Given a sequence of N integers where 1 <= N <= 500 and the numbers are between 1 and 50. In a step any two adjacent equal numbers x x can be replaced with a single x + 1. What is the maximum number achievable by such steps.
For example if given 2 3 1 1 2 2 then the maximum possible is 4:
2 3 1 1 2 2 ---> 2 3 2 2 2 ---> 2 3 3 2 ---> 2 4 2.
It is evident that I should try to do better than the maximum number available in the sequence. But I can't figure out a good algorithm.
Each substring of the input can make at most one single number (invariant: the log base two of the sum of two to the power of each entry). For every x, we can find the set of substrings that can make x. For each x, this is (1) every occurrence of x (2) the union of two contiguous substrings that can make x - 1. The resulting algorithm is O(N^2)-time.
An algorithm could work like this:
Convert the input to an array where every element has a frequency attribute, collapsing repeated consecutive values in the input into one single node. For example, this input:
1 2 2 4 3 3 3 3
Would be represented like this:
{val: 1, freq: 1} {val: 2, freq: 2} {val: 4, freq: 1} {val: 3, freq: 4}
Then find local minima nodes, like the node (3 3 3 3) in 1 (2 2) 4 (3 3 3 3) 4, i.e. nodes whose neighbours both have higher values. For those local minima that have an even frequency, "lift" those by applying the step. Repeat this until no such local minima (with even frequency) exist any more.
Start of the recursive part of the algorithm:
At both ends of the array, work inwards to "lift" values as long as the more inner neighbour has a higher value. With this rule, the following:
1 2 2 3 5 4 3 3 3 1 1
will completely resolve. First from the left side inward:
1 4 5 4 3 3 3 1 1
Then from the right side:
1 4 6 3 2
Note that when there is an odd frequency (like for the 3s above), there will be a "remainder" that cannot be incremented. The remainder should in this rule always be left on the outward side, so to maximise the potential towards the inner part of the array.
At this point the remaining local minima have odd frequencies. Applying the step to such a node will always leave a "remainder" (like above) with the original value. This remaining node can appear anywhere, but it only makes sense to look at solutions where this remainder is on the left side or the right side of the lift (not in the middle). So for example:
4 1 1 1 1 1 2 3 4
Can resolve to one of these:
4 2 2 1 2 3 4
Or:
4 1 2 2 2 3 4
The 1 in either second or fourth position, is the above mentioned "remainder". Obviously, the second way of resolving is more promising in this example. In general, the choice is obvious when on one side there is a value that is too high to merge with, like the left-most 4 is too high for five 1 values to get to. The 4 is like a wall.
When the frequency of the local minimum is one, there is nothing we can do with it. It actually separates the array in a left and right side that do not influence each other. The same is true for the remainder element discussed above: it separates the array into two parts that do not influence each other.
So the next step in the algorithm is to find such minima (where the choice is obvious), apply that kind of step and separate the problem into two distinct problems which should be solved recursively (from the top). So in the last example, the following two problems would be solved separately:
4
2 2 3 4
Then the best of both solutions will count as the overall solution. In this case that is 5.
The most challenging part of the algorithm is to deal with those local minima for which the choice of where to put the remainder is not obvious. For instance;
3 3 1 1 1 1 1 2 3
This can go to either:
3 3 2 2 1 2 3
3 3 1 2 2 2 3
In this example the end result is the same for both options, but in bigger arrays it would be less and less obvious. So here both options have to be investigated. In general you can have many of them, like 2 in this example:
3 1 1 1 2 3 1 1 1 1 1 3
Each of these two minima has two options. This seems like to explode into too many possibilities for larger arrays. But it is not that bad. The algorithm can take opposite choices in neighbouring minima, and go alternating like this through the whole array. This way alternating sections are favoured, and get the most possible value drawn into them, while the other sections are deprived of value. Now the algorithm turns the tables, and toggles all choices so that the sections that were previously favoured are now deprived, and vice versa. The solution of both these alternatives is derived by resolving each section recursively, and then comparing the two "grand" solutions to pick the best one.
Snippet
Here is a live JavaScript implementation of the above algorithm.
Comments are provided which hopefully should make it readable.
