Binary search of a Matrix - algorithm

Write an efficient algorithm that searches for a value in an m x n
matrix.
This matrix has the following properties:
-Integers in each row are sorted from left to right. -The first integer
of each row is greater than or equal to the last integer of the
previous row. Example:
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20], [23, 30, 34, 50] ] Given
target = 3, return 1 ( 1 corresponds to true )
Return 0 / 1 ( 0 if the element is not present, 1 if the element is
present ) for this problem
My solution works on NetBeans but fails on the website. Any help will be appreciated.
https://www.interviewbit.com/problems/matrix-search/
public class Solution {
public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
int r = a.size();
int c = a.get(0).size();
int start = 0;
int end = r - 1;
// default value is last row for edge case
int biRow = r -1; // row to search column
//binary search 1st value of rows
while (start <= end) {
int mid = (start + end) / 2;
if (b == a.get(mid).get(0)) {
return 1;
}
if (a.get(mid).get(0) < b && b < a.get(end).get(0)) {
if (mid + 1 >= end) {
biRow = mid;
break;
}
} if (b < a.get(mid).get(0)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
//binary search column of biRow
start = 0;
end = c-1;
while (start <= end) {
int mid = (start + end) / 2;
if (b == a.get(biRow).get(mid)) {
return 1;
}
if (b < a.get(biRow).get(mid)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return 0;
}
}

Okay, the first thing you MUST NOT do is that, you cannot physically concat the matrix into a 1D vector, as this is O(N*M) which is linear and not what we want.
// Easy but TLE code
int Solution::searchMatrix(vector<vector<int> > &A, int B) {
vector<int> v;
for(auto a : A) v.insert(v.end(), a.begin(), a.end());
return binary_search(v.begin(), v.end(), B);
}
So the point is, you have to do binary search directly on the matrix, and that is pretty much the same (except you have to write binary search your own now).
As you did not really access all of the elements, this is O(lg (N*M))
// Less Easy but AC code
int Solution::searchMatrix(vector<vector<int> > &A, int B) {
int m = A.size(), n = A[0].size(), lo = 0, hi = m*n-1, mi, row, col;
while(lo <= hi){
mi = lo + ((hi-lo) >> 1);
row = mi / n;
col = mi % n;
if(A[row][col] == B) return 1;
else if(A[row][col] > B) hi = mi - 1;
else lo = mi + 1;
}
return 0;
}

I think the shared program seems to have a logical error.
When updating the end value in the first while loop, if the end value is equal to start, biRow can not be updated.
It worked well when I updated the code like below.
public class Solution {
public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
int r = a.size();
int c = a.get(0).size();
int start = 0;
int end = r - 1;
// default value is last row for edge case
int biRow = r -1; // row to search column
//binary search 1st value of rows
int mid = 0;
while (start <= end) {
mid = (start + end) / 2;
if ( b >= a.get(mid).get(0) && b <= a.get(mid).get(c-1)) {
break;
}
if (b < a.get(mid).get(0)) {
end = mid-1;
} else {
start = mid+1;
}
}
biRow = mid;
//binary search column of biRow
start = 0;
end = c-1;
while (start <= end) {
mid = (start + end) / 2;
if (b == a.get(biRow).get(mid)) {
return 1;
}
if (b < a.get(biRow).get(mid)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return 0;
}
}

There is logical error in your row search loop. I made a correction and I also added the boundary conditions.Time complexity of this algorithm is O(logN).
public class Solution {
public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
int r = a.size();
int c = a.get(0).size();
// return 0 if b is less than 1st element or greater than last element
if (b < a.get(0).get(0) || b > a.get(r - 1).get(c - 1))
return 0;
int start = 0;
int end = r - 1;
// default value is last row for edge case
int biRow = r - 1; // row to search column
// binary search 1st value of rows
while (start <= end) {
int mid = (start + end) / 2;
if (b == a.get(mid).get(0)) {
return 1;
}
if (b >= a.get(mid).get(0) && b <= a.get(mid).get(c - 1)) {
{
biRow = mid;
break;
}
}
if (b < a.get(mid).get(0)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
// binary search column of biRow
start = 0;
end = c - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (b == a.get(biRow).get(mid)) {
return 1;
}
if (b < a.get(biRow).get(mid)) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return 0;
}
}

Since the rows and columns are sorted, a binary search will be proper as you said. This is a binary search(on a matrix) implementation in Ruby
def binary_search_on_matrix(matrix,target)
row_size = matrix.size
column_size = matrix[0].size
left_index = 0
right_index = (row_size * column_size) - 1
while (left_index <= right_index)
mid_point = left_index + ((right_index - left_index) / 2)
row = mid_point / column_size
col = mid_point % column_size
value = matrix[row][col]
if (value == target)
return true
elsif (value > target)
right_index = mid_point - 1
else
left_index = mid_point + 1
end
end
return false
end

You can first convert your 2D array into 1D array and perform binary search operation. You can refer the code given below:
void search(int a[][10],int search,int m,int n)
{
int arr[100],i=0,j=0,k=-1;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
arr[++k] = a[i][j];
int first = 0 , last = k-1 , middle = (first+last)/2;
while (first <= last)
{
if(arr[middle] < search)
{
first = middle + 1;
}
else if(arr[middle] == search)
{
printf("\n Element found at position:( %d , %d")",(middle/n)+1,(middle%n)+1);
printf(" \n Row : %d",(middle/n)+1);
printf("\n column : %d",(middle%n)+1);
break;
}
else
{
last = middle - 1;
}
middle = (first + last)/2;
}
if(first > last)
{
printf("\n Element not found! ");
}
}
This function prints the row and column of element to be searched if it exists.You can modify this code, if You want the function to return a value depending on the search operation.

Related

Challenge from "leetcode162" to find peak element. Why this case can't pass?

My code is as follows. I used binary search for the problem.
38/58 test cases can pass, yet this can not pass any way:
[3,4,3,2,1], I calculated it on paper, and the process is as below:
end=4,mid=0+(4-0)/2=2,start=2.
But in the case above, start should be 1.
I don't know how to revise my code.
class Solution {
public int findPeakElement(int[] nums) {
if (nums.length == 0 ||nums.length == 1) {
return 0;
}
int start = 0;
int end = nums.length - 1;
int mid;
//binary search
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (nums[start] < nums[start + 1]) {
start = mid;
}
else {
end = mid;
}
}
System.out.println ("start=" + start);
System.out.println ("end=" + end);
System.out.println ("nums[start]=" + nums[start]);
System.out.println ("nums[end]=" + nums[end]);
if (nums[start] < nums[end]) {
return end;
}
else {
return start;
}
}
}
Your solution is a little over complicated, here is a more concise solution. If you go line by line you will realize that it needs less checks than you think.
mid is tricky, but if start and end set properly, you don't need to add the start val.
class Solution {
public int findPeakElement(int[] nums) {
if (nums.length <= 1){
return 0;
}
int start = 0;
int end = nums.length - 1;
int mid = 0;
//binary search
while (start < end) {
mid = (start + end) / 2;
if (nums[mid] > nums[mid + 1]) {
end = mid;
}
else if (nums[mid] < nums[mid + 1]) {
start = mid+1;
}
}
return start;
}
}
You should change
nums[start] < nums[start + 1]
to
nums[mid] < nums[mid + 1]
but, your algorithm will work only when the array 'nums' is stricktly increasing and decreasing.
and
mid = start + (end - start) / 2;
is totally same with
mid = (start + end) / 2;
so i think it's better and simpler

getting compilation error in binary search implementation

I am implementing binary search algorithm , but i am facing return statement issue.Here is my method binarySearch() implementation
public static int binarySearch(int[] a, int n, int x) {
int start = 0;
int end = n - 1;
while (start <= end) {
int mid = (start + end) / 2;
if (a[mid] == x) {
return mid;
} else if (a[mid] < x) {
start = mid + 1;
} else {
end = mid - 1;
}
}
}
Any help would be appreciated.
Thanks
It is possible that this method is called on an array that does not have the value.
You can
return 0;
at the end of the method call. Or...
throw new RuntimeException("Value not found in array");
Whichever is right for this code.
Your function does not return all the paths.
public static int binarySearch(int[] a, int n, int x) {
int start = 0;
int end = n - 1;
int result = 0; // Or something you define yourself for not found case
while (start <= end) {
int mid = (start + end) / 2;
if (a[mid] == x) {
result = mid;
} else if (a[mid] < x) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return result;
}

For a given string which contains only digits , what's the optimal approach to return all valid ip address combinations?

