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Minimize difference between maximum and minimum Subarray sum by splitting Array into 4 parts

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  • Difficulty Level : Hard
  • Last Updated : 19 May, 2022
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Given an array arr[] of size N, the task is to find the minimum difference between the maximum and the minimum subarray sum when the given array is divided into 4 non-empty subarrays.

Examples:

Input: N = 5, arr[] = {3, 2, 4, 1, 2}
Output: 2
Explanation: Divide the array into four parts as {3}, {2}, {4} and {1, 2}
The sum of all the elements of these parts is 3, 2, 4, and 3
The difference between the maximum and minimum is (4 – 2) = 2.

Input: N = 4, arr[] = {14, 6, 1, 7}
Output: 13
Explanation:  Divide the array into four parts {14}, {6}, {1} and {7}
The sum of all the elements of these four parts is 14, 6, 1, and 7
The difference between the maximum and minimum (14 – 1) = 13.
It is the only possible way to divide the array into 4 possible parts

 

Naive Approach: The simplest way is to check for all possible combinations of three cuts and for each possible value check the subarray sums. Then calculate the minimum difference among all the possible combinations.

Time Complexity: O(N4)
Auxiliary Space: O(1) 

Efficient Approach: The problem can be solved using the concept of prefix sum and two-pointer based on the below observation:

To divide the array into 4 subarrays three splits are required. 

  • If the second split is fixed (say in between index i and i+1) there will be one split to the left and one split to the right. 
  • The difference will be minimized when the two subarrays on left will have sum as close to each other as possible and same for the two subarrays on the right side of the split. 
  • The overall sum of the left part and of the right part can be obtained  in constant time with the help of prefix sum calculation.

Now the split on the left part and on the right part can be decided optimally using the two-pointer technique. 

  • When the second split is fixed decide the left split by iterating through the left part till the difference between the sum of two parts is minimum. 
  • It can be found by minimizing the difference between the overall sum and twice the sum of any of the part. [The minimum value of this signifies that the difference between both the parts is minimum]

Do the same for the right part also.

Follow the below steps to solve this problem:

  • Firstly pre-compute the prefix sum array of the given array.
  • Create three variables i = 1, j = 2, and k = 3 each representing the cuts.(1 based indexing)
  • Iterate through possible values of j from 2 to N – 1.
    • For each value of j try to increase the value of i until the absolute difference between the Left_Sum_1 and Left_Sum_2 decreases and i is less than j (Left_Sum_1 and Left_Sum_2 are the sums of the two subarrays on the left).
    • For each value of j, try to increase the value of k, until the absolute difference between the Right_Sum_1 and Right_Sum_2 decreases and k is less than N + 1 (Right_Sum_1 and Right_Sum_2 are the sums of the two subarrays of the right).
    • Use prefix sum to directly calculate the values of Left_Sum_1, Left_Sum_2, Right_Sum_1 and Right_Sum_2.
  • For each valid value of i, j and k, find the difference between the maximum and minimum value of the sum of elements of these parts
  • The minimum among them is the answer.

Below is the implementation of the above approach:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum difference
// between maximum and minimum subarray sum
// after dividing the array into 4 subarrays
long long int minSum(vector<int>& v, int n)
{
    vector<long long int> a(n + 1);
 
    // Precompute the prefix sum
    a[0] = 0;
    for (int i = 1; i <= n; i++) {
        a[i] = a[i - 1] + v[i - 1];
    }
 
    // Initialize the ans with large value
    long long int ans = 1e18;
 
    // There are total four parts means 3 cuts.
    // Here i, j, k represent those 3 cuts
    for (int i = 1, j = 2, k = 3; j < n; j++) {
        while (i + 1 < j
               && abs(a[j] - 2 * a[i])
                      > abs(a[j]
                            - 2 * a[i + 1])) {
            i++;
        }
        while (k + 1 < n
               && abs(a[n] + a[j] - 2 * a[k])
                      > abs(a[n] + a[j]
                            - 2 * a[k + 1])) {
            k++;
        }
        ans = min(ans,
                  max({ a[i], a[j] - a[i],
                        a[k] - a[j],
                        a[n] - a[k] })
                      - min({ a[i], a[j] - a[i],
                              a[k] - a[j],
                              a[n] - a[k] }));
    }
    return ans;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 3, 2, 4, 1, 2 };
    int N = arr.size();
 
    // Function call
    cout << minSum(arr, N);
    return 0;
}


Java




// Java program for the above approach
public class GFG
{
 
  // Function to find the minimum difference
  // between maximum and minimum subarray sum
  // after dividing the array into 4 subarrays
  static int minCost(int arr[], int n)
  {
 
    // Precompute the prefix sum
    int a[] = new int[n + 1];
    a[0] = 0;
    for (int i = 1; i <= n; i++) {
      a[i] = a[i - 1] + arr[i - 1];
    }
 
    // Initialize the ans with large value
    int ans = Integer.MAX_VALUE;
 