"use strict";
function Node(val, freq) {
// Immutable plain object
return Object.freeze({
val: val,
freq: freq || 1, // Default frequency is 1.
// Max attainable value when merged:
reduced: val + (freq || 1).toString(2).length - 1
});
}
function compress(a) {
// Put repeated elements in a single node
var result = [], i, j;
for (i = 0; i < a.length; i = j) {
for (j = i + 1; j < a.length && a[j] == a[i]; j++);
result.push(Node(a[i], j - i));
}
return result;
}
function decompress(a) {
// Expand nodes into separate, repeated elements
var result = [], i, j;
for (i = 0; i < a.length; i++) {
for (j = 0; j < a[i].freq; j++) {
result.push(a[i].val);
}
}
return result;
}
function str(a) {
return decompress(a).join(' ');
}
function unstr(s) {
s = s.replace(/\D+/g, ' ').trim();
return s.length ? compress(s.split(/\s+/).map(Number)) : [];
}
/*
The function merge modifies an array in-place, performing a "step" on
the indicated element.
The array will get an element with an incremented value
and decreased frequency, unless a join occurs with neighboring
elements with the same value: then the frequencies are accumulated
into one element. When the original frequency was odd there will
be a "remainder" element in the modified array as well.
*/
function merge(a, i, leftWards, stats) {
var val = a[i].val+1,
odd = a[i].freq % 2,
newFreq = a[i].freq >> 1,
last = i;
// Merge with neighbouring nodes of same value:
if ((!odd || !leftWards) && a[i+1] && a[i+1].val === val) {
newFreq += a[++last].freq;
}
if ((!odd || leftWards) && i && a[i-1].val === val) {
newFreq += a[--i].freq;
}
// Replace nodes
a.splice(i, last-i+1, Node(val, newFreq));
if (odd) a.splice(i+leftWards, 0, Node(val-1));
// Update statistics and trace: this is not essential to the algorithm
if (stats) {
stats.total_applied_merges++;
if (stats.trace) stats.trace.push(str(a));
}
return i;
}
/* Function Solve
Parameters:
a: The compressed array to be reduced via merges. It is changed in-place
and should not be relied on after the call.
stats: Optional plain object that will be populated with execution statistics.
Return value:
The array after the best merges were applied to achieve the highest
value, which is stored in the maxValue custom property of the array.
*/
function solve(a, stats) {
var maxValue, i, j, traceOrig, skipLeft, skipRight, sections, goLeft,
b, choice, alternate;
if (!a.length) return a;
if (stats && stats.trace) {
traceOrig = stats.trace;
traceOrig.push(stats.trace = [str(a)]);
}
// Look for valleys of even size, and "lift" them
for (i = 1; i < a.length - 1; i++) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val && (a[i].freq % 2) < 1) {
// Found an even valley
i = merge(a, i, false, stats);
if (i) i--;
}
}
// Check left-side elements with always increasing values
for (i = 0; i < a.length-1 && a[i].val < a[i+1].val; i++) {
if (a[i].freq > 1) i = merge(a, i, false, stats) - 1;
};
// Check right-side elements with always increasing values, right-to-left
for (j = a.length-1; j > 0 && a[j-1].val > a[j].val; j--) {
if (a[j].freq > 1) j = merge(a, j, true, stats) + 1;
};
// All resolved?
if (i == j) {
while (a[i].freq > 1) merge(a, i, true, stats);
a.maxValue = a[i].val;
} else {
skipLeft = i;
skipRight = a.length - 1 - j;
// Look for other valleys (odd sized): they will lead to a split into sections
sections = [];
for (i = a.length - 2 - skipRight; i > skipLeft; i--) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val) {
// Odd number of elements: if more than one, there
// are two ways to merge them, but maybe
// one of both possibilities can be excluded.
