# Longest subarray having difference in the count of 1’s and 0’s equal to k

Given a binary array arr[] of size n and a value k. The task is to find the length of the longest subarray having difference in the count of 1’s and 0’s equal to k. The count of 1’s should be equal to or greater than the count of 0’s in the subarray according to the value of k.

Examples:

```Input: arr[] = {0, 1, 1, 0, 1}, k = 2
Output: 4
The highlighted portion is the required subarray
{0, 1, 1, 0, 1}. In the subarray count of 1's is 3
and count of 0's is 1.
Therefore, difference in count = 3 - 1 = 2.

Input: arr[] = {1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1}, k = 0
Output: 6
The highlighted portion is the required subarray
{1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1}. In the subarray
count of 1's is 3 and count of 0's is 3.
Therefore, difference in count = 3 - 3 = 0.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Consider the difference in the count of 1’s and 0’s of all the sub-arrays and return the length of the longest sub-array having required difference equal to ‘k’. Time Complexity will be O(n^2).

Efficient Approach: This problem is a variation of finding the longest sub-array having sum k. Replace all the 0’s in the arr[] with -1 and then find the longest subarray of ‘arr’ having sum equal to ‘k’.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the length of the longest ` `// subarray having difference in the count ` `// of 1's and 0's equal to k ` `int` `lenOfLongSubarr(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` ` `  `    ``// unordered_map 'um' implemented ` `    ``// as hash table ` `    ``unordered_map<``int``, ``int``> um; ` `    ``int` `sum = 0, maxLen = 0; ` ` `  `    ``// traverse the given array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// accumulate sum ` `        ``sum += ((arr[i] == 0) ? -1 : arr[i]); ` ` `  `        ``// when subarray starts from index '0' ` `        ``if` `(sum == k) ` `            ``maxLen = i + 1; ` ` `  `        ``// make an entry for 'sum' if it is ` `        ``// not present in 'um' ` `        ``if` `(um.find(sum) == um.end()) ` `            ``um[sum] = i; ` ` `  `        ``// check if 'sum-k' is present in 'um' ` `        ``// or not ` `        ``if` `(um.find(sum - k) != um.end()) { ` ` `  `            ``// update maxLength ` `            ``if` `(maxLen < (i - um[sum - k])) ` `                ``maxLen = i - um[sum - k]; ` `        ``} ` `    ``} ` ` `  `    ``// required maximum length ` `    ``return` `maxLen; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 1, 0, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `k = 2; ` `    ``cout << ``"Length = "` `         ``<< lenOfLongSubarr(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach. ` `import` `java.util.HashMap; ` `import` `java.util.Map; ` ` `  `class` `GfG ` `{ ` ` `  `    ``// Function to find the length of the longest  ` `    ``// subarray having difference in the count  ` `    ``// of 1's and 0's equal to k  ` `    ``static` `int` `lenOfLongSubarr(``int` `arr[], ``int` `n, ``int` `k)  ` `    ``{  ` `        ``// unordered_map 'um' implemented  ` `        ``// as hash table  ` `        ``HashMap um = ``new` `HashMap<>();  ` `        ``int` `sum = ``0``, maxLen = ``0``;  ` `     `  `        ``// traverse the given array  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `     `  `            ``// accumulate sum  ` `            ``sum += ((arr[i] == ``0``) ? -``1` `: arr[i]);  ` `     `  `            ``// when subarray starts from index '0'  ` `            ``if` `(sum == k)  ` `                ``maxLen = i + ``1``;  ` `     `  `            ``// make an entry for 'sum' if  ` `            ``// it is not present in 'um'  ` `            ``if` `(!um.containsKey(sum))  ` `                ``um.put(sum, i);  ` `     `  `            ``// check if 'sum-k' is present  ` `            ``// in 'um' or not  ` `            ``if` `(um.containsKey(sum - k))  ` `            ``{  ` `     `  `                ``// update maxLength  ` `                ``if` `(maxLen < (i - um.get(sum - k)))  ` `                    ``maxLen = i - um.get(sum - k);  ` `            ``}  ` `        ``}  ` `     `  `        ``// required maximum length  ` `        ``return` `maxLen;  ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `         `  `        ``int` `arr[] = { ``0``, ``1``, ``1``, ``0``, ``1` `};  ` `        ``int` `n = arr.length;  ` `        ``int` `k = ``2``;  ` ` `  `        ``System.out.println(``"Length = "` `+ lenOfLongSubarr(arr, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by Rituraj Jain `