Example:
Given “25525511135”
Output : [“255.255.11.135”, “255.255.111.35”]. (sorted order)
Kindly let me know if we could do a depth first search over here ?(that's the only thing striking me )
Why is it important to have an 'optimal' approach for answering this?
There are not many permutations so the simple approach of checking every combination that fits into the IP format and then filtering out those that have out of range numbers will easily work.
It's unlikely to be a bottle neck for whatever this is part of.
You probably want a dynamic programming algorithm for the general case (something like
http://www.geeksforgeeks.org/dynamic-programming-set-32-word-break-problem/).
Instead of testing whether prefixes can be segmented into words in the dictionary, you'd be testing to see whether the prefixes are prefixes of some valid IPv4 address.
Brutal DFS is acceptable in this problem:
class Solution{
private:
vector<string> ans;
int len;
string cur, rec, str;
bool IsOk(string s) {
if(s[0] == '0' && s.size() > 1) return false;
int sum = 0;
for(int i = 0; i < s.size(); i ++) {
if(s[i] == '.') return false;
sum = sum * 10 + s[i] - '0';
}
if(sum >= 0 && sum <= 255) return true;
return false;
}
void dfs(int x, int cnt) {
if(x == len) {
if(str.size() != len + 4) return ;
string tmp(str);
tmp.erase(tmp.size() - 1, 1);
if(cnt == 4) ans.push_back(tmp);
return ;
}
if(cnt > 4 || str.size() > len + 4) return ;
string tmp = cur;
cur += rec[x];
if(!IsOk(cur)) {
cur = tmp;
return ;
}
dfs(x + 1, cnt);
string tmp2 = cur + '.';
str += tmp2;
cur = "";
dfs(x + 1, cnt + 1);
str.erase(str.size() - tmp2.size(), tmp2.size());
cur = tmp;
}
public:
vector<string> restoreIpAddresses(string s) {
this->len = s.size();
this->rec = s;
cur = str = "";
ans.clear();
dfs(0, 0);
return ans;
}
};
Here is a recursive solution on JavaScript. The result is not sorted.
// Task from https://www.geeksforgeeks.org/program-generate-possible-valid-ip-addresses-given-string/
// Given a string containing only digits, restore it by returning all possible valid IP address combinations.
//
// Example:
// Input : 25525511135
// Output : [“255.255.11.135”, “255.255.111.35”]
//
(function () {
function getValidIP(str) {
const result = [];
const length = str.length;
check(0, 0, '');
function check(start, level, previous){
let i = 0;
let num;
if (level === 3) {
num = str.substring(start);
if (num && num < 256) {
result.push(`${previous}.${num}`);
}
return;
}
num = str.substring(start, start + 1);
if (num == 0) {
check(start + 1, level + 1, level === 0 ? `${num}`: `${previous}.${num}`);
} else {
while (num.length < 4 && num < 256 && start + i + 1 < length) {
check(start + i + 1, level + 1, level === 0 ? `${num}`: `${previous}.${num}`);
i++;
num = str.substring(start, start + i + 1);
}
}
}
return result;
}
console.log('12345:')
console.time('1-1');
console.log(getValidIP('12345'));
console.timeEnd('1-1');
console.log('1234:')
console.time('1-2');
console.log(getValidIP('1234'));
console.timeEnd('1-2');
console.log('2555011135:')
console.time('1-3');
console.log(getValidIP('2555011135'));
console.timeEnd('1-3');
console.log('222011135:')
console.time('1-4');
console.log(getValidIP('222011135'));
console.timeEnd('1-4');
})();

Find order statistic in union of 2 sorted lists on logarithmic time [duplicate]

