Minimize maximum array element possible by at most K splits on the given array

Given an array arr[] consisting of N positive integers and a positive integer K, the task is to minimize the maximum element present in the array by splitting at most K array elements into two numbers equal to their value.

Examples:

Input: arr[] = {2, 4, 8, 2}, K = 4
Output: 2
Explanation:
Following sequence of operations are required to be performed:
Operation 1: Splitting arr[1] (= 4) to {2, 2} modifies the array to {2, 2, 2, 8, 2}.
Operation 2: Splitting arr[3] (= 8) to {2, 6} modifies the array to {2, 2, 2, 2, 6, 2}.
Operation 3: Splitting arr[4] (= 6) to {2, 4} modifies the array to {2, 2, 2, 2, 2, 4, 2}.
Operation 4: Splitting arr[5] (= 4) to {2, 2} modifies the array to {2, 2, 2, 2, 2, 2, 2, 2}.
After completing the above operations, the maximum element present in the array is 2.

Input: arr[] = {7, 17}, K = 2
Output: 7

Approach: The given problem can be solved based on the following observations:

• If X can be the maximum element in the array arr[] by performing at most K operations, then there exists some value K (K > X), which can also be the maximum element present in the array arr[] by performing at most K splitting of the array elements.
• If X can’t be the maximum element in the array A[] by performing at most K operations, then there exists some value K (K < X) which can also not be the maximum element in the array arr[] by performing at most K splits of array elements.
• Therefore, the idea is to use Binary Search for finding the value over the range [1, INT_MAX] which can be the possible maximum value after at most K splits.

Follow the steps below to solve the problem:

• Initialize two variables, say low and high as 1 and the maximum element in the array arr[] respectively.
• Iterate until low is less than high and perform the following steps:
  • Find the middle value of the range [low, high] as mid = (low + high)/2.
  • Initialize a variable, say count, to store the maximum number of splits of array elements required to make the maximum element equal to mid.
  • Traverse the given array arr[] and update the value of count as (arr[i] – 1) / mid to count the number of splits required.
  • If the value of count is at most K, then update the value of high as mid.
  • Otherwise, update the value of low as (mid + 1).
• After completing the above steps, print the value of high as the resultant maximum element present in the array obtained.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
int possible(int A[], int N,
             int mid, int K)
{
    // Stores the number
    // of splits required
    int count = 0;

    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {

        // Update count
        count += (A[i] - 1) / mid;
    }

    // If possible, return true.
    // Otherwise return false
    return count <= K;
}

// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
int minimumMaximum(int A[], int N, int K)
{
    // Set lower and upper limits
    int lo = 1;
    int hi = *max_element(A, A + N);
    int mid;

    // Perform Binary Search
    while (lo < hi) {

        // Calculate mid
        mid = (lo + hi) / 2;

        // Check if all array elements
        // can be reduced to at most
        // mid value by at most K splits
        if (possible(A, N, mid, K)) {

            // Update the value of hi
            hi = mid;
        }

        // Otherwise
        else {

            // Update the value of lo
            lo = mid + 1;
        }
    }

    // Return the minimized maximum
    // element in the array
    return hi;
}

// Driver Code
int main()
{
    int arr[] = { 2, 4, 8, 2 };
    int K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);

    cout << minimumMaximum(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
static boolean possible(int A[], int N,
                        int mid, int K)
{

    // Stores the number
    // of splits required
    int count = 0;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Update count
        count += (A[i] - 1) / mid;
    }

    // If possible, return true.
    // Otherwise return false
    return count <= K;
}

// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
static int minimumMaximum(int A[], int N, int K)
{

    // Set lower and upper limits
    int lo = 1;
    Arrays.sort(A);

    int hi = A[N - 1];
    int mid;

    // Perform Binary Search
    while (lo < hi)
    {

        // Calculate mid
        mid = (lo + hi) / 2;

        // Check if all array elements
        // can be reduced to at most
        // mid value by at most K splits
        if (possible(A, N, mid, K))
        {

            // Update the value of hi
            hi = mid;
        }

        // Otherwise
        else
        {

            // Update the value of lo
            lo = mid + 1;
        }
    }

    // Return the minimized maximum
    // element in the array
    return hi;
}

// Driver Code
public static void main (String[] args)
{
    int arr[] = { 2, 4, 8, 2 };
    int K = 4;
    int N = arr.length;