goLeft = a[i+1].val > a[i].reduced;
if (a[i-1].val > a[i].reduced || goLeft) {
if (a[i].freq > 1) i = merge(a, i, goLeft, stats) + goLeft;
// i is the index of the element which has become a 1-sized valley
// Split off the right part of the array, and store the solution
sections.push(solve(a.splice(i--), stats));
}
}
}
if (sections.length) {
// Solve last remaining section
sections.push(solve(a, stats));
sections.reverse();
// Combine the solutions of all sections into one
maxValue = sections[0].maxValue;
for (i = sections.length - 1; i >= 0; i--) {
maxValue = Math.max(sections[i].maxValue, maxValue);
}
} else {
// There is no more valley that can be resolved without branching into two
// directions. Look for the remaining valleys.
sections = [];
b = a.slice(0); // take copy
for (choice = 0; choice < 2; choice++) {
if (choice) a = b; // restore from copy on second iteration
alternate = choice;
for (i = a.length - 2 - skipRight; i > skipLeft; i--) {
if (a[i-1].val > a[i].val && a[i].val < a[i+1].val) {
// Odd number of elements
alternate = !alternate
i = merge(a, i, alternate, stats) + alternate;
sections.push(solve(a.splice(i--), stats));
}
}
// Solve last remaining section
sections.push(solve(a, stats));
}
sections.reverse(); // put in logical order
// Find best section:
maxValue = sections[0].maxValue;
for (i = sections.length - 1; i >= 0; i--) {
maxValue = Math.max(sections[i].maxValue, maxValue);
}
for (i = sections.length - 1; i >= 0 && sections[i].maxValue < maxValue; i--);
// Which choice led to the highest value (choice = 0 or 1)?
choice = (i >= sections.length / 2)
// Discard the not-chosen version
sections = sections.slice(choice * sections.length/2);
}
// Reconstruct the solution from the sections.
a = [].concat.apply([], sections);
a.maxValue = maxValue;
}
if (traceOrig) stats.trace = traceOrig;
return a;
}
function randomValues(len) {
var a = [];
for (var i = 0; i < len; i++) {
// 50% chance for a 1, 25% for a 2, ... etc.
a.push(Math.min(/\.1*/.exec(Math.random().toString(2))[0].length,5));
}
return a;
}
// I/O
var inputEl = document.querySelector('#inp');
var randEl = document.querySelector('#rand');
var lenEl = document.querySelector('#len');
var goEl = document.querySelector('#go');
var outEl = document.querySelector('#out');
goEl.onclick = function() {
// Get the input and structure it
var a = unstr(inputEl.value),
stats = {
total_applied_merges: 0,
trace: a.length < 100 ? [] : undefined
};
// Apply algorithm
a = solve(a, stats);
// Output results
var output = {
value: a.maxValue,
compact: str(a),
total_applied_merges: stats.total_applied_merges,
trace: stats.trace || 'no trace produced (input too large)'
};
outEl.textContent = JSON.stringify(output, null, 4);
}
randEl.onclick = function() {
// Get input (count of numbers to generate):
len = lenEl.value;
// Generate
var a = randomValues(len);
// Output
inputEl.value = a.join(' ');
// Simulate click to find the solution immediately.
goEl.click();
}
// Tests
var tests = [
' ', '',
'1', '1',
'1 1', '2',
'2 2 1 2 2', '3 1 3',
'3 2 1 1 2 2 3', '5',
'3 2 1 1 2 2 3 1 1 1 1 3 2 2 1 1 2', '6',
'3 1 1 1 3', '3 2 1 3',
'2 1 1 1 2 1 1 1 2 1 1 1 1 1 2', '3 1 2 1 4 1 2',
'3 1 1 2 1 1 1 2 3', '4 2 1 2 3',
'1 4 2 1 1 1 1 1 1 1', '1 5 1',
];
var res;
for (var i = 0; i < tests.length; i+=2) {
var res = str(solve(unstr(tests[i])));
if (res !== tests[i+1]) throw 'Test failed: ' + tests[i] + ' returned ' + res + ' instead of ' + tests[i+1];
}
Enter series (space separated):<br>
<input id="inp" size="60" value="2 3 1 1 2 2"><button id="go">Solve</button>
<br>
<input id="len" size="4" value="30"><button id="rand">Produce random series of this size and solve</button>
<pre id="out"></pre>
As you can see the program produces a reduced array with the maximum value included. In general there can be many derived arrays that have this maximum; only one is given.
