) solvedSubsets subsetOtherProducts
where
n = length a
-- Subset size. Require that 1 < s < n.
s = ceiling $ sqrt $ fromIntegral n
solvedSubsets = map otherProducts subsets
subsetOtherProducts = otherProducts $ map product subsets
subsets = reverse $ loop a []
where loop [] acc = acc
loop a acc = loop (drop s a) ((take s a):acc)
Here is my code:
int multiply(int a[],int n,int nextproduct,int i)
{
int prevproduct=1;
if(i>=n)
return prevproduct;
prevproduct=multiply(a,n,nextproduct*a[i],i+1);
printf(" i=%d > %d\n",i,prevproduct*nextproduct);
return prevproduct*a[i];
}
int main()
{
int a[]={2,4,1,3,5};
multiply(a,5,1,0);
return 0;
}
Here's a slightly functional example, using C#:
Func<long>[] backwards = new Func<long>[input.Length];
Func<long>[] forwards = new Func<long>[input.Length];
for (int i = 0; i < input.Length; ++i)
{
var localIndex = i;
backwards[i] = () => (localIndex > 0 ? backwards[localIndex - 1]() : 1) * input[localIndex];
forwards[i] = () => (localIndex < input.Length - 1 ? forwards[localIndex + 1]() : 1) * input[localIndex];
}
var output = new long[input.Length];
for (int i = 0; i < input.Length; ++i)
{
if (0 == i)
{
output[i] = forwards[i + 1]();
}
else if (input.Length - 1 == i)
{
output[i] = backwards[i - 1]();
}
else
{
output[i] = forwards[i + 1]() * backwards[i - 1]();
}
}
I'm not entirely certain that this is O(n), due to the semi-recursion of the created Funcs, but my tests seem to indicate that it's O(n) in time.
To be complete here is the code in Scala:
val list1 = List(1, 2, 3, 4, 5)
for (elem <- list1) println(list1.filter(_ != elem) reduceLeft(_*_))
This will print out the following:
120
60
40
30
24
The program will filter out the current elem (_ != elem); and multiply the new list with reduceLeft method. I think this will be O(n) if you use scala view or Iterator for lazy eval.
// This is the recursive solution in Java
// Called as following from main product(a,1,0);
public static double product(double[] a, double fwdprod, int index){
double revprod = 1;
if (index < a.length){
revprod = product2(a, fwdprod*a[index], index+1);
double cur = a[index];
a[index] = fwdprod * revprod;
revprod *= cur;
}
return revprod;
}
A neat solution with O(n) runtime:
For each element calculate the product of all the elements that occur before that and it store in an array "pre".
For each element calculate the product of all the elements that occur after that element and store it in an array "post"
Create a final array "result", for an element i,
result[i] = pre[i-1]*post[i+1];
Here is the ptyhon version
# This solution use O(n) time and O(n) space
def productExceptSelf(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
N = len(nums)
if N == 0: return
# Initialzie list of 1, size N
l_prods, r_prods = [1]*N, [1]*N
for i in range(1, N):
l_prods[i] = l_prods[i-1] * nums[i-1]
for i in reversed(range(N-1)):
r_prods[i] = r_prods[i+1] * nums[i+1]
result = [x*y for x,y in zip(l_prods,r_prods)]
return result
# This solution use O(n) time and O(1) space
def productExceptSelfSpaceOptimized(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
N = len(nums)
if N == 0: return
# Initialzie list of 1, size N
result = [1]*N
for i in range(1, N):
result[i] = result[i-1] * nums[i-1]
r_prod = 1
for i in reversed(range(N)):
result[i] *= r_prod
r_prod *= nums[i]
return result
I'm use to C#:
public int[] ProductExceptSelf(int[] nums)
{
int[] returnArray = new int[nums.Length];
List<int> auxList = new List<int>();
int multTotal = 0;
// If no zeros are contained in the array you only have to calculate it once
if(!nums.Contains(0))
{
multTotal = nums.ToList().Aggregate((a, b) => a * b);
for (int i = 0; i < nums.Length; i++)
{
returnArray[i] = multTotal / nums[i];
}
}
else
{
for (int i = 0; i < nums.Length; i++)
{
auxList = nums.ToList();
auxList.RemoveAt(i);
if (!auxList.Contains(0))
{
returnArray[i] = auxList.Aggregate((a, b) => a * b);
}
else
{
returnArray[i] = 0;
}
}
}
return returnArray;
}
Here is simple Scala version in Linear O(n) time:
def getProductEff(in:Seq[Int]):Seq[Int] = {
//create a list which has product of every element to the left of this element
val fromLeft = in.foldLeft((1, Seq.empty[Int]))((ac, i) => (i * ac._1, ac._2 :+ ac._1))._2
//create a list which has product of every element to the right of this element, which is the same as the previous step but in reverse
val fromRight = in.reverse.foldLeft((1,Seq.empty[Int]))((ac,i) => (i * ac._1,ac._2 :+ ac._1))._2.reverse
//merge the two list by product at index
in.indices.map(i => fromLeft(i) * fromRight(i))
}
This works because essentially the answer is an array which has product of all elements to the left and to the right.
import java.util.Arrays;
public class Pratik
{
public static void main(String[] args)
{
int[] array = {2, 3, 4, 5, 6}; // OUTPUT: 360 240 180 144 120
int[] products = new int[array.length];
arrayProduct(array, products);
System.out.println(Arrays.toString(products));
}
public static void arrayProduct(int array[], int products[])
{
double sum = 0, EPSILON = 1e-9;
for(int i = 0; i < array.length; i++)
sum += Math.log(array[i]);
for(int i = 0; i < array.length; i++)
products[i] = (int) (EPSILON + Math.exp(sum - Math.log(array[i])));
}
}
OUTPUT:
[360, 240, 180, 144, 120]
Time complexity : O(n)
Space complexity: O(1)
I am given an input, "N", i have to find the number of list of length N, which starts with 1, such that the next number to be added is at most 1 more than the max number added till now. For Example,
N = 3, possible lists => (111, 112, 121, 122, 123), [113, or 131 is not possible as while adding '3' to the list, the maximum number present in the list would be '1', thus we can add only 1 or 2].