This is a homework question, binary search has already been introduced:
Given two arrays, respectively N and M elements in ascending order, not necessarily unique:
What is a time efficient algorithm to find the kth smallest element in the union of both arrays?
They say it takes O(logN + logM) where N and M are the arrays lengths.
Let's name the arrays a and b. Obviously we can ignore all a[i] and b[i] where i > k.
First let's compare a[k/2] and b[k/2]. Let b[k/2] > a[k/2]. Therefore we can discard also all b[i], where i > k/2.
Now we have all a[i], where i < k and all b[i], where i < k/2 to find the answer.
What is the next step?
I hope I am not answering your homework, as it has been over a year since this question was asked. Here is a tail recursive solution that will take log(len(a)+len(b)) time.
Assumption: The inputs are correct, i.e., k is in the range [0, len(a)+len(b)].
Base cases:
If length of one of the arrays is 0, the answer is kth element of the second array.
Reduction steps:
If mid index of a + mid index of b is less than k:
If mid element of a is greater than mid element of b, we can ignore the first half of b, adjust k.
Otherwise, ignore the first half of a, adjust k.
If k is less than sum of mid indices of a and b:
If mid element of a is greater than mid element of b, we can safely ignore second half of a.
Otherwise, we can ignore second half of b.
Code:
def kthlargest(arr1, arr2, k):
if len(arr1) == 0:
return arr2[k]
elif len(arr2) == 0:
return arr1[k]
mida1 = len(arr1) // 2 # integer division
mida2 = len(arr2) // 2
if mida1 + mida2 < k:
if arr1[mida1] > arr2[mida2]:
return kthlargest(arr1, arr2[mida2+1:], k - mida2 - 1)
else:
return kthlargest(arr1[mida1+1:], arr2, k - mida1 - 1)
else:
if arr1[mida1] > arr2[mida2]:
return kthlargest(arr1[:mida1], arr2, k)
else:
return kthlargest(arr1, arr2[:mida2], k)
Please note that my solution is creating new copies of smaller arrays in every call, this can be easily eliminated by only passing start and end indices on the original arrays.
You've got it, just keep going! And be careful with the indexes...
To simplify a bit I'll assume that N and M are > k, so the complexity here is O(log k), which is O(log N + log M).
Pseudo-code:
i = k/2
j = k - i
step = k/4
while step > 0
if a[i-1] > b[j-1]
i -= step
j += step
else
i += step
j -= step
step /= 2
if a[i-1] > b[j-1]
return a[i-1]
else
return b[j-1]
For the demonstration you can use the loop invariant i + j = k, but I won't do all your homework :)
Many people answered this "kth smallest element from two sorted array" question, but usually with only general ideas, not a clear working code or boundary conditions analysis.
Here I'd like to elaborate it carefully with the way I went though to help some novices to understand, with my correct working Java code. A1 and A2 are two sorted ascending arrays, with size1 and size2 as length respectively. We need to find the k-th smallest element from the union of those two arrays. Here we reasonably assume that (k > 0 && k <= size1 + size2), which implies that A1 and A2 can't be both empty.
First, let's approach this question with a slow O(k) algorithm. The method is to compare the first element of both array, A1[0] and A2[0]. Take the smaller one, say A1[0] away into our pocket. Then compare A1[1] with A2[0], and so on. Repeat this action until our pocket reached k elements. Very important: In the first step, we can only commit to A1[0] in our pocket. We can NOT include or exclude A2[0]!!!
The following O(k) code gives you one element before the correct answer. Here I use it to show my idea, and analysis boundary condition. I have correct code after this one:
private E kthSmallestSlowWithFault(int k) {
int size1 = A1.length, size2 = A2.length;
int index1 = 0, index2 = 0;
// base case, k == 1
if (k == 1) {
if (size1 == 0) {
return A2[index2];
} else if (size2 == 0) {
return A1[index1];
} else if (A1[index1].compareTo(A2[index2]) < 0) {
return A1[index1];
} else {
return A2[index2];
}
}
/* in the next loop, we always assume there is one next element to compare with, so we can
* commit to the smaller one. What if the last element is the kth one?
*/
if (k == size1 + size2) {
if (size1 == 0) {
return A2[size2 - 1];
} else if (size2 == 0) {
return A1[size1 - 1];
} else if (A1[size1 - 1].compareTo(A2[size2 - 1]) < 0) {
return A1[size1 - 1];
} else {
return A2[size2 - 1];
}
}
/*
* only when k > 1, below loop will execute. In each loop, we commit to one element, till we
* reach (index1 + index2 == k - 1) case. But the answer is not correct, always one element
* ahead, because we didn't merge base case function into this loop yet.
*/
int lastElementFromArray = 0;
while (index1 + index2 < k - 1) {
if (A1[index1].compareTo(A2[index2]) < 0) {
index1++;
lastElementFromArray = 1;
// commit to one element from array A1, but that element is at (index1 - 1)!!!
} else {
index2++;
lastElementFromArray = 2;
}
}
if (lastElementFromArray == 1) {
return A1[index1 - 1];
} else {
return A2[index2 - 1];
}
}
The most powerful idea is that in each loop, we always use the base case approach. After committed to the current smallest element, we get one step closer to the target: the k-th smallest element. Never jump into the middle and make yourself confused and lost!
By observing the above code base case k == 1, k == size1+size2, and combine with that A1 and A2 can't both be empty. We can turn the logic into below more concise style.
Here is a slow but correct working code:
private E kthSmallestSlow(int k) {
// System.out.println("this is an O(k) speed algorithm, very concise");
int size1 = A1.length, size2 = A2.length;
int index1 = 0, index2 = 0;
while (index1 + index2 < k - 1) {
if (size1 > index1 && (size2 <= index2 || A1[index1].compareTo(A2[index2]) < 0)) {
index1++; // here we commit to original index1 element, not the increment one!!!
} else {
index2++;
}
}
// below is the (index1 + index2 == k - 1) base case
// also eliminate the risk of referring to an element outside of index boundary
if (size1 > index1 && (size2 <= index2 || A1[index1].compareTo(A2[index2]) < 0)) {
return A1[index1];
} else {
return A2[index2];
}
}
Now we can try a faster algorithm runs at O(log k). Similarly, compare A1[k/2] with A2[k/2]; if A1[k/2] is smaller, then all the elements from A1[0] to A1[k/2] should be in our pocket. The idea is to not just commit to one element in each loop; the first step contains k/2 elements. Again, we can NOT include or exclude A2[0] to A2[k/2] anyway. So in the first step, we can't go more than k/2 elements. For the second step, we can't go more than k/4 elements...
After each step, we get much closer to k-th element. At the same time each step get smaller and smaller, until we reach (step == 1), which is (k-1 == index1+index2). Then we can refer to the simple and powerful base case again.
Here is the working correct code:
private E kthSmallestFast(int k) {
// System.out.println("this is an O(log k) speed algorithm with meaningful variables name");
int size1 = A1.length, size2 = A2.length;
int index1 = 0, index2 = 0, step = 0;
while (index1 + index2 < k - 1) {
step = (k - index1 - index2) / 2;
int step1 = index1 + step;
int step2 = index2 + step;
if (size1 > step1 - 1
