Minimum K such that sum of array elements after division by K does not exceed S

Given an array arr[] of N elements and an integer S. The task is to find the minimum number K such that the sum of the array elements does not exceed S after dividing all the elements by K.
Note: Consider integer division.

Examples:

Input: arr[] = {10, 7, 8, 10, 12, 19}, S = 27
Output: 3
After dividing by 3, the array becomes
{3, 2, 2, 3, 4, 6} and the new sum is 20.



Input: arr[] = {19, 17, 11, 10}, S = 40
Output: 2

Naive approach: Iterate for all values of K from 1 to the maximum element in the array and then sum up the array elements by dividing with K, if the sum does not exceed S then the current value will be the answer. The time complexity of this approach will be O(M * N) where M is the maximum element in the array.

Efficient approach: An efficient approach is to find the value of K by performing binary search on the answer. Initiate a binary search on the value of K and a check is done inside it to see if the sum exceeds K then the binary search is performed on the second half or the first half accordingly.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum value of k
// that satisfies the given condition
int findMinimumK(int a[], int n, int s)
{
    // Find the maximum element
    int maximum = a[0];
    for (int i = 0; i < n; i++) {
        maximum = max(maximum, a[i]);
    }
  
    // Lowest answer can be 1 and the
    // highest answer can be (maximum + 1)
    int low = 1, high = maximum + 1;
  
    int ans = high;
  
    // Binary search
    while (low <= high) {
  
        // Get the mid element
        int mid = (low + high) / 2;
        int sum = 0;
  
        // Calculate the sum after dividing
        // the array by new K which is mid
        for (int i = 0; i < n; i++) {
            sum += (int)(a[i] / mid);
        }
  
        // Search in the second half
        if (sum > s)
            low = mid + 1;
  
        // First half
        else {
            ans = min(ans, mid);
            high = mid - 1;
        }
    }
  
    return ans;
}
  
// Driver code
int main()
{
    int a[] = { 10, 7, 8, 10, 12, 19 };
    int n = sizeof(a) / sizeof(a[0]);
    int s = 27;
  
    cout << findMinimumK(a, n, s);
  
    return 0;
}

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Java

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// Java implementation of the approach 
class GFG 
{
      
    // Function to return the minimum value of k 
    // that satisfies the given condition 
    static int findMinimumK(int a[], 
                            int n, int s) 
    
        // Find the maximum element 
        int maximum = a[0]; 
          
        for (int i = 0; i < n; i++) 
        
            maximum = Math.max(maximum, a[i]); 
        
      
        // Lowest answer can be 1 and the 
        // highest answer can be (maximum + 1) 
        int low = 1, high = maximum + 1
      
        int ans = high; 
      
        // Binary search 
        while (low <= high)
        
      
            // Get the mid element 
            int mid = (low + high) / 2
            int sum = 0
      
            // Calculate the sum after dividing 
            // the array by new K which is mid 
            for (int i = 0; i < n; i++)
            
                sum += (int)(a[i] / mid); 
            
      
            // Search in the second half 
            if (sum > s) 
                low = mid + 1
      
            // First half 
            else
            
                ans = Math.min(ans, mid); 
                high = mid - 1
            
        
        return ans; 
    
      
    // Driver code 
    public static void main (String[] args)
    
        int a[] = { 10, 7, 8, 10, 12, 19 }; 
        int n = a.length; 
        int s = 27
      
        System.out.println(findMinimumK(a, n, s)); 
    
}    
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
  
# Function to return the minimum value of k
# that satisfies the given condition
def findMinimumK(a, n, s):
      
    # Find the maximum element
    maximum = a[0]
    for i in range(n):
        maximum = max(maximum, a[i])
  
    # Lowest answer can be 1 and the
    # highest answer can be (maximum + 1)
    low = 1
    high = maximum + 1
  
    ans = high
  
    # Binary search
    while (low <= high):
  
        # Get the mid element
        mid = (low + high) // 2
        sum = 0
  
        # Calculate the sum after dividing
        # the array by new K which is mid
        for i in range(n):
            sum += (a[i] // mid)
  
        # Search in the second half
        if (sum > s):
            low = mid + 1
  
        # First half
        else:
            ans = min(ans, mid)
            high = mid - 1
  
    return ans
  
# Driver code
a = [10, 7, 8, 10, 12, 19]
n = len(a)
s = 27
  
print(findMinimumK(a, n, s))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
      
    // Function to return the minimum value of k 
    // that satisfies the given condition 
    static int findMinimumK(int []a, 
                            int n, int s) 
    
        // Find the maximum element 
        int maximum = a[0]; 
          
        for (int i = 0; i < n; i++) 
        
            maximum = Math.Max(maximum, a[i]); 
        
      
        // Lowest answer can be 1 and the 
        // highest answer can be (maximum + 1) 
        int low = 1, high = maximum + 1; 
      
        int ans = high; 
      
        // Binary search 
        while (low <= high) 
        
      
            // Get the mid element 
            int mid = (low + high) / 2; 
            int sum = 0; 
      
            // Calculate the sum after dividing 
            // the array by new K which is mid 
            for (int i = 0; i < n; i++) 
            
                sum += (int)(a[i] / mid); 
            
      
            // Search in the second half 
            if (sum > s) 
                low = mid + 1; 
      
            // First half 
            else
            
                ans = Math.Min(ans, mid); 
                high = mid - 1; 
            
        
        return ans; 
    
      
    // Driver code 
    public static void Main () 
    
        int []a = { 10, 7, 8, 10, 12, 19 }; 
        int n = a.Length; 
        int s = 27; 
      
        Console.WriteLine(findMinimumK(a, n, s)); 
    
  
// This code is contributed by AnkitRai01 

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Output:

3


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Improved By : mohit kumar 29, AnkitRai01