## Python

 `# Python3 implementation of above approach ` ` `  `# function to find the length of the longest ` `# subarray having difference in the count ` `# of 1's and 0's equal to k ` `def` `lenOfLongSubarr(arr, n, k): ` ` `  `    ``# unordered_map 'um' implemented ` `    ``# as hash table ` `    ``um ``=` `dict``() ` ` `  `    ``Sum``, maxLen ``=` `0``, ``0` ` `  `    ``# traverse the given array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# accumulate sum ` `        ``if` `arr[i] ``=``=` `0``: ` `            ``Sum` `+``=` `-``1` `        ``else``: ` `            ``Sum``+``=``arr[i] ` ` `  `        ``# when subarray starts from index '0' ` `        ``if` `(``Sum` `=``=` `k): ` `            ``maxLen ``=` `i ``+` `1` ` `  `        ``# make an entry for 'Sum' if it is ` `        ``# not present in 'um' ` `        ``if` `(``Sum` `not` `in` `um.keys()): ` `            ``um[``Sum``] ``=` `i ` ` `  `        ``# check if 'Sum-k' is present in 'um' ` `        ``# or not ` `        ``if` `((``Sum` `-` `k) ``in` `um.keys()): ` ` `  `            ``# update maxLength ` `            ``if` `(maxLen < (i ``-` `um[``Sum` `-` `k])): ` `                ``maxLen ``=` `i ``-` `um[``Sum` `-` `k] ` ` `  `    ``# required maximum length ` `    ``return` `maxLen ` ` `  `# Driver Code ` `arr ``=` `[``0``, ``1``, ``1``, ``0``, ``1``] ` `n ``=` `len``(arr) ` `k ``=` `2` `print``(``"Length = "``,lenOfLongSubarr(arr, n, k)) ` ` `  `# This code is contributed by mohit kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to find the length of the longest  ` `    ``// subarray having difference in the count  ` `    ``// of 1's and 0's equal to k  ` `    ``static` `int` `lenOfLongSubarr(``int` `[]arr, ` `                               ``int` `n, ``int` `k)  ` `    ``{  ` `        ``// unordered_map 'um' implemented  ` `        ``// as hash table  ` `        ``Dictionary<``int``,  ` `                   ``int``> um = ``new` `Dictionary<``int``,  ` `                                            ``int``>(); ` `        ``int` `sum = 0, maxLen = 0;  ` `     `  `        ``// traverse the given array  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{  ` `     `  `            ``// accumulate sum  ` `            ``sum += ((arr[i] == 0) ? -1 : arr[i]);  ` `     `  `            ``// when subarray starts from index '0'  ` `            ``if` `(sum == k)  ` `                ``maxLen = i + 1;  ` `     `  `            ``// make an entry for 'sum' if  ` `            ``// it is not present in 'um'  ` `            ``if` `(!um.ContainsKey(sum))  ` `                ``um.Add(sum, i);  ` `     `  `            ``// check if 'sum-k' is present  ` `            ``// in 'um' or not  ` `            ``if` `(um.ContainsKey(sum - k))  ` `            ``{  ` `     `  `                ``// update maxLength  ` `                ``if` `(maxLen < (i - um[sum - k]))  ` `                    ``maxLen = i - um[sum - k];  ` `            ``}  ` `        ``}  ` `     `  `        ``// required maximum length  ` `        ``return` `maxLen;  ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args) ` `    ``{ ` `         `  `        ``int` `[]arr = { 0, 1, 1, 0, 1 };  ` `        ``int` `n = arr.Length;  ` `        ``int` `k = 2;  ` ` `  `        ``Console.WriteLine(``"Length = "` `+  ` `                ``lenOfLongSubarr(arr, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```Length = 4
```

Time Complexity:
O(n)
Auxiliary Space: O(n)

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