    // There are total four parts means 3 cuts.
    // Here i, j, k represent those 3 cuts
    for (int i = 1, j = 2, k = 3; j < n; j++) {
      while (i + 1 < j
             && Math.abs(a[j] - 2 * a[i])
             > Math.abs(a[j] - 2 * a[i + 1])) {
        i++;
      }
      while (k + 1 < n
             && Math.abs(a[n] + a[j] - 2 * a[k])
             > Math.abs(a[n] + a[j]
                        - 2 * a[k + 1])) {
        k++;
      }
      ans = Math.min(
        ans,
        Math.max(a[i],
                 Math.max(a[j] - a[i],
                          Math.max(a[k] - a[j],
                                   a[n] - a[k])))
        - Math.min(
          a[i],
          Math.min(a[j] - a[i],
                   Math.min(a[k] - a[j],
                            a[n] - a[k]))));
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    int arr[] = { 3, 2, 4, 1, 2 };
    int N = arr.length;
 
    System.out.println(minCost(arr, N));
  }
}
 
// This code is contributed by dwivediyash


Python3




# Python3 code to implement the approach
 
# Function to find the minimum difference
# between maximum and minimum subarray sum
# after dividing the array into 4 subarrays
def minSum(v, n):
    a = [0]
 
    # Precompute the prefix sum
    for i in range(1, n + 1):
        a.append(a[-1] + v[i - 1])
 
    # Initialize the ans with large value
    ans = 10 ** 18
 
    # There are total four parts means 3 cuts.
    # Here i, j, k represent those 3 cuts
    i = 1
    j = 2
    k = 3
    while (j < n):
        while (i + 1 < j and abs(a[j] - 2 * a[i]) > abs(a[j] - 2 * a[i + 1])):
            i += 1
        while (k + 1 < n and abs(a[n] + a[j] - 2 * a[k]) > abs(a[n] + a[j] - 2 * a[k + 1])):
            k += 1
        ans = min(ans, max([a[i], a[j] - a[i], a[k] - a[j], a[n] - a[k]]
                           ) - min([a[i], a[j] - a[i], a[k] - a[j], a[n] - a[k]]))
        j += 1
 
    return ans
 
# Driver Code
arr = [3, 2, 4, 1, 2]
N = len(arr)
 
# Function call
print(minSum(arr, N))
 
# this code is contributed by phasing17


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the minimum difference
// between maximum and minimum subarray sum
// after dividing the array into 4 subarrays
static int minCost(int[] arr, int n)
{
 
    // Precompute the prefix sum
    int[] a = new int[n + 1];
    a[0] = 0;
    for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] + arr[i - 1];
    }
 
    // Initialize the ans with large value
    int ans = Int16.MaxValue;
 
    // There are total four parts means 3 cuts.
    // Here i, j, k represent those 3 cuts
    for (int i = 1, j = 2, k = 3; j < n; j++) {
    while (i + 1 < j
            && Math.Abs(a[j] - 2 * a[i])
            > Math.Abs(a[j] - 2 * a[i + 1])) {
        i++;
    }
    while (k + 1 < n
            && Math.Abs(a[n] + a[j] - 2 * a[k])
            > Math.Abs(a[n] + a[j]
                        - 2 * a[k + 1])) {
        k++;
    }
    ans = Math.Min(
        ans,
        Math.Max(a[i],
                Math.Max(a[j] - a[i],
                        Math.Max(a[k] - a[j],
                                a[n] - a[k])))
        - Math.Min(
        a[i],
        Math.Min(a[j] - a[i],
                Math.Min(a[k] - a[j],
                            a[n] - a[k]))));
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 3, 2, 4, 1, 2 };
    int N = arr.Length;
     
    // Function call
    Console.WriteLine(minCost(arr, N));
}
}
 
// This code is contributed by Pushpesh Raj


Javascript




<script>
    // JavaScript code to implement the approach
 
    // Function to find the minimum difference
    // between maximum and minimum subarray sum
    // after dividing the array into 4 subarrays
    const minSum = (v, n) => {
        let a = new Array(n + 1).fill(0);
 
        // Precompute the prefix sum
        a[0] = 0;
        for (let i = 1; i <= n; i++) {
            a[i] = a[i - 1] + v[i - 1];
        }
 
        // Initialize the ans with large value
        let ans = 1e18;
 
        // There are total four parts means 3 cuts.
        // Here i, j, k represent those 3 cuts
        for (let i = 1, j = 2, k = 3; j < n; j++) {
            while (i + 1 < j
                && Math.abs(a[j] - 2 * a[i])
                > Math.abs(a[j]
                    - 2 * a[i + 1])) {
                i++;
            }
            while (k + 1 < n
                && Math.abs(a[n] + a[j] - 2 * a[k])
                > Math.abs(a[n] + a[j]
                    - 2 * a[k + 1])) {
                k++;
            }
            ans = Math.min(ans,
                Math.max(...[a[i], a[j] - a[i],
                a[k] - a[j],
                a[n] - a[k]])
                - Math.min(...[a[i], a[j] - a[i],
                a[k] - a[j],
                a[n] - a[k]]));
        }
        return ans;
    }
 
    // Driver Code
    let arr = [3, 2, 4, 1, 2];
    let N = arr.length;
 
    // Function call
    document.write(minSum(arr, N));
 
// This code is contributed by rakeshsahni
 
</script>


Output

2

Time Complexity: O(N)
Auxiliary Space: O(N)


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