&& (size2 <= step2 - 1 || A1[step1 - 1].compareTo(A2[step2 - 1]) < 0)) {
index1 = step1; // commit to element at index = step1 - 1
} else {
index2 = step2;
}
}
// the base case of (index1 + index2 == k - 1)
if (size1 > index1 && (size2 <= index2 || A1[index1].compareTo(A2[index2]) < 0)) {
return A1[index1];
} else {
return A2[index2];
}
}
Some people may worry what if (index1+index2) jump over k-1? Could we miss the base case (k-1 == index1+index2)? That's impossible. You can add up 0.5+0.25+0.125..., and you will never go beyond 1.
Of course, it is very easy to turn the above code into recursive algorithm:
private E kthSmallestFastRecur(int k, int index1, int index2, int size1, int size2) {
// System.out.println("this is an O(log k) speed algorithm with meaningful variables name");
// the base case of (index1 + index2 == k - 1)
if (index1 + index2 == k - 1) {
if (size1 > index1 && (size2 <= index2 || A1[index1].compareTo(A2[index2]) < 0)) {
return A1[index1];
} else {
return A2[index2];
}
}
int step = (k - index1 - index2) / 2;
int step1 = index1 + step;
int step2 = index2 + step;
if (size1 > step1 - 1 && (size2 <= step2 - 1 || A1[step1 - 1].compareTo(A2[step2 - 1]) < 0)) {
index1 = step1;
} else {
index2 = step2;
}
return kthSmallestFastRecur(k, index1, index2, size1, size2);
}
Hope the above analysis and Java code could help you to understand. But never copy my code as your homework! Cheers ;)
Here's a C++ iterative version of #lambdapilgrim's solution (see the explanation of the algorithm there):
#include <cassert>
#include <iterator>
template<class RandomAccessIterator, class Compare>
typename std::iterator_traits<RandomAccessIterator>::value_type
nsmallest_iter(RandomAccessIterator firsta, RandomAccessIterator lasta,
RandomAccessIterator firstb, RandomAccessIterator lastb,
size_t n,
Compare less) {
assert(issorted(firsta, lasta, less) && issorted(firstb, lastb, less));
for ( ; ; ) {
assert(n < static_cast<size_t>((lasta - firsta) + (lastb - firstb)));
if (firsta == lasta) return *(firstb + n);
if (firstb == lastb) return *(firsta + n);
size_t mida = (lasta - firsta) / 2;
size_t midb = (lastb - firstb) / 2;
if ((mida + midb) < n) {
if (less(*(firstb + midb), *(firsta + mida))) {
firstb += (midb + 1);
n -= (midb + 1);
}
else {
firsta += (mida + 1);
n -= (mida + 1);
}
}
else {
if (less(*(firstb + midb), *(firsta + mida)))
lasta = (firsta + mida);
else
lastb = (firstb + midb);
}
}
}
It works for all 0 <= n < (size(a) + size(b)) indexes and has O(log(size(a)) + log(size(b))) complexity.
Example
#include <functional> // greater<>
#include <iostream>
#define SIZE(a) (sizeof(a) / sizeof(*a))
int main() {
int a[] = {5,4,3};
int b[] = {2,1,0};
int k = 1; // find minimum value, the 1st smallest value in a,b
int i = k - 1; // convert to zero-based indexing
int v = nsmallest_iter(a, a + SIZE(a), b, b + SIZE(b),
SIZE(a)+SIZE(b)-1-i, std::greater<int>());
std::cout << v << std::endl; // -> 0
return v;
}
My attempt for first k numbers, kth number in 2 sorted arrays, and in n sorted arrays:
// require() is recognizable by node.js but not by browser;
// for running/debugging in browser, put utils.js and this file in <script> elements,
if (typeof require === "function") require("./utils.js");
// Find K largest numbers in two sorted arrays.
function k_largest(a, b, c, k) {
var sa = a.length;
var sb = b.length;
if (sa + sb < k) return -1;
var i = 0;
var j = sa - 1;
var m = sb - 1;
while (i < k && j >= 0 && m >= 0) {
if (a[j] > b[m]) {
c[i] = a[j];
i++;
j--;
} else {
c[i] = b[m];
i++;
m--;
}
}
debug.log(2, "i: "+ i + ", j: " + j + ", m: " + m);
if (i === k) {
return 0;
} else if (j < 0) {
while (i < k) {
c[i++] = b[m--];
}
} else {
while (i < k) c[i++] = a[j--];
}
return 0;
}
// find k-th largest or smallest number in 2 sorted arrays.
function kth(a, b, kd, dir){
sa = a.length; sb = b.length;
if (kd<1 || sa+sb < kd){
throw "Mission Impossible! I quit!";
}
var k;
//finding the kd_th largest == finding the smallest k_th;
if (dir === 1){ k = kd;
} else if (dir === -1){ k = sa + sb - kd + 1;}
else throw "Direction has to be 1 (smallest) or -1 (largest).";
return find_kth(a, b, k, sa-1, 0, sb-1, 0);
}
// find k-th smallest number in 2 sorted arrays;
function find_kth(c, d, k, cmax, cmin, dmax, dmin){
sc = cmax-cmin+1; sd = dmax-dmin+1; k0 = k; cmin0 = cmin; dmin0 = dmin;
debug.log(2, "=k: " + k +", sc: " + sc + ", cmax: " + cmax +", cmin: " + cmin + ", sd: " + sd +", dmax: " + dmax + ", dmin: " + dmin);
c_comp = k0-sc;
if (c_comp <= 0){
cmax = cmin0 + k0-1;
} else {
dmin = dmin0 + c_comp-1;
k -= c_comp-1;
}
d_comp = k0-sd;
if (d_comp <= 0){
dmax = dmin0 + k0-1;
} else {
cmin = cmin0 + d_comp-1;
k -= d_comp-1;
}
sc = cmax-cmin+1; sd = dmax-dmin+1;
debug.log(2, "#k: " + k +", sc: " + sc + ", cmax: " + cmax +", cmin: " + cmin + ", sd: " + sd +", dmax: " + dmax + ", dmin: " + dmin + ", c_comp: " + c_comp + ", d_comp: " + d_comp);
if (k===1) return (c[cmin]<d[dmin] ? c[cmin] : d[dmin]);
if (k === sc+sd) return (c[cmax]>d[dmax] ? c[cmax] : d[dmax]);
m = Math.floor((cmax+cmin)/2);
n = Math.floor((dmax+dmin)/2);
debug.log(2, "m: " + m + ", n: "+n+", c[m]: "+c[m]+", d[n]: "+d[n]);
if (c[m]<d[n]){
if (m === cmax){ // only 1 element in c;
return d[dmin+k-1];
}
k_next = k-(m-cmin+1);
return find_kth(c, d, k_next, cmax, m+1, dmax, dmin);
} else {
if (n === dmax){
return c[cmin+k-1];
}
k_next = k-(n-dmin+1);
return find_kth(c, d, k_next, cmax, cmin, dmax, n+1);
}
}
function traverse_at(a, ae, h, l, k, at, worker, wp){
var n = ae ? ae.length : 0;
var get_node;
switch (at){
case "k": get_node = function(idx){
var node = {};
var pos = l[idx] + Math.floor(k/n) - 1;
if (pos<l[idx]){ node.pos = l[idx]; }
else if (pos > h[idx]){ node.pos = h[idx];}
else{ node.pos = pos; }
node.idx = idx;
node.val = a[idx][node.pos];
debug.log(6, "pos: "+pos+"\nnode =");
debug.log(6, node);
return node;
};
break;
case "l": get_node = function(idx){
debug.log(6, "a["+idx+"][l["+idx+"]]: "+a[idx][l[idx]]);
return a[idx][l[idx]];
};
break;
case "h": get_node = function(idx){
debug.log(6, "a["+idx+"][h["+idx+"]]: "+a[idx][h[idx]]);
return a[idx][h[idx]];
};
break;
case "s": get_node = function(idx){
debug.log(6, "h["+idx+"]-l["+idx+"]+1: "+(h[idx] - l[idx] + 1));
return h[idx] - l[idx] + 1;
};
break;
default: get_node = function(){
debug.log(1, "!!! Exception: get_node() returns null.");
return null;
};
break;
}
worker.init();
debug.log(6, "--* traverse_at() *--");
var i;
if (!wp){
for (i=0; i<n; i++){
worker.work(get_node(ae[i]));
}
} else {
for (i=0; i<n; i++){
worker.work(get_node(ae[i]), wp);
}
}
return worker.getResult();
}
sumKeeper = function(){
var res = 0;
return {
init : function(){ res = 0;},
getResult: function(){
debug.log(5, "## sumKeeper.getResult: returning: "+res);
return res;
},
work : function(node){ if (node!==null) res += node;}
};
}();
maxPicker = function(){
var res = null;
return {
init : function(){ res = null;},
getResult: function(){
debug.log(5, "## maxPicker.getResult: returning: "+res);
return res;
},
work : function(node){
if (res === null){ res = node;}
else if (node!==null && node > res){ res = node;}
}
};
}();
minPicker = function(){
var res = null;
return {
init : function(){ res = null;},
getResult: function(){
debug.log(5, "## minPicker.getResult: returning: ");
debug.log(5, res);
