Split array into two subarrays such that difference of their sum is minimum
Given an integer array arr[], the task is to split the given array into two subarrays such that the difference between their sum is minimum.
Examples:
Input: arr[] = {7, 9, 5, 10}
Output: 1
Explanation: The difference between the sum of the subarrays {7, 9} and {5, 10} is equal to [16 – 15] = 1, which is the minimum possible.Input: arr[] = {6, 6, 6}
Output: 6
Naive Approach: The idea is to use the Prefix and Suffix Sum array technique. Generate the prefix sum array and the suffix sum array of the given array. Now, iterate over the array and print the minimum difference between prefix_sum[i] and suffix_sum[i+1], for any index i ( 0 <= i <= N – 1) from the array.
Time Complexity: O(N)
Auxiliary Space: O(N)
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to return minimum difference // between two subarray sums int minDiffSubArray(int arr[], int n) { // To store prefix sums int prefix_sum[n]; // Generate prefix sum array prefix_sum[0] = arr[0]; for(int i = 1; i < n; i++) prefix_sum[i] = prefix_sum[i - 1] + arr[i]; // To store suffix sums int suffix_sum[n]; // Generate suffix sum array suffix_sum[n - 1] = arr[n - 1]; for(int i = n - 2; i >= 0; i--) suffix_sum[i] = suffix_sum[i + 1] + arr[i]; // Stores the minimum difference int minDiff = INT_MAX; // Traverse the given array for(int i = 0; i < n - 1; i++) { // Calculate the difference int diff = abs(prefix_sum[i] - suffix_sum[i + 1]); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver Code int main() { // Given array int arr[] = { 7, 9, 5, 10 }; // Length of the array int n = sizeof(arr) / sizeof(arr[0]); cout << minDiffSubArray(arr, n); } // This code is contributed by code_hunt |
Java
// Java Program for above approach import java.util.*; import java.lang.*; class GFG { // Function to return minimum difference // between two subarray sums static int minDiffSubArray(int arr[], int n) { // To store prefix sums int[] prefix_sum = new int[n]; // Generate prefix sum array prefix_sum[0] = arr[0]; for (int i = 1; i < n; i++) prefix_sum[i] = prefix_sum[i - 1] + arr[i]; // To store suffix sums int[] suffix_sum = new int[n]; // Generate suffix sum array suffix_sum[n - 1] = arr[n - 1]; for (int i = n - 2; i >= 0; i--) suffix_sum[i] = suffix_sum[i + 1] + arr[i]; // Stores the minimum difference int minDiff = Integer.MAX_VALUE; // Traverse the given array for (int i = 0; i < n - 1; i++) { // Calculate the difference int diff = Math.abs(prefix_sum[i] - suffix_sum[i + 1]); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver Code public static void main(String[] args) { // Given array int[] arr = { 7, 9, 5, 10 }; // Length of the array int n = arr.length; System.out.println( minDiffSubArray(arr, n)); } } |
Python3
# Python3 program for the above approach import sys # Function to return minimum difference # between two subarray sums def minDiffSubArray(arr, n): # To store prefix sums prefix_sum = [0] * n # Generate prefix sum array prefix_sum[0] = arr[0] for i in range(1, n): prefix_sum[i] = (prefix_sum[i - 1] + arr[i]) # To store suffix sums suffix_sum = [0] * n # Generate suffix sum array suffix_sum[n - 1] = arr[n - 1] for i in range(n - 2, -1, -1): suffix_sum[i] = (suffix_sum[i + 1] + arr[i]) # Stores the minimum difference minDiff = sys.maxsize # Traverse the given array for i in range(n - 1): # Calculate the difference diff = abs(prefix_sum[i] - suffix_sum[i + 1]) # Update minDiff if (diff < minDiff): minDiff = diff # Return minDiff return minDiff # Driver Code if __name__ == '__main__': # Given array arr = [ 7, 9, 5, 10 ] # Length of the array n = len(arr) print(minDiffSubArray(arr, n)) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approach using System; class GFG{ // Function to return minimum difference // between two subarray sums static int minDiffSubArray(int []arr, int n) { // To store prefix sums int[] prefix_sum = new int[n]; // Generate prefix sum array prefix_sum[0] = arr[0]; for(int i = 1; i < n; i++) prefix_sum[i] = prefix_sum[i - 1] + arr[i]; // To store suffix sums int[] suffix_sum = new int[n]; // Generate suffix sum array suffix_sum[n - 1] = arr[n - 1]; for(int i = n - 2; i >= 0; i--) suffix_sum[i] = suffix_sum[i + 1] + arr[i]; // Stores the minimum difference int minDiff = int.MaxValue; // Traverse the given array for(int i = 0; i < n - 1; i++) { // Calculate the difference int diff = Math.Abs(prefix_sum[i] - suffix_sum[i + 1]); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver Code public static void Main(String[] args) { // Given array int[] arr = {7, 9, 5, 10}; // Length of the array int n = arr.Length; Console.WriteLine(minDiffSubArray(arr, n)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program for the above approach // Function to return minimum difference // between two subarray sums function minDiffSubArray(arr, n) { // To store prefix sums let prefix_sum = new Array(n); // Generate prefix sum array prefix_sum[0] = arr[0]; for(let i = 1; i < n; i++) prefix_sum[i] = prefix_sum[i - 1] + arr[i]; // To store suffix sums let suffix_sum = new Array(n); // Generate suffix sum array suffix_sum[n - 1] = arr[n - 1]; for(let i = n - 2; i >= 0; i--) suffix_sum[i] = suffix_sum[i + 1] + arr[i]; // Stores the minimum difference let minDiff = Number.MAX_VALUE; // Traverse the given array for(let i = 0; i < n - 1; i++) { // Calculate the difference let diff = Math.abs(prefix_sum[i] - suffix_sum[i + 1]); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Given array let arr = [7, 9, 5, 10]; // Length of the array let n = arr.length; document.write(minDiffSubArray(arr, n)); // This code is contributed by vaibhavrabadiya117. </script> |
Output:
