Skip to content
Related Articles

Related Articles

Split the given array into K sub-arrays such that maximum sum of all sub arrays is minimum
  • Difficulty Level : Hard
  • Last Updated : 08 Feb, 2021

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 :
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 :

Approach : 

  • Idea is to use Binary Search to find an optimal solution.
  • For binary search minimum sum can be 1 and the maximum sum can be the sum of all the elements.
  • To check if mid is maximum subarray sum possible. Maintain a count of sub – arrays, include all possible elements in sub array until their sum is less than mid. After this evaluation, if the count is less than or equal to K, then mid is achievable else not. (Since if the count is less than K, we can further divide any subarray its sum will never increase mid ).
  • Find the minimum possible value of mid which satisfies the condition.

Below is the implementation of the above approach: 

C++




// 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* max = max_element(array, array + n);
    int start = *max;
    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);
}

Java




// 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;
        for (int i = 0; i < n; ++i) {
            if (array[i] > start)
                start = array[i];
        }
        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

Python3




# 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 = max(array)
    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

C#




// 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;
        for (int i = 0; i < n; ++i) {
            if (array[i] > start)
                start = array[i];
        }
        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
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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :