# Maximize length of subarray having equal elements by adding at most K

Given an array arr[] consisting of N positive integers and an integer K, which represents the maximum number that can be added to the array elements. The task is to maximize the length of the longest possible subarray of equal elements by adding at most K.

Examples:

Input: arr[] = {3, 0, 2, 2, 1}, k = 3
Output: 4
Explanation:
Step 1: Adding 2 to arr[1] modifies array to {3, 2, 2, 2, 1}
Step 2: Adding 1 to arr[4] modifies array to {3, 2, 2, 2, 2}
Therefore, answer will be 4 ({arr[1], …, arr[4]}).

Input: arr[] = {1, 1, 1}, k = 7
Output: 3
Explanation:
All array elements are already equal. Therefore, the length is 3.

Approach: Follow the steps below to solve the problem:

• Sort the array arr[]. Then, use Binary Search to pick a possible value for the maximum indices having the same element.
• For each picked_value, use the Sliding Window technique to check if it is possible to make all elements equal for any subarray of size picked_value.
• Finally, print the longest possible length of subarray obtained.

Below is the implementation for the above approach:

## C++14

 // C++14 program for above approach #include using namespace std;   // Function to check if a subarray of // length len consisting of equal elements // can be obtained or not bool check(vector pSum, int len, int k,            vector a) {           // Sliding window     int i = 0;     int j = len;           while (j <= a.size())     {                   // Last element of the sliding window         // will be having the max size in the         // current window         int maxSize = a[j - 1];           int totalNumbers = maxSize * len;           // The current number of element in all         // indices of the current sliding window         int currNumbers = pSum[j] - pSum[i];           // If the current number of the window,         // added to k exceeds totalNumbers         if (currNumbers + k >= totalNumbers)         {             return true;         }         else         {             i++;             j++;         }     }     return false; }   // Function to find the maximum number of // indices having equal elements after // adding at most k numbers int maxEqualIdx(vector arr, int k) {           // Sort the array in     // ascending order     sort(arr.begin(), arr.end());       // Make prefix sum array     vector prefixSum(arr.size());     prefixSum[1] = arr[0];       for(int i = 1;             i < prefixSum.size() - 1; ++i)     {         prefixSum[i + 1] = prefixSum[i] +                                  arr[i];     }       // Initialize variables     int max = arr.size();     int min = 1;     int ans = 1;       while (min <= max)     {                   // Update mid         int mid = (max + min) / 2;           // Check if any subarray         // can be obtained of length         // mid having equal elements         if (check(prefixSum, mid, k, arr))         {             ans = mid;             min = mid + 1;         }         else         {                           // Decrease max to mid             max = mid - 1;         }     }     return ans; }   // Driver Code int main() {     vector arr = { 1, 1, 1 };     int k = 7;       // Function call     cout << (maxEqualIdx(arr, k)); }   // This code is contributed by mohit kumar 29

## Java

 // Java program for above approach   import java.util.*;   class GFG {       // Function to find the maximum number of     // indices having equal elements after     // adding at most k numbers     public static int maxEqualIdx(int[] arr,                                   int k)     {         // Sort the array in         // ascending order         Arrays.sort(arr);           // Make prefix sum array         int[] prefixSum             = new int[arr.length + 1];         prefixSum[1] = arr[0];           for (int i = 1; i < prefixSum.length - 1;              ++i) {               prefixSum[i + 1]                 = prefixSum[i] + arr[i];         }           // Initialize variables         int max = arr.length;         int min = 1;         int ans = 1;           while (min <= max) {               // Update mid             int mid = (max + min) / 2;               // Check if any subarray             // can be obtained of length             // mid having equal elements             if (check(prefixSum, mid, k, arr)) {                   ans = mid;                 min = mid + 1;             }             else {                   // Decrease max to mid                 max = mid - 1;             }         }           return ans;     }       // Function to check if a subarray of     // length len consisting of equal elements     // can be obtained or not     public static boolean check(int[] pSum,                                 int len, int k,                                 int[] a)     {           // Sliding window         int i = 0;         int j = len;         while (j <= a.length) {               // Last element of the sliding window             // will be having the max size in the             // current window             int maxSize = a[j - 1];               int totalNumbers = maxSize * len;               // The current number of element in all             // indices of the current sliding window             int currNumbers = pSum[j] - pSum[i];               // If the current number of the window,             // added to k exceeds totalNumbers             if (currNumbers + k >= totalNumbers) {                   return true;             }             else {                 i++;                 j++;             }         }         return false;     }       // Driver Code     public static void main(String[] args)     {           int[] arr = { 1, 1, 1 };         int k = 7;           // Function call         System.out.println(maxEqualIdx(arr, k));     } }