1
Efficient Approach: To optimize the above approach, the idea is to calculate the total sum of the array elements. Now, iterate over the array and calculate the prefix sum of the array and find the minimum difference between the prefix sum and the difference between total sum and the prefix sum for any index of the array, i.e.,
Maximum difference between sum of subarrays = ( prefix_sum – (total_sum – prefix_sum) )
Below is the implementation of the above approach:
C++
// C++ program for above approach #include <bits/stdc++.h> using namespace std; // Function to return minimum difference // between sum of two subarrays int minDiffSubArray(int arr[], int n) { // To store total sum of array int total_sum = 0; // Calculate the total sum // of the array for(int i = 0; i < n; i++) total_sum += arr[i]; // Stores the prefix sum int prefix_sum = 0; // Stores the minimum difference int minDiff = INT_MAX; // Traverse the given array for(int i = 0; i < n - 1; i++) { prefix_sum += arr[i]; // To store minimum difference int diff = abs((total_sum - prefix_sum) - prefix_sum); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver code int main() { // Given array int arr[] = { 7, 9, 5, 10 }; // Length of the array int n = sizeof(arr) / sizeof(arr[0]); cout << minDiffSubArray(arr, n) << endl; return 0; } // This code is contributed by divyeshrabadiya07 |
Java
// Java Program for above approach import java.util.*; import java.lang.*; class GFG { // Function to return minimum difference // between sum of two subarrays static int minDiffSubArray(int arr[], int n) { // To store total sum of array int total_sum = 0; // Calculate the total sum // of the array for (int i = 0; i < n; i++) total_sum += arr[i]; // Stores the prefix sum int prefix_sum = 0; // Stores the minimum difference int minDiff = Integer.MAX_VALUE; // Traverse the given array for (int i = 0; i < n - 1; i++) { prefix_sum += arr[i]; // To store minimum difference int diff = Math.abs((total_sum - prefix_sum) - prefix_sum); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver Code public static void main(String[] args) { // Given array int[] arr = { 7, 9, 5, 10 }; // Length of the array int n = arr.length; System.out.println( minDiffSubArray(arr, n)); } } |
Python3
# Python3 program for the above approach import sys # Function to return minimum difference # between sum of two subarrays def minDiffSubArray(arr, n): # To store total sum of array total_sum = 0 # Calculate the total sum # of the array for i in range(n): total_sum += arr[i] # Stores the prefix sum prefix_sum = 0 # Stores the minimum difference minDiff = sys.maxsize # Traverse the given array for i in range(n - 1): prefix_sum += arr[i] # To store minimum difference diff = abs((total_sum - prefix_sum) - prefix_sum) # Update minDiff if (diff < minDiff): minDiff = diff # Return minDiff return minDiff # Driver code # Given array arr = [ 7, 9, 5, 10 ] # Length of the array n = len(arr) print(minDiffSubArray(arr, n)) # This code is contributed by code_hunt |
C#
// C# Program for above approach using System; class GFG{ // Function to return minimum difference // between sum of two subarrays static int minDiffSubArray(int []arr, int n) { // To store total sum of array int total_sum = 0; // Calculate the total sum // of the array for (int i = 0; i < n; i++) total_sum += arr[i]; // Stores the prefix sum int prefix_sum = 0; // Stores the minimum difference int minDiff = int.MaxValue; // Traverse the given array for (int i = 0; i < n - 1; i++) { prefix_sum += arr[i]; // To store minimum difference int diff = Math.Abs((total_sum - prefix_sum) - prefix_sum); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver Code public static void Main(String[] args) { // Given array int[] arr = {7, 9, 5, 10}; // Length of the array int n = arr.Length; Console.WriteLine( minDiffSubArray(arr, n)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program for above approach // Function to return minimum difference // between sum of two subarrays function minDiffSubArray(arr, n) { // To store total sum of array var total_sum = 0; // Calculate the total sum // of the array for(var i = 0; i < n; i++) total_sum += arr[i]; // Stores the prefix sum var prefix_sum = 0; // Stores the minimum difference var minDiff = 1000000000; // Traverse the given array for(var i = 0; i < n - 1; i++) { prefix_sum += arr[i]; // To store minimum difference var diff = Math.abs((total_sum - prefix_sum) - prefix_sum); // Update minDiff if (diff < minDiff) minDiff = diff; } // Return minDiff return minDiff; } // Driver code // Given array var arr = [7, 9, 5, 10]; // Length of the array var n = arr.length; document.write( minDiffSubArray(arr, n)); // This code is contributed by importantly. </script> |
Output:
1
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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