Split given arrays into subarrays to maximize the sum of maximum and minimum in each subarrays

Last Updated : 03 Mar, 2022

Given an array arr[] consisting of N integers, the task is to maximize the sum of the difference of the maximum and minimum in each subarrays by splitting the given array into non-overlapping subarrays.

Examples:

Input: arr[] = {8, 1, 7, 9, 2}
Output: 14
Explanation:
Split the given array arr[] as {8, 1}, {7, 9, 2}. Now, the sum of the difference maximum and minimum for all the subarrays is (8 - 7) + (9 - 2) = 7 + 7 = 14, which is maximum among all possible combinations.

Input: arr[] = {1, 2, 1, 0, 5}
Output: 6

Approach: The given problem can be solved by considering all possible breaking of subarrays and find the sum of the difference of the maximum and minimum in each subarray and maximize this value. This idea can be implemented using Dynamic Programming. Follow the steps below to solve the given problem:

Initialize an array, dp[] with all elements as 0 such that dp[i] stores the sum of all possible splitting of the first i elements such that the sum of the difference of the maximum and minimum in each subarray is maximum.
Traverse the given array arr[] using the variable i and perform the following steps:
- Choose the current value as the maximum and the minimum element for the subarrays and store it in variables, say minValue and maxValue respectively.
- Iterate a loop over the range [0, i] using the variable j and perform the following steps:
  - Update the minValue and maxValue with the array element arr[j].
  - If the value of j is positive then update dp[i] as the maximum of dp[i] and (maxValue - minValue + dp[j - 1]). Otherwise, update dp[i] as the maximum of dp[i] and (maxValue - minValue).
After completing the above steps, print the value of dp[N - 1] as the resultant maximum value.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // Function to find the maximum sum of // differences of subarrays by splitting // array into non-overlapping subarrays int maxDiffSum(int arr[], int n) {     // Stores the answer for prefix     // and initialize with zero     int dp[n];     memset(dp, 0, sizeof(dp));      // Assume i-th index as right endpoint     for (int i = 0; i < n; i++) {          // Choose the current value as         // the maximum and minimum         int maxVal = arr[i], minVal = arr[i];          // Find the left endpoint and         // update the array dp[]         for (int j = i; j >= 0; j--) {             minVal = min(minVal, arr[j]);             maxVal = max(maxVal, arr[j]);              if (j - 1 >= 0)                 dp[i] = max(                     dp[i], maxVal - minVal + dp[j - 1]);             else                 dp[i] = max(                     dp[i], maxVal - minVal);         }     }      // Return answer     return dp[n - 1]; }  // Driver Code int main() {     int arr[] = { 8, 1, 7, 9, 2 };     int N = sizeof(arr) / sizeof(arr[0]);      cout << maxDiffSum(arr, N);      return 0; }

Java

// Java program for the above approach import java.io.*; class GFG {      // Function to find the maximum sum of     // differences of subarrays by splitting     // array into non-overlapping subarrays     static int maxDiffSum(int[] arr, int n)     {                // Stores the answer for prefix         // and initialize with zero         int[] dp = new int[n];          // Assume i-th index as right endpoint         for (int i = 0; i < n; i++) {              // Choose the current value as             // the maximum and minimum             int maxVal = arr[i], minVal = arr[i];              // Find the left endpoint and             // update the array dp[]             for (int j = i; j >= 0; j--) {                 minVal = Math.min(minVal, arr[j]);                 maxVal = Math.max(maxVal, arr[j]);                  if (j - 1 >= 0)                     dp[i] = Math.max(                         dp[i], maxVal - minVal + dp[j - 1]);                 else                     dp[i]                         = Math.max(dp[i], maxVal - minVal);             }         }          // Return answer         return dp[n - 1];     }      // Driver Code     public static void main(String []args)     {         int[] arr = { 8, 1, 7, 9, 2 };         int N = arr.length;          System.out.print(maxDiffSum(arr, N));     } }  // This code is contributed by shivanisinghss2110

Python3

# Python Program to implement # the above approach  # Function to find the maximum sum of # differences of subarrays by splitting # array into non-overlapping subarrays def maxDiffSum(arr, n):      # Stores the answer for prefix     # and initialize with zero     dp = [0] * n      # Assume i-th index as right endpoint     for i in range(n):          # Choose the current value as         # the maximum and minimum         maxVal = arr[i]         minVal = arr[i]          # Find the left endpoint and         # update the array dp[]         for j in range(i, -1, -1):             minVal = min(minVal, arr[j])             maxVal = max(maxVal, arr[j])              if (j - 1 >= 0):                 dp[i] = max(                     dp[i], maxVal - minVal + dp[j - 1])             else:                 dp[i] = max(                     dp[i], maxVal - minVal)      # Return answer     return dp[n - 1]  # Driver Code arr = [8, 1, 7, 9, 2] N = len(arr)  print(maxDiffSum(arr, N))  # This code is contributed by _saurabh_jaiswal

// C# program for the above approach using System; class GFG {      // Function to find the maximum sum of     // differences of subarrays by splitting     // array into non-overlapping subarrays     static int maxDiffSum(int[] arr, int n)     {                // Stores the answer for prefix         // and initialize with zero         int[] dp = new int[n];          // Assume i-th index as right endpoint         for (int i = 0; i < n; i++) {              // Choose the current value as             // the maximum and minimum             int maxVal = arr[i], minVal = arr[i];              // Find the left endpoint and             // update the array dp[]             for (int j = i; j >= 0; j--) {                 minVal = Math.Min(minVal, arr[j]);                 maxVal = Math.Max(maxVal, arr[j]);                  if (j - 1 >= 0)                     dp[i] = Math.Max(                         dp[i], maxVal - minVal + dp[j - 1]);                 else                     dp[i]                         = Math.Max(dp[i], maxVal - minVal);             }         }          // Return answer         return dp[n - 1];     }      // Driver Code     public static void Main()     {         int[] arr = { 8, 1, 7, 9, 2 };         int N = arr.Length;          Console.WriteLine(maxDiffSum(arr, N));     } }  // This code is contributed by ukasp.

JavaScript

 <script>         // JavaScript Program to implement         // the above approach          // Function to find the maximum sum of         // differences of subarrays by splitting         // array into non-overlapping subarrays         function maxDiffSum(arr, n)          {                      // Stores the answer for prefix             // and initialize with zero             let dp = new Array(n).fill(0);              // Assume i-th index as right endpoint             for (let i = 0; i < n; i++) {                  // Choose the current value as                 // the maximum and minimum                 let maxVal = arr[i], minVal = arr[i];                  // Find the left endpoint and                 // update the array dp[]                 for (let j = i; j >= 0; j--) {                     minVal = Math.min(minVal, arr[j]);                     maxVal = Math.max(maxVal, arr[j]);                      if (j - 1 >= 0)                         dp[i] = Math.max(                             dp[i], maxVal - minVal + dp[j - 1]);                     else                         dp[i] = Math.max(                             dp[i], maxVal - minVal);                 }             }              // Return answer             return dp[n - 1];         }          // Driver Code         let arr = [8, 1, 7, 9, 2];         let N = arr.length;          document.write(maxDiffSum(arr, N));       // This code is contributed by Potta Lokesh     </script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

Split array into K subsets to maximize their sum of maximums and minimums

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Split given arrays into subarrays to maximize the sum of maximum and minimum in each subarrays

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