return res;
},
work : function(node){
if (res === null && node !== null){ res = node;}
else if (node!==null &&
node.val !==undefined &&
node.val < res.val){ res = node; }
else if (node!==null && node < res){ res = node;}
}
};
}();
// find k-th smallest number in n sorted arrays;
// need to consider the case where some of the subarrays are taken out of the selection;
function kth_n(a, ae, k, h, l){
var n = ae.length;
debug.log(2, "------** kth_n() **-------");
debug.log(2, "n: " +n+", k: " + k);
debug.log(2, "ae: ["+ae+"], len: "+ae.length);
debug.log(2, "h: [" + h + "]");
debug.log(2, "l: [" + l + "]");
for (var i=0; i<n; i++){
if (h[ae[i]]-l[ae[i]]+1>k) h[ae[i]]=l[ae[i]]+k-1;
}
debug.log(3, "--after reduction --");
debug.log(3, "h: [" + h + "]");
debug.log(3, "l: [" + l + "]");
if (n === 1)
return a[ae[0]][k-1];
if (k === 1)
return traverse_at(a, ae, h, l, k, "l", minPicker);
if (k === traverse_at(a, ae, h, l, k, "s", sumKeeper))
return traverse_at(a, ae, h, l, k, "h", maxPicker);
var kn = traverse_at(a, ae, h, l, k, "k", minPicker);
debug.log(3, "kn: ");
debug.log(3, kn);
var idx = kn.idx;
debug.log(3, "last: k: "+k+", l["+kn.idx+"]: "+l[idx]);
k -= kn.pos - l[idx] + 1;
l[idx] = kn.pos + 1;
debug.log(3, "next: "+"k: "+k+", l["+kn.idx+"]: "+l[idx]);
if (h[idx]<l[idx]){ // all elements in a[idx] selected;
//remove a[idx] from the arrays.
debug.log(4, "All elements selected in a["+idx+"].");
debug.log(5, "last ae: ["+ae+"]");
ae.splice(ae.indexOf(idx), 1);
h[idx] = l[idx] = "_"; // For display purpose only.
debug.log(5, "next ae: ["+ae+"]");
}
return kth_n(a, ae, k, h, l);
}
function find_kth_in_arrays(a, k){
if (!a || a.length<1 || k<1) throw "Mission Impossible!";
var ae=[], h=[], l=[], n=0, s, ts=0;
for (var i=0; i<a.length; i++){
s = a[i] && a[i].length;
if (s>0){
ae.push(i); h.push(s-1); l.push(0);
ts+=s;
}
}
if (k>ts) throw "Too few elements to choose from!";
return kth_n(a, ae, k, h, l);
}
/////////////////////////////////////////////////////
// tests
// To show everything: use 6.
debug.setLevel(1);
var a = [2, 3, 5, 7, 89, 223, 225, 667];
var b = [323, 555, 655, 673];
//var b = [99];
var c = [];
debug.log(1, "a = (len: " + a.length + ")");
debug.log(1, a);
debug.log(1, "b = (len: " + b.length + ")");
debug.log(1, b);
for (var k=1; k<a.length+b.length+1; k++){
debug.log(1, "================== k: " + k + "=====================");
if (k_largest(a, b, c, k) === 0 ){
debug.log(1, "c = (len: "+c.length+")");
debug.log(1, c);
}
try{
result = kth(a, b, k, -1);
debug.log(1, "===== The " + k + "-th largest number: " + result);
} catch (e) {
debug.log(0, "Error message from kth(): " + e);
}
debug.log("==================================================");
}
debug.log(1, "################# Now for the n sorted arrays ######################");
debug.log(1, "####################################################################");
x = [[1, 3, 5, 7, 9],
[-2, 4, 6, 8, 10, 12],
[8, 20, 33, 212, 310, 311, 623],
[8],
[0, 100, 700],
[300],
[],
null];
debug.log(1, "x = (len: "+x.length+")");
debug.log(1, x);
for (var i=0, num=0; i<x.length; i++){
if (x[i]!== null) num += x[i].length;
}
debug.log(1, "totoal number of elements: "+num);
// to test k in specific ranges:
var start = 0, end = 25;
for (k=start; k<end; k++){
debug.log(1, "=========================== k: " + k + "===========================");
try{
result = find_kth_in_arrays(x, k);
debug.log(1, "====== The " + k + "-th smallest number: " + result);
} catch (e) {
debug.log(1, "Error message from find_kth_in_arrays: " + e);
}
debug.log(1, "=================================================================");
}
debug.log(1, "x = (len: "+x.length+")");
debug.log(1, x);
debug.log(1, "totoal number of elements: "+num);
The complete code with debug utils can be found at: https://github.com/brainclone/teasers/tree/master/kth
Most of the answers I found here focus on both arrays. while it's good but it's harder to implement as there are a lot of edge cases that we need to take care of. Besides that most of the implementations are recursive which adds the space complexity of recursion stack. So instead of focusing on both arrays I decided to just focus on the smaller array and do the binary search on just the smaller array and adjust the pointer for the second array based on the value of the pointer in the first Array. By the following implementation, we have the complexity of O(log(min(n,m)) with O(1) space complexity.
public static int kth_two_sorted(int []a, int b[],int k){
if(a.length > b.length){
return kth_two_sorted(b,a,k);
}
if(a.length + a.length < k){
throw new RuntimeException("wrong argument");
}
int low = 0;
int high = k;
if(a.length <= k){
high = a.length-1;
}
while(low <= high){
int sizeA = low+(high - low)/2;
int sizeB = k - sizeA;
boolean shrinkLeft = false;
boolean extendRight = false;
if(sizeA != 0){
if(sizeB !=b.length){
if(a[sizeA-1] > b[sizeB]){
shrinkLeft = true;
high = sizeA-1;
}
}
}
if(sizeA!=a.length){
if(sizeB!=0){
if(a[sizeA] < b[sizeB-1]){
extendRight = true;
low = sizeA;
}
}
}
if(!shrinkLeft && !extendRight){
return Math.max(a[sizeA-1],b[sizeB-1]) ;
}
}
throw new IllegalArgumentException("we can't be here");
}
We have a range of [low, high] for array a and we narrow this range as we go further through the algorithm. sizeA shows how many of items from k items are from array a and it derives from the value of low and high. sizeB is the same definition except we calculate the value such a way that sizeA+sizeB=k. The based on the values on those two borders with conclude that we have to extend to the right side in array a or shrink to the left side. if we stuck in the same position it means that we found the solution and we will return the max of values in the position of sizeA-1 from a and sizeB-1 from b.
Here's my code based on Jules Olleon's solution:
int getNth(vector<int>& v1, vector<int>& v2, int n)
{
int step = n / 4;
int i1 = n / 2;
int i2 = n - i1;
while(!(v2[i2] >= v1[i1 - 1] && v1[i1] > v2[i2 - 1]))
{
if (v1[i1 - 1] >= v2[i2 - 1])
{
i1 -= step;
i2 += step;
}
else
{
i1 += step;
i2 -= step;
}
step /= 2;
if (!step) step = 1;
}
if (v1[i1 - 1] >= v2[i2 - 1])
return v1[i1 - 1];
else
return v2[i2 - 1];
}
int main()
{
int a1[] = {1,2,3,4,5,6,7,8,9};
int a2[] = {4,6,8,10,12};
//int a1[] = {1,2,3,4,5,6,7,8,9};
//int a2[] = {4,6,8,10,12};
//int a1[] = {1,7,9,10,30};
//int a2[] = {3,5,8,11};
vector<int> v1(a1, a1+9);
vector<int> v2(a2, a2+5);
cout << getNth(v1, v2, 5);
return 0;
}
Here is my implementation in C, you can refer to #Jules Olléon 's explains for the algorithm: the idea behind the algorithm is that we maintain i + j = k, and find such i and j so that a[i-1] < b[j-1] < a[i] (or the other way round). Now since there are i elements in 'a' smaller than b[j-1], and j-1 elements in 'b' smaller than b[j-1], b[j-1] is the i + j-1 + 1 = kth smallest element. To find such i,j the algorithm does a dichotomic search on the arrays.
int find_k(int A[], int m, int B[], int n, int k) {
if (m <= 0 )return B[k-1];
else if (n <= 0) return A[k-1];
int i = ( m/double (m + n)) * (k-1);
if (i < m-1 && i<k-1) ++i;
int j = k - 1 - i;
int Ai_1 = (i > 0) ? A[i-1] : INT_MIN, Ai = (i<m)?A[i]:INT_MAX;
int Bj_1 = (j > 0) ? B[j-1] : INT_MIN, Bj = (j<n)?B[j]:INT_MAX;
if (Ai >= Bj_1 && Ai <= Bj) {
