# 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++ 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 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 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

 # 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# 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 um = new Dictionary();         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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