Minimize the max of Array by doing at most K increment and decrement

Given a positive array arr[] of size N (2 ≤ N ≤ 10⁵) and a positive integer K, the task is to minimize the maximum value of arr[] by performing at most K operations where in each operation, you can select any two distinct integers i and j (0 ≤ i, j < N), then increase the value of arr[i] by 1 and decrease the value of arr[j] by 1.

Example:

Input: arr = {3, 3, 7, 1, 6}, k = 2
Output: 6
Explanation:
One set of optimal operations is as follows:
For K = 1 => Select i = 2 and j = 0  and arr becomes {4, 3, 6, 1, 6}.
For K = 2 => Select i = 4 and j = 3, and arr becomes {4, 3, 6, 2, 5}.
The maximum integer of nums is . It can be shown that the maximum number cannot be less than 6.
Therefore, we return 6.

Input: nums = {10, 1, 1}, k = 3
Output: 7

Approach: The problem can be solved using binary search based on the following idea.

Choose the minimum and maximum possible value in start and end variable respectively. Find the mid value of start and end. Check if we can choose mid as the maximum possible value in array. If its true then this is one of our valid answer. So update our answer in result and shift the start to mid – 1 to minimise the maximum possible value. Otherwise shift the end to mid + 1. Finally return the result.

Follow the steps below to implement the above idea:

• Initialise a variable start = 0, end = max value in arr[] and result
• While start is less than equal to end, do the following
  • Calculate the mid by (start + end) / 2
  • Check if mid can be a valid max Value by the following
    • Initialize a variable operationCount = 0
    • Iterate over the array arr[]
      • Check if arr[i] is greater than our expected maximum is mid
        • operationCount += mid – arr[i]
      • Check if operationsCount is greater than K.
        •  If true, return false
    • Return true
  • If mid is the valid maximum possible value then,
    • Shift end to mid – 1
    • store mid into result
  • Otherwise, shift the start to mid + 1
• Return the result

Below is the implementation of the above approach.

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if maxx can be a valid value
bool isValid(int maxx, vector<int>& arr, int k)
{
    // Initialize a variable sum = 0
    long long operationCount = 0;

    // Iterate over the array arr[]
    for (long long i = 0; i < arr.size(); i++) {

        // Keep storing the maximum value in sum
        if (arr[i] > maxx)
            operationCount += arr[i] - maxx;

        // Check if number of operation required to
        // make maximum element at max maxx.
        // If true, return false
        if (operationCount > k)
            return false;
    }

    // Return true
    return true;
}

// Function to find the minimum possible maximum
int minimizeArrayMaxValue(vector<int>& arr, int k)
{
    long long n = arr.size();

    // Initialise a variable start = 0,
    // end = max value in, arr[] and result
    long long start = *min_element(arr.begin(), arr.end()),
              end = *max_element(arr.begin(), arr.end());

    long long result;

    // Do while start is less than equals to end
    while (start <= end) {

        // Calculate the mid by (start + end) / 2;
        long long mid = (start + end) / 2;

        // Check if mid can be a valid max Value.
        if (isValid(mid, arr, k)) {

            // Shift end to mid - 1
            end = mid - 1;

            // Store mid into result
            result = mid;
          cout<<result;
        }

        // Otherwise, shift start to mid + 1
        else {
            start = mid + 1;
        }
    }

    // Return the result
    return result;
}

// Driver code
int main()
{
    vector<int> arr = { 3, 3, 7, 1, 6 };
    int K = 3;

    // Function call
    cout << minimizeArrayMaxValue(arr, K);

    return 0;
}

Java

// java implementation
import java.io.*;
import java.util.ArrayList;
import java.util.Collections;

class GFG
{

  // Function to check if maxx can be a valid value
  public static boolean
    isValid(int maxx, ArrayList<Integer> arr, int k)
  {

    // Initialize a variable sum = 0
    int operationCount = 0;

    // Iterate over the array arr[]
    for (int i = 0; i < arr.size(); i++) {

      // Keep storing the maximum value in sum
      if (arr.get(i) > maxx)
        operationCount += arr.get(i) - maxx;

      // Check if number of operation required to
      // make maximum element at max maxx.
      // If true, return false
      if (operationCount > k)
        return false;
    }

    // Return true
    return true;
  }

  // Function to find the minimum possible maximum
  public static int
    minimizeArrayMaxValue(ArrayList<Integer> arr, int k)
  {
    int n = arr.size();

    // Initialise a variable start = 0,
    // end = max value in, arr[] and result
    int start = Collections.min(arr);
    int end = Collections.max(arr);

    int result = 0;

    // Do while start is less than equals to end
    while (start <= end) {

      // Calculate the mid by (start + end) / 2;
      int mid = (int)((start + end) / 2);

      // Check if mid can be a valid max Value.
      if (isValid(mid, arr, k)) {

        // Shift end to mid - 1
        end = mid - 1;

        // Store mid into result
        result = mid;
      }

      // Otherwise, shift start to mid + 1
      else {
        start = mid + 1;
      }
    }

