Given an array arr[] of size N, the task is to find the maximum alternating sum of a subarray possible for a given array.
Alternating Subarray Sum: Considering a subarray {arr[i], arr[j]}, alternating sum of the subarray is arr[i] – arr[i + 1] + arr[i + 2] – …….. (+ / -) arr[j].
Examples:
Input: arr[] = {-4, -10, 3, 5}
Output: 9
Explanation: Subarray {arr[0], arr[2]} = {-4, -10, 3}. Therefore, the sum of this subarray is 9.Input: arr[] = {-1, 2, -1, 4, 7}
Output: 7
Approach: The given problem can be solved by using Dynamic Programming. Follow the steps below to solve the problem:
- Initialize a variable, say sum as 0, which will hold a maximum alternating subarray sum and a variable, say sumSoFar, to store the sum of subarrays starting from even indices in the 1st loop and the sum starting from odd indices, in the 2nd loop.
- In every iteration of both the loops, update sum as max(sum, sumSoFar).
- Finally, return the maximum alternating sum stored in the sum variable.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum alternating// sum of a subarray for the given arrayint alternatingSum(int arr[],int n){ int sum = 0; int sumSoFar = 0; // Traverse the array for (int i = 0; i < n; i++) { // Store sum of subarrays // starting at even indices if (i % 2 == 1) { sumSoFar -= arr[i]; } else { sumSoFar = max( sumSoFar + arr[i], arr[i]); } // Update sum sum = max(sum, sumSoFar); } sumSoFar = 0; // Traverse the array for (int i = 1; i < n; i++) { // Store sum of subarrays // starting at odd indices if (i % 2 == 0) { sumSoFar -= arr[i]; } else { sumSoFar = max( sumSoFar + arr[i], arr[i]); } // Update sum sum = max(sum, sumSoFar); } return sum;}// Driver codeint main() { // Given Input int arr[] ={ -4, -10, 3, 5 }; int n = sizeof(arr)/sizeof(arr[0]); // Function call int ans = alternatingSum(arr,n); cout<<ans<<endl; return 0;}// This code is contributed by Potta Lokesh |
Java
// Java implementation for the above approachimport java.io.*;class GFG { // Function to find the maximum alternating // sum of a subarray for the given array public static int alternatingSum(int[] arr) { int sum = 0; int sumSoFar = 0; // Traverse the array for (int i = 0; i < arr.length; i++) { // Store sum of subarrays // starting at even indices if (i % 2 == 1) { sumSoFar -= arr[i]; } else { sumSoFar = Math.max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.max(sum, sumSoFar); } sumSoFar = 0; // Traverse the array for (int i = 1; i < arr.length; i++) { // Store sum of subarrays // starting at odd indices if (i % 2 == 0) { sumSoFar -= arr[i]; } else { sumSoFar = Math.max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.max(sum, sumSoFar); } return sum; } // Driver code public static void main(String[] args) { // Given Input int arr[] = new int[] { -4, -10, 3, 5 }; // Function call int ans = alternatingSum(arr); System.out.println(ans); }} |
Python3
# Python implementation for the above approach# Function to find the maximum alternating# sum of a subarray for the given arraydef alternatingSum(arr, n): sum_ = 0 sumSoFar = 0 # Traverse the array for i in range(n): # Store sum of subarrays # starting at even indices if i % 2 == 1: sumSoFar -= arr[i] else: sumSoFar = max(arr[i], sumSoFar + arr[i]) # Update sum sum_ = max(sum_, sumSoFar) sumSoFar = 0 # Traverse array for i in range(1, n): # Store sum of subarrays # starting at odd indices if i % 2 == 0: sumSoFar -= arr[i] else: sumSoFar = max(arr[i], sumSoFar + arr[i]) sum_ = max(sum_, sumSoFar) # update sum return sum_# given arrayarr = [-4, -10, 3, 5]n = len(arr)# return sumans = alternatingSum(arr, n)print(ans)# This code is contributed by Parth Manchanda |
C#
// C# implementation for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the maximum alternating// sum of a subarray for the given arraystatic int alternatingSum(int []arr,int n){ int sum = 0; int sumSoFar = 0; // Traverse the array for(int i = 0; i < n; i++) { // Store sum of subarrays // starting at even indices if (i % 2 == 1) { sumSoFar -= arr[i]; } else { sumSoFar = Math.Max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.Max(sum, sumSoFar); } sumSoFar = 0; // Traverse the array for(int i = 1; i < n; i++) { // Store sum of subarrays // starting at odd indices if (i % 2 == 0) { sumSoFar -= arr[i]; } else { sumSoFar = Math.Max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.Max(sum, sumSoFar); } return sum;}// Driver codepublic static void Main(){ // Given Input int []arr = { -4, -10, 3, 5 }; int n = arr.Length; // Function call int ans = alternatingSum(arr,n); Console.Write(ans);}}// This code is contributed by SURENDRA_GANGWAR |
Javascript
<script>// JavaScript implementation for the above approach // Function to find the maximum alternating // sum of a subarray for the given array function alternatingSum(arr) { var sum = 0; var sumSoFar = 0; // Traverse the array for (var i = 0; i < arr.length; i++) { // Store sum of subarrays // starting at even indices if (i % 2 == 1) { sumSoFar -= arr[i]; } else { sumSoFar = Math.max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.max(sum, sumSoFar); } sumSoFar = 0; // Traverse the array for (var i = 1; i < arr.length; i++) { // Store sum of subarrays // starting at odd indices if (i % 2 == 0) { sumSoFar -= arr[i]; } else { sumSoFar = Math.max( sumSoFar + arr[i], arr[i]); } // Update sum sum = Math.max(sum, sumSoFar); } return sum; } // Driver code // Given Input var arr = new Array ( -4, -10, 3, 5 ); // Function call var ans = alternatingSum(arr); document.write(ans); // This code is contributed by shivanisinghss2110</script> |
Output:
9
Time Complexity: O(N)
Auxiliary Space: O(1)
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