An O(n*m) time and space algorithm is possible, where, according to your stated limits, n <= 500 and m <= 58 (consider that even for a billion elements, m need only be about 60, representing the largest element ± log2(n)). m is representing the possible numbers 50 + floor(log2(500)):
Consider the condensed sequence, s = {[x, number of x's]}.
If M[i][j] = [num_j,start_idx] where num_j represents the maximum number of contiguous js ending at index i of the condensed sequence; start_idx, the index where the sequence starts or -1 if it cannot join earlier sequences; then we have the following relationship:
M[i][j] = [s[i][1] + M[i-1][j][0], M[i-1][j][1]]
when j equals s[i][0]
j's greater than s[i][0] but smaller than or equal to s[i][0] + floor(log2(s[i][1])), represent converting pairs and merging with an earlier sequence if applicable, with a special case after the new count is odd:
When M[i][j][0] is odd, we do two things: first calculate the best so far by looking back in the matrix to a sequence that could merge with M[i][j] or its paired descendants, and then set a lower bound in the next applicable cells in the row (meaning a merge with an earlier sequence cannot happen via this cell). The reason this works is that:
if s[i + 1][0] > s[i][0], then s[i + 1] could only possibly pair with the new split section of s[i]; and
if s[i + 1][0] < s[i][0], then s[i + 1] might generate a lower j that would combine with the odd j from M[i], potentially making a longer sequence.
At the end, return the largest entry in the matrix, max(j + floor(log2(num_j))), for all j.
JavaScript code (counterexamples would be welcome; the limit on the answer is set at 7 for convenient visualization of the matrix):
function f(str){
var arr = str.split(/\s+/).map(Number);
var s = [,[arr[0],0]];
for (var i=0; i<arr.length; i++){
if (s[s.length - 1][0] == arr[i]){
s[s.length - 1][1]++;
} else {
s.push([arr[i],1]);
}
}
var M = [new Array(8).fill([0,0])],
best = 0;
for (var i=1; i<s.length; i++){
M[i] = new Array(8).fill([0,i]);
var temp = s[i][1],
temp_odd,
temp_start,
odd = false;
for (var j=s[i][0]; temp>0; j++){
var start_idx = odd ? temp_start : M[i][j-1][1];
if (start_idx != -1 && M[start_idx - 1][j][0]){
temp += M[start_idx - 1][j][0];
start_idx = M[start_idx - 1][j][1];
}
if (!odd){
M[i][j] = [temp,start_idx];
temp_odd = temp;
} else {
M[i][j] = [temp_odd,-1];
temp_start = start_idx;
}
if (!odd && temp & 1 && temp > 1){
odd = true;
temp_start = start_idx;
}
best = Math.max(best,j + Math.floor(Math.log2(temp)));
temp >>= 1;
temp_odd >>= 1;
}
}
return [arr, s, best, M];
}
// I/O
var button = document.querySelector('button');
var input = document.querySelector('input');
var pre = document.querySelector('pre');
button.onclick = function() {
var val = input.value;
var result = f(val);
var text = '';
for (var i=0; i<3; i++){
text += JSON.stringify(result[i]) + '\n\n';
}
for (var i in result[3]){
text += JSON.stringify(result[3][i]) + '\n';
}
pre.textContent = text;
}
<input value ="2 2 3 3 2 2 3 3 5">
<button>Solve</button>
<pre></pre>
Here's a brute force solution:
function findMax(array A, int currentMax)
for each pair (i, i+1) of indices for which A[i]==A[i+1] do
currentMax = max(A[i]+1, currentMax)
replace A[i],A[i+1] by a single number A[i]+1
currentMax = max(currentMax, findMax(A, currentMax))
end for
return currentMax
Given the array A, let currentMax=max(A[0], ..., A[n])
print findMax(A, currentMax)
The algorithm terminates because in each recursive call the array shrinks by 1.
It's also clear that it is correct: we try out all possible replacement sequences.
The code is extremely slow when the array is large and there's lots of options regarding replacements, but actually works reasonbly fast on arrays with small number of replaceable pairs. (I'll try to quantify the running time in terms of the number of replaceable pairs.)