return Ai;
} else if (Bj >= Ai_1 && Bj <= Ai) {
return Bj;
}
if (Ai < Bj_1) { // the answer can't be within A[0,...,i]
return find_k(A+i+1, m-i-1, B, n, j);
} else { // the answer can't be within A[0,...,i]
return find_k(A, m, B+j+1, n-j-1, i);
}
}
Here's my solution. The C++ code prints the kth smallest value as well as the number of iterations to get the kth smallest value using a loop, which in my opinion is in the order of log(k). The code however requires k to be smaller than the length of the first array which is a limitation.
#include <iostream>
#include <vector>
#include<math.h>
using namespace std;
template<typename comparable>
comparable kthSmallest(vector<comparable> & a, vector<comparable> & b, int k){
int idx1; // Index in the first array a
int idx2; // Index in the second array b
comparable maxVal, minValPlus;
float iter = k;
int numIterations = 0;
if(k > a.size()){ // Checks if k is larger than the size of first array
cout << " k is larger than the first array" << endl;
return -1;
}
else{ // If all conditions are satisfied, initialize the indexes
idx1 = k - 1;
idx2 = -1;
}
for ( ; ; ){
numIterations ++;
if(idx2 == -1 || b[idx2] <= a[idx1] ){
maxVal = a[idx1];
minValPlus = b[idx2 + 1];
idx1 = idx1 - ceil(iter/2); // Binary search
idx2 = k - idx1 - 2; // Ensures sum of indices = k - 2
}
else{
maxVal = b[idx2];
minValPlus = a[idx1 + 1];
idx2 = idx2 - ceil(iter/2); // Binary search
idx1 = k - idx2 - 2; // Ensures sum of indices = k - 2
}
if(minValPlus >= maxVal){ // Check if kth smallest value has been found
cout << "The number of iterations to find the " << k << "(th) smallest value is " << numIterations << endl;
return maxVal;
}
else
iter/=2; // Reduce search space of binary search
}
}
int main(){
//Test Cases
vector<int> a = {2, 4, 9, 15, 22, 34, 45, 55, 62, 67, 78, 85};
vector<int> b = {1, 3, 6, 8, 11, 13, 15, 20, 56, 67, 89};
// Input k < a.size()
int kthSmallestVal;
for (int k = 1; k <= a.size() ; k++){
kthSmallestVal = kthSmallest<int>( a ,b ,k );
cout << k <<" (th) smallest Value is " << kthSmallestVal << endl << endl << endl;
}
}
Basically, via this approach you can discard k/2 elements at each step.
The K will recursively change from k => k/2 => k/4 => ... till it reaches 1.
So, Time Complexity is O(logk)
At k=1 , we get the lowest of the two arrays.
The following code is in JAVA. Please note that the we are subtracting 1 (-1) in the code from the indices because Java array's index starts from 0 and not 1, eg. k=3 is represented by the element in 2nd index of an array.
private int kthElement(int[] arr1, int[] arr2, int k) {
if (k < 1 || k > (arr1.length + arr2.length))
return -1;
return helper(arr1, 0, arr1.length - 1, arr2, 0, arr2.length - 1, k);
}
private int helper(int[] arr1, int low1, int high1, int[] arr2, int low2, int high2, int k) {
if (low1 > high1) {
return arr2[low2 + k - 1];
} else if (low2 > high2) {
return arr1[low1 + k - 1];
}
if (k == 1) {
return Math.min(arr1[low1], arr2[low2]);
}
int i = Math.min(low1 + k / 2, high1 + 1);
int j = Math.min(low2 + k / 2, high2 + 1);
if (arr1[i - 1] > arr2[j - 1]) {
return helper(arr1, low1, high1, arr2, j, high2, k - (j - low2));
} else {
return helper(arr1, i, high1, arr2, low2, high2, k - (i - low1));
}
}
The first pseudo code provided above, does not work for many values. For example,
here are two arrays.
int[] a = { 1, 5, 6, 8, 9, 11, 15, 17, 19 };
int[] b = { 4, 7, 8, 13, 15, 18, 20, 24, 26 };
It did not work for k=3 and k=9 in it. I have another solution. It is given below.
private static void traverse(int pt, int len) {
int temp = 0;
if (len == 1) {
int val = 0;
while (k - (pt + 1) - 1 > -1 && M[pt] < N[k - (pt + 1) - 1]) {
if (val == 0)
val = M[pt] < N[k - (pt + 1) - 1] ? N[k - (pt + 1) - 1]
: M[pt];
else {
int t = M[pt] < N[k - (pt + 1) - 1] ? N[k - (pt + 1) - 1]
: M[pt];
val = val < t ? val : t;
}
++pt;
}
if (val == 0)
val = M[pt] < N[k - (pt + 1) - 1] ? N[k - (pt + 1) - 1] : M[pt];
System.out.println(val);
return;
}
temp = len / 2;
if (M[pt + temp - 1] < N[k - (pt + temp) - 1]) {
traverse(pt + temp, temp);
} else {
traverse(pt, temp);
}
}
But... it is also not working for k=5. There is this even/odd catch of k which is not letting it to be simple.
public class KthSmallestInSortedArray {
public static void main(String[] args) {
int a1[] = {2, 3, 10, 11, 43, 56},
a2[] = {120, 13, 14, 24, 34, 36},
k = 4;
System.out.println(findKthElement(a1, a2, k));
}
private static int findKthElement(int a1[], int a2[], int k) {
/** Checking k must less than sum of length of both array **/
if (a1.length + a2.length < k) {
throw new IllegalArgumentException();
}
/** K must be greater than zero **/
if (k <= 0) {
throw new IllegalArgumentException();
}
/**
* Finding begin, l and end such that
* begin <= l < end
* a1[0].....a1[l-1] and
* a2[0]....a2[k-l-1] are the smallest k numbers
*/
int begin = Math.max(0, k - a2.length);
int end = Math.min(a1.length, k);
while (begin < end) {
int l = begin + (end - begin) / 2;
/** Can we include a1[l] in the k smallest numbers */
if ((l < a1.length) &&
(k - l > 0) &&
(a1[l] < a2[k - l - 1])) {
begin = l + 1;
} else if ((l > 0) &&
(k - l < a2.length) &&
(a1[l - 1] > a2[k - 1])) {
/**
* This is the case where we can discard
* a[l-1] from the set of k smallest numbers
*/
end = l;
} else {
/**
* We found our answer since both inequalities were
* false
*/
begin = l;
break;
}
}
if (begin == 0) {
return a2[k - 1];
} else if (begin == k) {
return a1[k - 1];
} else {
return Math.max(a1[begin - 1], a2[k - begin - 1]);
}
}
}
Here is mine solution in java . Will try to further optimize it
public class FindKLargestTwoSortedArray {
public static void main(String[] args) {
int[] arr1 = { 10, 20, 40, 80 };
int[] arr2 = { 15, 35, 50, 75 };
FindKLargestTwoSortedArray(arr1, 0, arr1.length - 1, arr2, 0,
arr2.length - 1, 6);
}
public static void FindKLargestTwoSortedArray(int[] arr1, int start1,
int end1, int[] arr2, int start2, int end2, int k) {
if ((start1 <= end1 && start1 >= 0 && end1 < arr1.length)
&& (start2 <= end2 && start2 >= 0 && end2 < arr2.length)) {
int midIndex1 = (start1 + (k - 1) / 2);
midIndex1 = midIndex1 >= arr1.length ? arr1.length - 1 : midIndex1;
int midIndex2 = (start2 + (k - 1) / 2);
midIndex2 = midIndex2 >= arr2.length ? arr2.length - 1 : midIndex2;
if (arr1[midIndex1] == arr2[midIndex2]) {
System.out.println("element is " + arr1[midIndex1]);
} else if (arr1[midIndex1] < arr2[midIndex2]) {
if (k == 1) {
System.out.println("element is " + arr1[midIndex1]);
return;
} else if (k == 2) {
System.out.println("element is " + arr2[midIndex2]);
return;
}else if (midIndex1 == arr1.length-1 || midIndex2 == arr2.length-1 ) {
if(k==(arr1.length+arr2.length)){
System.out.println("element is " + arr2[midIndex2]);
return;
}else if(k==(arr1.length+arr2.length)-1){
System.out.println("element is " + arr1[midIndex1]);
return;
}
}
int remainingElementToSearch = k - (midIndex1-start1);
FindKLargestTwoSortedArray(
arr1,
midIndex1,
(midIndex1 + remainingElementToSearch) >= arr1.length ? arr1.length-1
: (midIndex1 + remainingElementToSearch), arr2,
start2, midIndex2, remainingElementToSearch);
} else if (arr1[midIndex1] > arr2[midIndex2]) {
FindKLargestTwoSortedArray(arr2, start2, end2, arr1, start1,
end1, k);
}
} else {
return;
}
}