## Python3

 # Python3 program for above approach    # Function to find the maximum number of # indices having equal elements after # adding at most k numbers def maxEqualIdx(arr, k):           # Sort the array in     # ascending order     arr.sort()           # Make prefix sum array     prefixSum = [0] * (len(arr) + 1)     prefixSum[1] = arr[0]        for i in range(1, len(prefixSum) - 1 ,1):         prefixSum[i + 1] = prefixSum[i] + arr[i]         # Initialize variables     max = len(arr)     min = 1     ans = 1        while (min <= max):                   # Update mid         mid = (max + min) // 2            # Check if any subarray         # can be obtained of length         # mid having equal elements         if (check(prefixSum, mid, k, arr)):             ans = mid             min = mid + 1         else:                           # Decrease max to mid             max = mid - 1           return ans   # Function to check if a subarray of # length len consisting of equal elements # can be obtained or not def check(pSum, lenn, k, a):           # Sliding window     i = 0     j = lenn           while (j <= len(a)):                 # Last element of the sliding window         # will be having the max size in the         # current window         maxSize = a[j - 1]            totalNumbers = maxSize * lenn            # The current number of element in all         # indices of the current sliding window         currNumbers = pSum[j] - pSum[i]            # If the current number of the window,         # added to k exceeds totalNumbers         if (currNumbers + k >= totalNumbers):             return True               else:             i += 1             j += 1           return False   # Driver Code   arr = [ 1, 1, 1 ] k = 7    # Function call print(maxEqualIdx(arr, k))   # This code is contributed by code_hunt

## C#

 // C# program for // the above approach using System; class GFG{   // Function to find the maximum number of // indices having equal elements after // adding at most k numbers public static int maxEqualIdx(int[] arr,                               int k) {   // Sort the array in   // ascending order   Array.Sort(arr);     // Make prefix sum array   int[] prefixSum = new int[arr.Length + 1];   prefixSum[1] = arr[0];     for (int i = 1;            i < prefixSum.Length - 1; ++i)   {     prefixSum[i + 1] = prefixSum[i] + arr[i];   }     // Initialize variables   int max = arr.Length;   int min = 1;   int ans = 1;     while (min <= max)   {     // Update mid     int mid = (max + min) / 2;       // Check if any subarray     // can be obtained of length     // mid having equal elements     if (check(prefixSum, mid, k, arr))     {       ans = mid;       min = mid + 1;     }     else     {       // Decrease max to mid       max = mid - 1;     }   }   return ans; }   // Function to check if a subarray of // length len consisting of equal elements // can be obtained or not public static bool check(int[] pSum,                          int len, int k,                          int[] a) {   // Sliding window   int i = 0;   int j = len;   while (j <= a.Length)   {     // Last element of the sliding window     // will be having the max size in the     // current window     int maxSize = a[j - 1];       int totalNumbers = maxSize * len;       // The current number of element in all     // indices of the current sliding window     int currNumbers = pSum[j] - pSum[i];       // If the current number of the window,     // added to k exceeds totalNumbers     if (currNumbers + k >= totalNumbers)     {       return true;     }     else     {       i++;       j++;     }   }   return false; }   // Driver Code public static void Main(String[] args) {   int[] arr = {1, 1, 1};   int k = 7;     // Function call   Console.WriteLine(maxEqualIdx(arr, k)); } }   // This code is contributed by Rajput-Ji

## Javascript



Output:

3

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

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