A naive working code in Python:
def findMax(L, currMax):
for i in range(len(L)-1):
if L[i] == L[i+1]:
L[i] += 1
del L[i+1]
currMax = max(currMax, L[i])
currMax = max(currMax, findMax(L, currMax))
L[i] -= 1
L.insert(i+1, L[i])
return currMax
# entry point
if __name__ == '__main__':
L1 = [2, 3, 1, 1, 2, 2]
L2 = [2, 3, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2]
print findMax(L1, max(L1))
print findMax(L2, max(L2))
The result of the first call is 4, as expected.
The result of the second call is 5 as expected; the sequence that gives the result: 2,3,1,1,2,2,2,2,2,2,2,2, -> 2,3,1,1,3,2,2,2,2,2,2 -> 2,3,1,1,3,3,2,2,2,2, -> 2,3,1,1,3,3,3,2,2 -> 2,3,1,1,3,3,3,3 -> 2,3,1,1,4,3, -> 2,3,1,1,4,4 -> 2,3,1,1,5
I've been trying to solve this Hackerrank challenge: Link
This is what you have to do:
You have one large matrix:
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 0 1 1
and one small matrix:
1 1 1
1 1 1
1 1 0
You have to find out if the small matrix is present in the large matrix.
There are up to 5 testcases and each matrix can be of max 1000x1000 size and I need to solve this in under 4 seconds.
My code timeouts for the largest possible input, I thought that maybe how I'm scanning the matrix is too slow.
This is my code:
package main
import (
"fmt"
"strconv"
"strings"
)
func main() {
var t, rL, cL, rS, cS, temp int
var s string
var sl []string
var mxL, mxS [][]int
var found bool
fmt.Scanf("%d", &t)
for ; t > 0; t-- {
// Start scanning input
// Scanning large matrix
fmt.Scanf("%d%d", &rL, &cL)
mxL = make([][]int, rL)
for i := range mxL {
mxL[i] = make([]int, cL)
}
for i := 0; i < rL; i++ {
fmt.Scanf("%s", &s)
sl = strings.Split(s, "")
for j, v := range sl {
temp, _ = strconv.Atoi(v)
mxL[i][j] = temp
}
}
// Scanning small matrix
fmt.Scanf("%d%d", &rS, &cS)
mxS = make([][]int, rS)
for i := range mxS {
mxS[i] = make([]int, cS)
}
for i := 0; i < rS; i++ {
fmt.Scanf("%s", &s)
sl = strings.Split(s, "")
for j, v := range sl {
temp, _ = strconv.Atoi(v)
mxS[i][j] = temp
}
}
// Stop scanning input
// Start searching for small matrix in large matrix
found = true
for iL := 0; iL <= rL-rS; iL++ {
for jL := 0; jL <= cL-cS; jL++ {
found = true
if mxL[iL][jL] == mxS[0][0] {
for iS := 0; iS < rS; iS++ {
for jS := 1; jS < cS; jS++ {
if mxS[iS][jS] != mxL[iS+iL][jS+jL] {
found = false
break
}
}
if !found {
break
}
}
if found {
break
}
} else {
found = false
}
}
if found {
fmt.Println("YES")
break
}
}
if !found {
fmt.Println("NO")
}
// Stop searching for small matrix in large matrix
}
}
I'm using a slice of slices of ints to store the input.
mxL is the large matrix and mxS is the small matrix.
rL and cL stand for row and column of the large matrix.
rS and cS stand for row and column of the small matrix.
Well I am gonna point out an idea to you and then you can try to implement it. So create a new 2d array as large as your large array. Call it sumArray. Now let each cell in this sumArray represent the sum where the current cell is the most bottom-left cell. Now what you do is check only the cells that has the same sum as your small array instead of checking every element in the array.
So if those are your inputs
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 0 1 1
1 1 1
1 1 1
1 1 0
First sum your small array --> 8
Now let me show you how your sum array would look like
-1 -1 -1 -1 -1 -1 -1 means that we can't sum at this point because
-1 -1 -1 -1 -1 -1 the dimensions are just smaller than your small array
-1 -1 9 9 9 9 each other cell represent the sum of your original
9 9 9 9 9 9 matrix values.
