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

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// C++14 program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a subarray of
// length len consisting of equal elements
// can be obtained or not
bool check(vector<int> pSum, int len, int k,
           vector<int> 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<int> arr, int k)
{
     
    // Sort the array in
    // ascending order
    sort(arr.begin(), arr.end());
 
    // Make prefix sum array
    vector<int> 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<int> arr = { 1, 1, 1 };
    int k = 7;
 
    // Function call
    cout << (maxEqualIdx(arr, k));
}
 
// This code is contributed by mohit kumar 29

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Java

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// 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));
    }
}

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Python3

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# 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

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C#

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// 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

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Output: 

3







 

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

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