Split the given array into K sub-arrays such that maximum sum of all sub arrays is minimum

Given an Array[] of N elements and a number K. ( 1 <= K <= N ) . Split the given array into K subarrays (they must cover all the elements). The maximum subarray sum achievable out of K subarrays formed, must be minimum possible. Find that possible subarray sum.

Examples:

Input : Array[] = {1, 2, 3, 4}, K = 3
Output : 4
Optimal Split is {1, 2}, {3}, {4} . Maximum sum of all subarrays is 4, which is minimum possible for 3 splits.

Input : Array[] = {1, 1, 2} K = 2
Output : 2

Approach :



Below is the implementation of the above approach:

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// C++ implemenattion of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if mid can
// be maximum sub - arrays sum
bool check(int mid, int array[], int n, int K)
{
    int count = 0;
    int sum = 0;
    for (int i = 0; i < n; i++) {
  
        // If individual element is greater
        // maximum possible sum
        if (array[i] > mid)
            return false;
  
        // Increase sum of current sub - array
        sum += array[i];
  
        // If the sum is greater than 
        // mid increase count
        if (sum > mid) {
            count++;
            sum = array[i];
        }
    }
    count++;
  
    // Check condition
    if (count <= K)
        return true;
    return false;
}
  
// Function to find maximum subarray sum
// which is minimum
int solve(int array[], int n, int K)
{
    int start = 1;
    int end = 0;
  
    for (int i = 0; i < n; i++) {
        end += array[i];
    }
  
    // Answer stores possible 
    // maximum sub array sum
    int answer = 0;
    while (start <= end) {
        int mid = (start + end) / 2;
  
        // If mid is possible solution
        // Put answer = mid;
        if (check(mid, array, n, K)) {
            answer = mid;
            end = mid - 1;
        }
        else {
            start = mid + 1;
        }
    }
  
    return answer;
}
  
// Driver Code
int main()
{
    int array[] = { 1, 2, 3, 4 };
    int n = sizeof(array) / sizeof(array[0]);
    int K = 3;
    cout << solve(array, n, K);
}
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// Java implemenattion of the above approach 
class GFG 
{
      
    // Function to check if mid can 
    // be maximum sub - arrays sum 
    static boolean check(int mid, int array[], int n, int K) 
    {
          
        int count = 0
        int sum = 0
        for (int i = 0; i < n; i++) 
        
      
            // If individual element is greater 
            // maximum possible sum 
            if (array[i] > mid) 
                return false
      
            // Increase sum of current sub - array 
            sum += array[i]; 
      
            // If the sum is greater than 
            // mid increase count 
            if (sum > mid) 
            
                count++; 
                sum = array[i]; 
            
        
        count++; 
      
        // Check condition 
        if (count <= K) 
            return true
        return false
    
      
    // Function to find maximum subarray sum 
    // which is minimum 
    static int solve(int array[], int n, int K) 
    
        int start = 1
        int end = 0
      
        for (int i = 0; i < n; i++)
        
            end += array[i]; 
        
      
        // Answer stores possible 
        // maximum sub array sum 
        int answer = 0
        while (start <= end) 
        
            int mid = (start + end) / 2
      
            // If mid is possible solution 
            // Put answer = mid; 
            if (check(mid, array, n, K))
            
                answer = mid; 
                end = mid - 1
            
            else
            
                start = mid + 1
            
        
      
        return answer; 
    
      
    // Driver Code 
    public static void main (String[] args) 
    {
        int array[] = { 1, 2, 3, 4 }; 
        int n = array.length ; 
        int K = 3
        System.out.println(solve(array, n, K)); 
    }
}
  
// This code is contributed by AnkitRai01
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# Python 3 implemenattion of the above approach
  
# Function to check if mid can
# be maximum sub - arrays sum
def check(mid, array, n, K):
    count = 0
    sum = 0
    for i in range(n):
          
        # If individual element is greater
        # maximum possible sum
        if (array[i] > mid):
            return False
  
        # Increase sum of current sub - array
        sum += array[i]
  
        # If the sum is greater than 
        # mid increase count
        if (sum > mid):
            count += 1
            sum = array[i]
    count += 1
  
    # Check condition
    if (count <= K):
        return True
    return False
  
# Function to find maximum subarray sum
# which is minimum
def solve(array, n, K):
    start = 1
    end = 0
  
    for i in range(n):
        end += array[i]
  
    # Answer stores possible 
    # maximum sub array sum
    answer = 0
    while (start <= end):
        mid = (start + end) // 2
  
        # If mid is possible solution
        # Put answer = mid;
        if (check(mid, array, n, K)):
            answer = mid
            end = mid - 1
        else:
            start = mid + 1
  
    return answer
  
# Driver Code
if __name__ == '__main__':
    array = [1, 2, 3, 4]
    n = len(array)
    K = 3
    print(solve(array, n, K))
      
# This code is contributed by
# Surendra_Gangwar
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// C# implementation of the above approach
using System;
      
class GFG 
{
      
    // Function to check if mid can 
    // be maximum sub - arrays sum 
    static Boolean check(int mid, int []array, 
                                int n, int K) 
    {
          
        int count = 0; 
        int sum = 0; 
        for (int i = 0; i < n; i++) 
        
      
            // If individual element is greater 
            // maximum possible sum 
            if (array[i] > mid) 
                return false
      
            // Increase sum of current sub - array 
            sum += array[i]; 
      
            // If the sum is greater than 
            // mid increase count 
            if (sum > mid) 
            
                count++; 
                sum = array[i]; 
            
        
        count++; 
      
        // Check condition 
        if (count <= K) 
            return true
        return false
    
      
    // Function to find maximum subarray sum 
    // which is minimum 
    static int solve(int []array, int n, int K) 
    
        int start = 1; 
        int end = 0; 
      
        for (int i = 0; i < n; i++)
        
            end += array[i]; 
        
      
        // Answer stores possible 
        // maximum sub array sum 
        int answer = 0; 
        while (start <= end) 
        
            int mid = (start + end) / 2; 
      
            // If mid is possible solution 
            // Put answer = mid; 
            if (check(mid, array, n, K))
            
                answer = mid; 
                end = mid - 1; 
            
            else
            
                start = mid + 1; 
            
        
      
        return answer; 
    
      
    // Driver Code 
    public static void Main (String[] args) 
    {
        int []array = { 1, 2, 3, 4 }; 
        int n = array.Length ; 
        int K = 3; 
        Console.WriteLine(solve(array, n, K)); 
    }
}
  
// This code is contributed by Princi Singh
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Output:
4

Time Complexity : O(N*log(Sum))
Where N is the number of elements of the array and Sum is the sum of all the elements of the array.

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