Maximum elements that can be made equal with k updates

Given an array and a value k. We have to find the maximum number of equal elements possible for the array so that we can increase the elements of the array by incrementing a total of at-most k.

Examples:

Input : array = { 2, 4, 9 }, k = 3
Output : 2
We are allowed to do at most three increments. We can make two elements 4 by increasing 2 by 2. Note that we can not make two elements 9 as converting 4 to 9 requires 5 increments.

Input : array = { 5, 5, 3, 1 }, k = 5
Output : 3
Explanation: Here 1st and 2nd elements are equal. Then we can increase 3rd element 3 upto 5. Then k becomes (k-2) = 3. Now we can’t increase 1 to 5 because k value is 3 and we need 4 for the updation. Thus equal elements possible are 3. Here we can also increase 1 to 5. Then also we have 3 because we can’t update 3 to 5.

Input : array = { 5, 5, 3, 1 }, k = 6
Output : 4



Naive Approach: In the naive approach we have an algorithm in O(n^2) time in which we check for each element how many other elements can be incremented so that they will become equal to them.

Efficient Approach: In this approach, first we will sort the array. Then we maintain two arrays. First is prefix sum array which stores the prefix sum of the array and another is maxx[] array which stores the maximum element found till every point, i.e., max[i] means maximum element from 1 to i. After storing these values in prefix[] array and maxx[] array, we do the binary search from 1 to n(number of elements of the array) to calculate how many elements which can be incremented to make them equal. In the binary search, we use one function in which we determine what is the number of elements can be incremented to make them equal to a single value.

C++

// C++ program to find maximum elements that can 
// be made equal with k updates
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the maximum number of
// equal elements possible with atmost K increment
// of values .Here we have done sliding window
// to determine that whether there are x number of
// elements present which on increment will become
// equal. The loop here will run in fashion like
// 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1
bool ElementsCalculationFunc(int pre[], int maxx[],
                               int x, int k, int n)
{
    for (int i = 0, j = x; j <= n; j++, i++) {

        // It can be explained with the reasoning
        // that if for some x number of elements
        // we can update the values then the
        // increment to the segment (i to j having
        // length -> x) so that all will be equal is
        // (x*maxx[j]) this is the total sum of
        // segment and (pre[j]-pre[i]) is present sum
        // So difference of them should be less than k
        // if yes, then that segment length(x) can be 
        // possible return true
        if (x * maxx[j] - (pre[j] - pre[i]) <= k)
            return true;
    }
    return false;
}

void MaxNumberOfElements(int a[], int n, int k)
{
    // sort the array in ascending order
    sort(a, a + n);
    int pre[n + 1]; // prefix sum array
    int maxx[n + 1]; // maximum value array

    // Initializing the prefix array
    // and maximum array
    for (int i = 0; i <= n; ++i) {
        pre[i] = 0;
        maxx[i] = 0;
    }

    // set the first element of both
    // array
    maxx[0] = a[0];
    pre[0] = a[0];
    for (int i = 1; i < n; i++) {

        // Calculating prefix sum of the array
        pre[i] = pre[i - 1] + a[i];

        // Calculating max value upto that position
        // in the array
        maxx[i] = max(maxx[i - 1], a[i]);
    }

    // Binary seach applied for
    // computation here
    int l = 1, r = n, ans;
    while (l < r) {
        int mid = (l + r) / 2;

        if (ElementsCalculationFunc(pre, maxx,
                              mid - 1, k, n)) {
            ans = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }

    // printing result
    cout << ans << "\n";
}

int main()
{
    int arr[] = { 2, 4, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    MaxNumberOfElements(arr, n, k);
    return 0;
}

Java

// java program to find maximum elements that can 
// be made equal with k updates

import java.util.Arrays;
public class GFG {
    
    // Function to calculate the maximum number of
    // equal elements possible with atmost K increment
    // of values .Here we have done sliding window
    // to determine that whether there are x number of
    // elements present which on increment will become
    // equal. The loop here will run in fashion like
    // 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1
    static boolean ElementsCalculationFunc(int pre[], 
                      int maxx[], int x, int k, int n)
    {
        for (int i = 0, j = x; j <= n; j++, i++) {
     