9 9 9 8 9 9
Now if you scan trough this array only you can see that you will reduce your search space from every possible position to only the position where your sum is equal. This doesn't guarantee that the array are in this position you still have to add a verification step but it reduce your search space.
For a classwork problem I am doing, I am supposed to trace (check for bugs) the following algorithm (in pseudocode):
num <- 2
count <- 1
while count < 5
{
count <- count * num
if count / 2 < 2
print "Hello"
else
while count < 7
{
count <- count + 1
}
print "The count is " + count + "."
}
When i traced this code, I got
num count output
2 1 Hello The count is 1.
My question is, was my trace right? It looks like there is something else I have to add.
When you are tracing the problem, you need to note down all value changes in the program.
In your program, we have 2 variables to trace: count and num. From the program, we can figure out 2 facts:
There is no assignment of num;
All output statements are related to count.
Therefore, we should focus on tracing the changes on count.
Notice that this block:
while count < 7
{
count <- count + 1
}
can be replaced with
if count < 7
{
count = 7
}
The workflow of the program can be depicted in English like below:
Check if count is smaller than 5, YES go to 2, NO program ends;
Double count;
If count / 2 is smaller than 2, YES go to 4, NO go to 5;
Print "Hello", go to 6;
If count is smaller than 7, set count to 7;
Print "The count is +count+.`", go to 1;
Now the task is to use 1 as initial value of count and walk through the work flow until the program terminates.
Let's do it together:
count equals to 1, so go to 2;
Now count equals to 2;
count / 2 equals to 1, which is smaller than 2, so go to 4;
Hello is printed, go to 6;
"The count is 2." is printed, go to 1;
count equals to 2, so go to 2;
Now count equals to 4;
count / 2 equals to 2, which is NOT smaller than 2, so go to 5;
count is set to 7;
"The count is 7." is printed, go to 1;
count equals to 7, so program terminates.
Therefore the output will be:
HelloThe count is 2.The count is 7.
Here is how you should walk through this.
num = 2
count = 1
while 1 < 5
{
2 = 1 * 2
if 2 /2 < 2 //since 1 < 2 print Hello
print "Hello"
else //This is skipped because the if was true
while count < 7
{
count <- count + 1
}
print "The count is " + count + "." //This prints "The Count is 2
}
Then you continue through the while loop with count = 2.
Start of second iteration.
while 2 < 5
{
4 = 2 * 2
count changes each time through the loop.
Given an infinite positive integer array or say a stream of positive integers, find out the first five numbers whose sum is twenty.
By reading the problem statement, it first seems to be 0-1 Knapsack problem, but I am confused that can 0-1 Knapsack algo be used on a stream of integers. Let suppose I write a recursive program for the above problem.
int knapsack(int sum, int count, int idx)
{
if (sum == 0 && count == 0)
return 1;
if ((sum == 0 && count != 0) || (sum != 0 && count == 0))
return 0;
if (arr[idx] > 20) //element cann't be included.
return knapsack(sum, count idx + 1);
return max(knapsack(sum, count, idx +1), knapsack(sum - arr[idx], count -1, idx + 1));
}
Now when the above function will call on an infinite array, the first call in max function i.e. knapsack(sum, count, idx +1) will never return as it will keep on ignoring the current element. Even if we change the order of the call in max function, there is still possibility that the first call will never return. Is there any way to apply knapsack algo in such scenarios?
This works if you are working with only positive integers.
Basically keep a list of ways you can reach any of the first 20 numbers and whenever you process a new number process this list accordingly.
def update(dictlist, num):
dk = dictlist.keys()
for i in dk:
if i+num <=20:
for j in dictlist[i]:
listlen = len(dictlist[i][j]) + 1
if listlen >5:
continue
if i+num not in dictlist or listlen not in dictlist[i+num]:
dictlist[i+num][listlen] = dictlist[i][j]+[num]
if num not in dictlist:
dictlist[num]= {}
dictlist[num][1] = [num]
return dictlist
dictlist = {}
for x in infinite_integer_stream:
dictlist = update(dictlist,x)
if 20 in dictlist and 5 in dictlist[20]:
print dictlist[20][5]
break
This code might have some minor bugs and I do not have time now to debug it. But basically dictlist[i][j] stores a j length list that sums to i.