}
This is inspired from Algo at wonderful youtube video
Link to code complexity (log(n)+log(m))
Link to Code (log(n)*log(m))
Implementation of (log(n)+log(m)) solution
I would like to add my explanation to the problem.
This is a classic problem where we have to use the fact that the two arrays are sorted .
we have been given two sorted arrays arr1 of size sz1 and arr2 of size sz2
a)Lets suppose if
Checking If k is valid
k is > (sz1+sz2)
then we cannot find kth smallest element in union of both sorted arrays ryt So return Invalid data.
b)Now if above condition holds false and we have valid and feasible value of k,
Managing Edge Cases
We will append both the arrays by -infinity values at front and +infinity values at end to cover the edge cases of k = 1,2 and k = (sz1+sz2-1),(sz1+sz2)etc.
Now both the arrays have size (sz1+2) and (sz2+2) respectively
Main Algorithm
Now,we will do binary search on arr1 .We will do binary search on arr1 looking for an index i , startIndex <= i <= endIndex
such that if we find corresponding index j in arr2 using constraint {(i+j) = k},then if
if (arr2[j-1] < arr1[i] < arr2[j]),then arr1[i] is the kth smallest (Case 1)
else if (arr1[i-1] < arr2[j] < arr1[i]) ,then arr2[i] is the kth smallest (Case 2)
else signifies either arr1[i] < arr2[j-1] < arr2[j] (Case3)
or arr2[j-1] < arr2[j] < arr1[i] (Case4)
Since we know that the kth smallest element has (k-1) elements smaller than it in union of both the arrays ryt? So,
In Case1, what we did , we ensured that there are a total of (k-1) smaller elements to arr1[i] because elements smaller than arr1[i] in arr1 array are i-1 in number than we know (arr2[j-1] < arr1[i] < arr2[j]) and number of elements smaller than arr1[i] in arr2 is j-1 because j is found using (i-1)+(j-1) = (k-1) So kth smallest element will be arr1[i]
But answer may not always come from the first array ie arr1 so we checked for case2 which also satisfies similarly like case 1 because (i-1)+(j-1) = (k-1) . Now if we have (arr1[i-1] < arr2[j] < arr1[i]) we have a total of k-1 elements smaller than arr2[j] in union of both the arrays so its the kth smallest element.
In case3 , to form it to any of case 1 or case 2, we need to increment i and j will be found according using constraint {(i+j) = k} ie in binary search move to right part ie make startIndex = middleIndex
In case4, to form it to any of case 1 or case 2, we need to decrement i and j will be found according using constraint {(i+j) = k} ie in binary search move to left part ie make endIndex = middleIndex.
Now how to decide startIndex and endIndex at beginning of binary search over arr1
with startindex = 1 and endIndex = ??.We need to decide.
If k > sz1,endIndex = (sz1+1) , else endIndex = k;
Because if k is greater than the size of the first array we may have to do binary search over the entire array arr1 else we only need to take first k elements of it because sz1-k elements can never contribute in calculating kth smallest.
CODE Shown Below
// Complexity O(log(n)+log(m))
#include<bits/stdc++.h>
using namespace std;
#define f(i,x,y) for(int i = (x);i < (y);++i)
#define F(i,x,y) for(int i = (x);i > (y);--i)
int max(int a,int b){return (a > b?a:b);}
int min(int a,int b){return (a < b?a:b);}
int mod(int a){return (a > 0?a:((-1)*(a)));}
#define INF 1000000
int func(int *arr1,int *arr2,int sz1,int sz2,int k)
{
if((k <= (sz1+sz2))&&(k > 0))
{
int s = 1,e,i,j;
if(k > sz1)e = sz1+1;
else e = k;
while((e-s)>1)
{
i = (e+s)/2;
j = ((k-1)-(i-1));
j++;
if(j > (sz2+1)){s = i;}
else if((arr1[i] >= arr2[j-1])&&(arr1[i] <= arr2[j]))return arr1[i];
else if((arr2[j] >= arr1[i-1])&&(arr2[j] <= arr1[i]))return arr2[j];
else if(arr1[i] < arr2[j-1]){s = i;}
else if(arr1[i] > arr2[j]){e = i;}
else {;}
}
i = e,j = ((k-1)-(i-1));j++;
if((arr1[i] >= arr2[j-1])&&(arr1[i] <= arr2[j]))return arr1[i];
else if((arr2[j] >= arr1[i-1])&&(arr2[j] <= arr1[i]))return arr2[j];
else
{
i = s,j = ((k-1)-(i-1));j++;
if((arr1[i] >= arr2[j-1])&&(arr1[i] <= arr2[j]))return arr1[i];
else return arr2[j];
}
}
else
{
cout << "Data Invalid" << endl;
return -INF;
}
}
int main()
{
int n,m,k;
cin >> n >> m >> k;
int arr1[n+2];
int arr2[m+2];
f(i,1,n+1)
cin >> arr1[i];
f(i,1,m+1)
cin >> arr2[i];
arr1[0] = -INF;
arr2[0] = -INF;
arr1[n+1] = +INF;
arr2[m+1] = +INF;
int val = func(arr1,arr2,n,m,k);
if(val != -INF)cout << val << endl;
return 0;
}
For Solution of complexity (log(n)*log(m))
Just i missed using advantage of the fact that for each i the j can be found using constraint {(i-1)+(j-1)=(k-1)} So for each i i was further applying binary search on second array to find j such that arr2[j] <= arr1[i].So this solution can be optimized further
#include <bits/stdc++.h>
using namespace std;
int findKthElement(int a[],int start1,int end1,int b[],int start2,int end2,int k){
if(start1 >= end1)return b[start2+k-1];
if(start2 >= end2)return a[start1+k-1];
if(k==1)return min(a[start1],b[start2]);
int aMax = INT_MAX;
int bMax = INT_MAX;
if(start1+k/2-1 < end1) aMax = a[start1 + k/2 - 1];
if(start2+k/2-1 < end2) bMax = b[start2 + k/2 - 1];
if(aMax > bMax){
return findKthElement(a,start1,end1,b,start2+k/2,end2,k-k/2);
}
else{
return findKthElement(a,start1 + k/2,end1,b,start2,end2,k-k/2);
}
}
int main(void){
int t;
scanf("%d",&t);
while(t--){
int n,m,k;
cout<<"Enter the size of 1st Array"<<endl;
cin>>n;
int arr[n];
cout<<"Enter the Element of 1st Array"<<endl;
for(int i = 0;i<n;i++){
cin>>arr[i];
}
cout<<"Enter the size of 2nd Array"<<endl;
cin>>m;
int arr1[m];
cout<<"Enter the Element of 2nd Array"<<endl;
for(int i = 0;i<m;i++){
cin>>arr1[i];
}
cout<<"Enter The Value of K";
cin>>k;
sort(arr,arr+n);
sort(arr1,arr1+m);
cout<<findKthElement(arr,0,n,arr1,0,m,k)<<endl;
}
return 0;
}
Time Complexcity is O(log(min(n,m)))
Below C# code to Find the k-th Smallest Element in the Union of Two Sorted Arrays. Time Complexity : O(logk)
public static int findKthSmallestElement1(int[] A, int startA, int endA, int[] B, int startB, int endB, int k)
{
int n = endA - startA;
int m = endB - startB;
if (n <= 0)
return B[startB + k - 1];
if (m <= 0)
return A[startA + k - 1];
if (k == 1)
return A[startA] < B[startB] ? A[startA] : B[startB];
int midA = (startA + endA) / 2;
int midB = (startB + endB) / 2;
if (A[midA] <= B[midB])
{
if (n / 2 + m / 2 + 1 >= k)
return findKthSmallestElement1(A, startA, endA, B, startB, midB, k);
else
return findKthSmallestElement1(A, midA + 1, endA, B, startB, endB, k - n / 2 - 1);
}
else
{
if (n / 2 + m / 2 + 1 >= k)
return findKthSmallestElement1(A, startA, midA, B, startB, endB, k);
else
return findKthSmallestElement1(A, startA, endA, B, midB + 1, endB, k - m / 2 - 1);
}
}
Check this code.
import math
def findkthsmallest():
A=[1,5,10,22,30,35,75,125,150,175,200]
B=[15,16,20,22,25,30,100,155,160,170]
lM=0
lN=0
hM=len(A)-1
hN=len(B)-1
k=17
while True:
if k==1:
return min(A[lM],B[lN])
cM=hM-lM+1
cN=hN-lN+1
tmp = cM/float(cM+cN)
iM=int(math.ceil(tmp*k))
iN=k-iM
iM=lM+iM-1
iN=lN+iN-1
if A[iM] >= B[iN]:
if iN == hN or A[iM] < B[iN+1]:
return A[iM]
else:
k = k - (iN-lN+1)
lN=iN+1
hM=iM-1
if B[iN] >= A[iM]:
if iM == hM or B[iN] < A[iM+1]:
return B[iN]
else:
k = k - (iM-lM+1)
lM=iM+1
hN=iN-1
if hM < lM:
return B[lN+k-1]
if hN < lN:
return A[lM+k-1]
if __name__ == '__main__':
print findkthsmallest();