    // Return the result
    return result;
  }

  public static void main(String[] args)
  {
    int n = 5;
    ArrayList<Integer> arr = new ArrayList<Integer>(n);

    arr.add(3);
    arr.add(3);
    arr.add(7);
    arr.add(1);
    arr.add(6);

    int K = 3;

    // Function call
    System.out.println(minimizeArrayMaxValue(arr, K));
  }
}
// This code is contributed by ksam24000

Python3

# Function to check if maxx can be a valid value
def isValid(maxx, arr, k):
    # Initialize a variable sum = 0
    operationCount = 0

    # Iterate over the array arr[]
    for i in range(len(arr)):

        # Keep storing the maximum value in sum
        if arr[i] > maxx:
            operationCount += arr[i] - maxx

        # Check if number of operation required to
        # make maximum element at max maxx.
        # If true, return false
        if operationCount > k:
            return False

    # Return true
    return True

# Function to find the minimum possible maximum
def minimizeArrayMaxValue(arr, k):

    # Initialise a variable start = 0,
    # end = max value in, arr[] and result
    start = min(arr)
    end = max(arr)

    result = 0

    # Do while start is less than equals to end
    while start <= end:
        # Calculate the mid by (start + end) / 2
        mid = (start + end) // 2

        # Check if mid can be a valid max Value.
        if isValid(mid, arr, k):
            # Shift end to mid - 1
            end = mid - 1

            # Store mid into result
            result = mid

        # Otherwise, shift start to mid + 1
        else:
            start = mid + 1

    # Return the result
    return result


# Driver code
arr = [3, 3, 7, 1, 6]
K = 3

# Function call
print(minimizeArrayMaxValue(arr, K))

# This code is contributed by Tapesh(tapeshdua420)

C#

// Include namespace system
using System;
using System.Linq;
using System.Collections.Generic;

public class GFG
{

  // Function to check if maxx can be a valid value
  public static bool isValid(int maxx, int[] arr, int k)
  {

    // Initialize a variable sum = 0
    var operationCount = 0;

    // Iterate over the array arr[]
    for (int i = 0; i < arr.Length; i++)
    {

      // Keep storing the maximum value in sum
      if (arr[i] > maxx)
      {
        operationCount += arr[i] - maxx;
      }

      // Check if number of operation required to
      // make maximum element at max maxx.
      // If true, return false
      if (operationCount > k)
      {
        return false;
      }
    }

    // Return true
    return true;
  }

  // Function to find the minimum possible maximum
  public static int minimizeArrayMaxValue(int[] arr, int k)
  {
    var n = arr.Length;

    // Initialise a variable start = 0,
    // end = max value in, arr[] and result
    var start = arr.Min();
    var end = arr.Max();
    var result = 0;

    // Do while start is less than equals to end
    while (start <= end)
    {

      // Calculate the mid by (start + end) / 2;
      var mid = (int)((int)((start + end) / 2));

      // Check if mid can be a valid max Value.
      if (GFG.isValid(mid, arr, k))
      {

        // Shift end to mid - 1
        end = mid - 1;

        // Store mid into result
        result = mid;
      }
      else
      {
        start = mid + 1;
      }
    }

    // Return the result
    return result;
  }
  public static void Main(String[] args)
  {
    var arr = new int[]{3,3,7,1,6};

    var K = 3;

    // Function call
    Console.WriteLine(GFG.minimizeArrayMaxValue(arr, K));
  }
}

// This code is contributed by aadityaburujwale.

Javascript

// Include namespace system
// JS implementation
  // Function to check if maxx can be a valid value
function isValid(maxx, arr, k)
  {

    // Initialize a variable sum = 0
    let operationCount = 0;

    // Iterate over the array arr[]
    for (let i = 0; i < arr.length; i++)
    {

      // Keep storing the maximum value in sum
      if (arr[i] > maxx)
      {
        operationCount += arr[i] - maxx;
      }

      // Check if number of operation required to
      // make maximum element at max maxx.
      // If true, return false
      if (operationCount > k)
      {
        return false;
      }
    }

    // Return true
    return true;
  }

  // Function to find the minimum possible maximum
 function minimizeArrayMaxValue(arr,k)
  {
    let n = arr.length;

    // Initialise a variable start = 0,
    // end = max value in, arr[] and result
    let start = Math.min.apply(Math, arr);
    let end = Math.max.apply(Math, arr);
    let result = 0;

    // Do while start is less than equals to end
    while (start <= end)
    {

      // Calculate the mid by (start + end) / 2;
      let mid = Math.floor((start + end) / 2);

      // Check if mid can be a valid max Value.
      if (isValid(mid, arr, k))
      {

        // Shift end to mid - 1
        end = mid - 1;

        // Store mid into result
        result = mid;
      }
      else
      {
        start = mid + 1;
      }
    }

    // Return the result
    return result;
  }

    let arr = [3,3,7,1,6];

    let K = 3;

    // Function call
    console.log(minimizeArrayMaxValue(arr, K));


// This code is contributed by ksam24000

Output

5

Time Complexity: O(N* log(max(arr)))
Auxiliary Space: O(1)