Delphi code:
var
PossibleSums: array[1..4, 0..20] of Integer;
Value, i, j: Integer;
s: string;
begin
s := '';
for j := 1 to 4 do
for i := 0 to 20 do
PossibleSums[j, i] := -1;
while True do begin
Value := 1 + Random(20); // stream emulation
Memo1.Lines.Add(IntToStr(Value));
if PossibleSums[4, 20 - Value] <> -1 then begin
//we just have found 5th number to make the full sum
s := IntToStr(Value);
i := 20 - Value;
for j := 4 downto 1 do begin
//unwind storage chain
s := IntToStr(PossibleSums[j, i]) + ' ' + s;
i := i - PossibleSums[j, i];
end;
Memo1.Lines.Add(s);
Break;
end;
for j := 3 downto 1 do
for i := 0 to 20 - Value do
if (PossibleSums[j, i] <> -1) and (PossibleSums[j + 1, i + Value] = -1) then
PossibleSums[j + 1, i + Value] := Value;
if PossibleSums[1, Value] = -1 then
PossibleSums[1, Value] := Value;
end;
end;
output:
4
8
9
2
10
2
17
2
4 2 10 2 2
Problem
Each row of an n x n matrix consists of 1's and 0's such that in any row, all 1's come before any 0's. Find row containing most no of 1's in O(n).
Example
1 1 1 1 1 0 <- Contains maximum number of 1s, return index 1
1 1 1 0 0 0
1 0 0 0 0 0
1 1 1 1 0 0
1 1 1 1 0 0
1 1 0 0 0 0
I found this question in my algorithms book. The best I could do took O(n logn) time.
How to do this in O(n)?
Start at 1,1.
If the cell contains 1, you're on the longest row so far; write it down and go right.
If the cell contains 0, go down.
If the cell is out of bounds, you're done.
You can do it in O(N) as follows:
Start at A[i][j] with i=j=0.
1, keep moving to the right by doing j++
A[i][j] =
0, move down to the next row by doing i++
When you reach the last row or the last column, the value of j will be the answer.
Pseudo code:
Let R be number of rows
Let C be number of columns
Let i = 0
Let j = 0
Let max1Row = 0
while ( i<R && j<C )
if ( matrix[i][j] == 1 )
j++
max1Row = i
else
i++
end-while
print "Max 1's = j"
print "Row number with max 1's = max1Row"
Start with the first row. Keep the row R that has the most numbers of 1s and the index i of the last 1 of R. in each iteration compare the current row with the row R on the index i. if the current row has a 0 on position i, the row R is still the answer.
Otherwise, return the index of the current row. Now we just have to find the last 1 of the current row. Iterate from index i up to the last 1 of the current row, set R to this row and i to this new index.
i
|
v
R-> 1 1 1 1 1 0
|
v 1 1 1 0 0 0 (Compare ith index of this row)
1 0 0 0 0 0 Repeat
1 1 1 1 0 0 "
1 1 1 1 0 0 "
1 1 0 0 0 0 "
Some C code to do this.
int n = 6;
int maxones = 0, maxrow = -1, row = 0, col = 0;
while(row < n) {
while(col < n && matrix[row][col] == 1) col++;
if(col == n) return row;
if(col > maxones){
maxrow = row;
maxones = col;
}
row++;
}
int [] getMax1withRow(int [][] matrix){
int [] result=new int[2];
int rows=matrix.length;
int cols=matrix[0].length;
int i=0, j=0;
int max_row=0;// This will show row with maximum 1. Intialing pointing to 0th row.
int max_one=0;// max one
while(i< rows){
while(matrix[i][j]==1){
j++;
}
if(j==n){
result[0]=n;
result[1]=i;
return result;
}
if(j>max_one){
max_one=j;
max_row=i;
}
j=0;// Again start from the first column
i++;// increase row number
}
result[0]=max_one;
result[1]=max_row;
return result;
}
Time complexity => O(row+col), In worse case If every row has n-1 one except last row which have n 1s then we have be travel till last row.