N = 4, the list 1213 is possible as while adding 3, the maximum number in the list is '2', thus 3 can be added.
Problem is to count the number of such lists possible for a given input "N".
My code is :-
public static void Main(string[] args)
{
var noOfTestCases = Convert.ToInt32(Console.ReadLine());
var listOfOutput = new List<long>();
for (int i = 0; i < noOfTestCases; i++)
{
var requiredSize = Convert.ToInt64(Console.ReadLine());
long result;
const long listCount = 1;
const long listMaxTillNow = 1;
if (requiredSize < 3)
result = requiredSize;
else
{
SeqCount.Add(requiredSize, 0);
AddElementToList(requiredSize, listCount, listMaxTillNow);
result = SeqCount[requiredSize];
}
listOfOutput.Add(result);
}
foreach (var i in listOfOutput)
{
Console.WriteLine(i);
}
}
private static Dictionary<long, long> SeqCount = new Dictionary<long, long>();
private static void AddElementToList(long requiredSize, long listCount, long listMaxTillNow)
{
if (listCount == requiredSize)
{
SeqCount[requiredSize] = SeqCount[requiredSize] + 1;
return;
}
var listMaxTillNowNew = listMaxTillNow + 1;
for(var i = listMaxTillNowNew; i > 0; i--)
{
AddElementToList(requiredSize, listCount + 1,
i == listMaxTillNowNew ? listMaxTillNowNew : listMaxTillNow);
}
return;
}
Which is the brute force method. I wish to know what might be the best algorithm for the problem?
PS : I only wish to know the number of such lists, so i am sure creating all the list won't be required. (The way i am doing in the code)
I am not at all good in algorithms, so please excuse for the long question.
This problem is a classic example of a dynamic programming problem:
If you define a function dp(k, m) to be the number of lists of length k for which the maximum number is m, then you have a recurrence relation:
dp(1, 1) = 1
dp(1, m) = 0, for m > 1
dp(k, m) = dp(k-1, m) * m + dp(k-1, m-1)
Indeed, there is only one list of length 1 and its maximum element is 1.
When you are building a list of length k with max element m, you can take any of the (k-1)-lists with max = m and append 1 or 2 or .... or m. Or you can take a (k-1)-list with max element m-1 and append m. If you take a (k-1)-list with max element less than m-1 then by your rule you can't get a max of m by appending just one element.
You can compute dp(k,m) for all k = 1,...,N and m = 1,...,N+1 using dynamic programming in O(N^2) and then the answer to your question would be
dp(N,1) + dp(N,2) + ... + dp(N,N+1)
Thus the algorithm is O(N^2).
See below for the implementation of dp calculation in C#:
int[] arr = new int[N + 2];
for (int m = 1; m < N + 2; m++)
arr[m] = 0;
arr[1] = 1;
int[] newArr = new int[N + 2];
int[] tmp;
for (int k = 1; k < N; k++)
{
for (int m = 1; m < N + 2; m++)
newArr[m] = arr[m] * m + arr[m - 1];
tmp = arr;
arr = newArr;
newArr = tmp;
}
int answer = 0;strong text
for (int m = 1; m < N + 2; m++)
answer += arr[m];
Console.WriteLine("The answer for " + N + " is " + answer);
Well, I got interrupted by a fire this afternoon (really!) but FWIW, here's my contribution:
/*
* Counts the number of possible integer list on langth N, with the
* property that no integer in a list(starting with one) may be more
* than one greater than the greatest integer preceeding it in the list.
*
* I am calling this "Semi-Factorial" since it is somewhat similar to
* the factorial function and its constituent integer combinations.
*/
public int SemiFactorial(int N)
{
int sumCounts = 0;
// get a list of the counts of all valid lists of length N,
//whose maximum integer is listCounts[maxInt].
List<int> listCounts = SemiFactorialCounts(N);
for (int maxInt = 1; maxInt <= N; maxInt++)
{
// Get the number of lists, of length N-1 whose maximum integer
//is (maxInt):
int maxIntCnt = listCounts[maxInt];
// just sum them up
sumCounts += maxIntCnt;
}
return sumCounts;
}
// Returns a list of the counts of all valid lists of length N, and
//whose maximum integer is [i], where [i] is also its index in this
//returned list. (0 is not used).
public List<int> SemiFactorialCounts(int N)
{
List<int> cnts;
if (N == 0)
{
// no valid lists,
cnts = new List<int>();
// (zero isn't used)
cnts.Add(0);
}
else if (N == 1)
{
// the only valid list is {1},
cnts = new List<int>();
// (zero isn't used)
cnts.Add(0);
//so that's one list of length 1
cnts.Add(1);
}
else
{
// start with the maxInt counts of lists whose length is N-1:
cnts = SemiFactorialCounts(N - 1);
// add an entry for (N)
cnts.Add(0);
// (reverse order because we overwrite the list using values
// from the next lower index.)
for (int K = N; K > 0; K--)
{
// The number of lists of length N and maxInt K { SF(N,K) }
// Equals K times # of lists one shorter, but same maxInt,
// Plus, the number of lists one shorter with maxInt-1.
cnts[K] = K * cnts[K] + cnts[K - 1];
}
}
return cnts;
}
pretty similar to the others. Though I wouldn't call this "classic dynamic programming" so much as just "classic recursion".