Fast code to determine if any two subsets of columns have the same sum

For a given n and m I iterate over all n by m partial circulant matrices with entries that are either 0 or 1. I want to find if there is a matrix such that there are no two subsets of the columns that give the same sum. Here when we add columns we just do it elementwise. My current code uses constraint programming via ortools. However it is not as fast I would like. For n = 7 and m = 12 it takes over 3 minutes and for n = 10, m = 18 it doesn't terminate even though there are only 2^18 = 262144 different matrices to consider. Here is my code.
from scipy.linalg import circulant
import numpy as np
import itertools
from ortools.constraint_solver import pywrapcp as cs
n = 7
m = 12
def isdetecting(matrix):
X = np.array([solver.IntVar(values) for i in range(matrix.shape[1])])
X1 = X.tolist()
for row in matrix:
x = X[row].tolist()
solver.Add(solver.Sum(x) == 0)
db = solver.Phase(X1, solver.INT_VAR_DEFAULT, solver.INT_VALUE_DEFAULT)
solver.NewSearch(db)
count = 0
while (solver.NextSolution() and count < 2):
solution = [x.Value() for x in X1]
count += 1
solver.EndSearch()
if (count < 2):
return True
values = [-1,0,1]
solver = cs.Solver("scip")
for row in itertools.product([0,1],repeat = m):
M = np.array(circulant(row)[0:n], dtype=bool)
if isdetecting(M):
print M.astype(int)
break
Can this problem be solved fast enough so that n = 10, m = 18 can be solved?
One problem is that you are declaring the "solver" variable globally and it seems to confuse or-tools to reuse it many times. When moving it inside "isdetecting", then the (7,12) problem is solved much faster, in about 7 seconds (compared to 2:51 minutes for the original model). I haven't checked it for the larger problem, though.
Also, it might be good idea to test different labelings (instead of solver.INT_VAR_DEFAULT and solver.INT_VALUE_DEFAULT), though binary value tend to be not very sensitive to different labelings. See the code for another labeling.
def isdetecting(matrix):
solver = cs.Solver("scip") # <----
X = np.array([solver.IntVar(values) for i in range(matrix.shape[1])])
X1 = X.tolist()
for row in matrix:
x = X[row].tolist()
solver.Add(solver.Sum(x) == 0)
# db = solver.Phase(X1, solver.INT_VAR_DEFAULT, solver.INT_VALUE_DEFAULT)
db = solver.Phase(X1, solver.CHOOSE_FIRST_UNBOUND, solver.ASSIGN_CENTER_VALUE)
solver.NewSearch(db)
count = 0
while (solver.NextSolution() and count < 2):
solution = [x.Value() for x in X1]
count += 1
solver.EndSearch()
if (count < 2):
print "FOUND"
return True
Edit: Here are constraints to remove the all-0 solutions as mentioned in the comments. What I know of, it require a separate list. It now takes a little longer (10.4s vs 7s).
X1Abs = [solver.IntVar(values, 'X1Abs[%i]' % i) for i in range(X1_len)]
for i in range(X1_len):
solver.Add(X1Abs[i] == abs(X1[i]))
solver.Add(solver.Sum(X1Abs) > 0)
Something like this is what I had in mind. I'd estimate the running time for command line parameters 10 18 at less than 8 hours on my machine.
public class Search {
public static void main(String[] args) {
int n = Integer.parseInt(args[0]);
int m = Integer.parseInt(args[1]);
int row = search(n, m);
if (row >= 0) {
printRow(m, row);
}
}
private static int search(int n, int m) {
if (n < 0 || m < n || m >= 31 || powOverflows(m + 1, n)) {
throw new IllegalArgumentException();
}
long[] column = new long[m];
long[] sums = new long[1 << m];
int row = 1 << m;
while (row-- > 0) {
System.err.println(row);
for (int j = 0; j < m; j++) {
column[j] = 0;
for (int i = 0; i < n; i++) {
column[j] = (column[j] * (m + 1)) + ((row >> ((i + j) % m)) & 1);
}
}
for (int subset = 0; subset < (1 << m); subset++) {
long sum = 0;
for (int j = 0; j < m; j++) {
if (((subset >> j) & 1) == 1) {
sum += column[j];
}
}
sums[subset] = sum;
}
java.util.Arrays.sort(sums);
boolean duplicate = false;
for (int k = 1; k < (1 << m); k++) {
if (sums[k - 1] == sums[k]) {
duplicate = true;
break;
}
}
if (!duplicate) {
break;
}
}
return row;
}
private static boolean powOverflows(long b, int e) {
if (b <= 0 || e < 0) {
throw new IllegalArgumentException();
}
if (e == 0) {
return false;
}
long max = Long.MAX_VALUE;
while (e > 1) {
if (b > Integer.MAX_VALUE) {
return true;
}
if ((e & 1) == 1) {
max /= b;
}
b *= b;
e >>= 1;
}
return b > max;
}
private static void printRow(int m, int row) {
for (int j = 0; j < m; j++) {
System.out.print((row >> j) & 1);
}
System.out.println();
}
}

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