            // It can be explained with the reasoning
            // that if for some x number of elements
            // we can update the values then the
            // increment to the segment (i to j having
            // length -> x) so that all will be equal is
            // (x*maxx[j]) this is the total sum of
            // segment and (pre[j]-pre[i]) is present sum
            // So difference of them should be less than k
            // if yes, then that segment length(x) can be 
            // possible return true
            if (x * maxx[j] - (pre[j] - pre[i]) <= k)
                return true;
        }
        return false;
    }
     
    static void MaxNumberOfElements(int a[], int n, int k)
    {
        // sort the array in ascending order
        Arrays.sort(a);
        int []pre = new int[n + 1]; // prefix sum array
        int []maxx = new int[n + 1]; // maximum value array
     
        // Initializing the prefix array
        // and maximum array
        for (int i = 0; i <= n; ++i) {
            pre[i] = 0;
            maxx[i] = 0;
        }
     
        // set the first element of both
        // array
        maxx[0] = a[0];
        pre[0] = a[0];
        for (int i = 1; i < n; i++) {
     
            // Calculating prefix sum of the array
            pre[i] = pre[i - 1] + a[i];
     
            // Calculating max value upto that position
            // in the array
            maxx[i] = Math.max(maxx[i - 1], a[i]);
        }
     
        // Binary seach applied for
        // computation here
        int l = 1, r = n, ans=0;
        while (l < r) {
             
            int mid = (l + r) / 2;
     
            if (ElementsCalculationFunc(pre, maxx,
                                   mid - 1, k, n))
            {
                ans = mid;
                l = mid + 1;
            }
            else
                r = mid - 1;
        }
     
        // printing result
        System.out.print((int)ans + "\n");
    } 
     
    public static void main(String args[]) {
        
        int arr[] = { 2, 4, 9 };
        int n = arr.length;
        int k = 3;
         
        MaxNumberOfElements(arr, n, k);
         
    }
}

// This code is contributed by Sam007

C#

// C# program to find maximum elements that can 
// be made equal with k updates
using System;

class GFG {
    
    // Function to calculate the maximum number of
    // equal elements possible with atmost K increment
    // of values .Here we have done sliding window
    // to determine that whether there are x number of
    // elements present which on increment will become
    // equal. The loop here will run in fashion like
    // 0...x-1, 1...x, 2...x+1, ...., n-x-1...n-1
    static bool ElementsCalculationFunc(int []pre, 
                      int []maxx, int x, int k, int n)
    {
        for (int i = 0, j = x; j <= n; j++, i++) {
    
            // It can be explained with the reasoning
            // that if for some x number of elements
            // we can update the values then the
            // increment to the segment (i to j having
            // length -> x) so that all will be equal is
            // (x*maxx[j]) this is the total sum of
            // segment and (pre[j]-pre[i]) is present sum
            // So difference of them should be less than k
            // if yes, then that segment length(x) can be 
            // possible return true
            if (x * maxx[j] - (pre[j] - pre[i]) <= k)
                return true;
        }
        return false;
    }
    
    static void MaxNumberOfElements(int []a, int n, int k)
    {
        // sort the array in ascending order
        Array.Sort(a);
        int []pre = new int[n + 1]; // prefix sum array
        int []maxx = new int[n + 1]; // maximum value array
    
        // Initializing the prefix array
        // and maximum array
        for (int i = 0; i <= n; ++i) {
            pre[i] = 0;
            maxx[i] = 0;
        }
    
        // set the first element of both
        // array
        maxx[0] = a[0];
        pre[0] = a[0];
        for (int i = 1; i < n; i++) {
    
            // Calculating prefix sum of the array
            pre[i] = pre[i - 1] + a[i];
    
            // Calculating max value upto that position
            // in the array
            maxx[i] = Math.Max(maxx[i - 1], a[i]);
        }
    
        // Binary seach applied for
        // computation here
        int l = 1, r = n, ans=0;
        while (l < r) {
            
            int mid = (l + r) / 2;
    
            if (ElementsCalculationFunc(pre, maxx,
                                   mid - 1, k, n))
            {
                ans = mid;
                l = mid + 1;
            }
            else
                r = mid - 1;
        }
    
        // printing result
        Console.Write ((int)ans + "\n");
    } 
    
    // Driver code
    public static void Main()
    {
        int []arr = { 2, 4, 9 };
        int n = arr.Length;
        int k = 3;
        
        MaxNumberOfElements(arr, n, k);
    }
}

// This code is contributed by Sam007
Output:

2

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




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