    System.out.println(minimumMaximum(arr, N, K));
}
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to check if all array
# elements can be reduced to at
# most mid by at most K splits
def possible(A, N, mid, K):

    # Stores the number
    # of splits required
    count = 0

    # Traverse the array arr[]
    for i in range(N):

        # Update count
        count += (A[i] - 1) // mid

    # If possible, return true.
    # Otherwise return false
    return count <= K

# Function to find the minimum possible
# value of maximum array element that
# can be obtained by at most K splits
def minimumMaximum(A, N, K):

    # Set lower and upper limits
    lo = 1
    hi = max(A)

    # Perform Binary Search
    while (lo < hi):

        # Calculate mid
        mid = (lo + hi) // 2

        # Check if all array elements
        # can be reduced to at most
        # mid value by at most K splits
        if (possible(A, N, mid, K)):

            # Update the value of hi
            hi = mid

        # Otherwise
        else:

            # Update the value of lo
            lo = mid + 1

    # Return the minimized maximum
    # element in the array
    return hi

# Driver Code
if __name__ == '__main__':

    arr =  [ 2, 4, 8, 2 ]
    K = 4
    N = len(arr)

    print(minimumMaximum(arr, N, K))

# This code is contributed by ipg2016107

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits
static bool possible(int[] A, int N,
                     int mid, int K)
{

    // Stores the number
    // of splits required
    int count = 0;

    // Traverse the array arr[]
    for(int i = 0; i < N; i++)
    {

        // Update count
        count += (A[i] - 1) / mid;
    }

    // If possible, return true.
    // Otherwise return false
    return count <= K;
}

// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
static int minimumMaximum(int[] A, int N, int K)
{

    // Set lower and upper limits
    int lo = 1;
    Array.Sort(A);

    int hi = A[N - 1];
    int mid;

    // Perform Binary Search
    while (lo < hi)
    {

        // Calculate mid
        mid = (lo + hi) / 2;

        // Check if all array elements
        // can be reduced to at most
        // mid value by at most K splits
        if (possible(A, N, mid, K))
        {

            // Update the value of hi
            hi = mid;
        }

        // Otherwise
        else
        {

            // Update the value of lo
            lo = mid + 1;
        }
    }

    // Return the minimized maximum
    // element in the array
    return hi;
}

// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 2, 4, 8, 2 };
    int K = 4;
    int N = arr.Length;

    Console.WriteLine(minimumMaximum(arr, N, K));
}
}

// This code is contributed by ukasp

Javascript

<script>

// javascript program for the above approach

// Function to check if all array
// elements can be reduced to at
// most mid by at most K splits

function possible(A, N, mid, K)
{

    // Stores the number
    // of splits required
    var count = 0;

     var i;
    // Traverse the array arr[]
    for (i = 0; i < N; i++) {

        // Update count
        count += Math.floor((A[i] - 1) / mid);
    }

    // If possible, return true.
    // Otherwise return false
    if(count <= K)
      return true;
    else
      return false
}

// Function to find the minimum possible
// value of maximum array element that
// can be obtained by at most K splits
function minimumMaximum(A, N, K)
{

    // Set lower and upper limits
    var lo = 1;
    var hi = Math.max.apply(Math,A);
    var mid;

    // Perform Binary Search
    while (lo < hi) {

        // Calculate mid
        mid = Math.floor((lo + hi) / 2);

        // Check if all array elements
        // can be reduced to at most
        // mid value by at most K splits
        if (possible(A, N, mid, K)) {

            // Update the value of hi
            hi = mid;
        }

        // Otherwise
        else {

            // Update the value of lo
            lo = mid + 1;
        }
    }

    // Return the minimized maximum
    // element in the array
    return hi;
}

// Driver Code
    var arr = [2, 4, 8, 2];
    var K = 4;
    var N = arr.length;

    document.write(minimumMaximum(arr, N, K));

// This code is contributed by SURENDRA_GANGWAR.
</script>

Output:

2

Time Complexity: O(N * log M), where M is the maximum element of the array.
Auxiliary Space: O(1)