Maximize array sum by alternating the signs of adjacent elements
Given an array, arr[] of size N, the task is to find the maximum possible sum of array elements by alternating the signs of adjacent array elements.
Examples:
Input: arr[] = { -2, 1, 0 }
Output: 3
Explanation:
Alternating the signs of (arr[0], arr[1]) modifies arr[] to {2, -1, 0}.
Alternating the signs of (arr[1], arr[2]) modifies arr[] to {2, 1, 0}.
Therefore, the required output = (2 + 1 + 0) = 3, which is the maximum sum possible.Input: arr[] = { 1, 1, -2, -4, 5 }
Output: 13
Explanation:
Alternating the signs of (arr[2], arr[3]) modifies arr[] to { 1, 1, 2, 4, 5 }
Therefore, the required output = (1 + 1 + 2 + 4 + 5) = 13, which is the maximum sum possible.
Approach: The problem can be solved using Greedy technique. The idea is based on the fact that the maximum count of negative elements in the array after alternating the signs of adjacent elements can’t be greater than 1. Follow the steps below to solve the problem:
- Initialize a variable, say MaxAltSum, to store the maximum possible sum of array elements by alternating the signs of adjacent elements.
- Traverse the array and count the number of negative elements in the array.
- If count of negative elements in the array is even, then maximum possible sum possible by alternating the signs of adjacent array elements is equal to the sum of absolute value of array elements, i.e. MaxAltSum = Σabs(arr[i])
- Otherwise, maximum possible sum obtained from the array by alternating the signs of adjacent array elements equal to the sum of the absolute value of all possible array elements, except the smallest absolute value of array elements. i.e, MaxAltSum = ((Σabs(arr[i])) – 2 * X), where X is the smallest absolute value of array elements.
- Finally, print the value MaxAltSum.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum sum by alternating// the signs of adjacent elements of the arrayint findMaxSumByAlternatingSign(int arr[], int N){ // Stores count of negative // elements in the array int cntNeg = 0; // Stores maximum sum by alternating // the signs of adjacent elements int MaxAltSum = 0; // Stores smallest absolute // value of array elements int SmValue = 0; // Stores sum of absolute // value of array elements int sum = 0; // Traverse the array for (int i = 0; i < N; i++) { // If arr[i] is // a negative number if (arr[i] < 0) { // Update cntNeg cntNeg += 1; } // Update sum sum += abs(arr[i]); // Update SmValue SmValue = min(SmValue, abs(arr[i])); } // Update MaxAltSum MaxAltSum = sum; // If cntNeg is // an odd number if (cntNeg & 1) { // Update MaxAltSum MaxAltSum -= 2 * SmValue; } return MaxAltSum;}// Drivers Codeint main(){ int arr[] = { 1, 1, -2, -4, 5 }; int N = sizeof(arr) / sizeof(arr[0]); cout << findMaxSumByAlternatingSign( arr, N);} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find the maximum sum by alternating// the signs of adjacent elements of the arraystatic int findMaxSumByAlternatingSign(int arr[], int N){ // Stores count of negative // elements in the array int cntNeg = 0; // Stores maximum sum by alternating // the signs of adjacent elements int MaxAltSum = 0; // Stores smallest absolute // value of array elements int SmValue = 0; // Stores sum of absolute // value of array elements int sum = 0; // Traverse the array for(int i = 0; i < N; i++) { // If arr[i] is // a negative number if (arr[i] < 0) { // Update cntNeg cntNeg += 1; } // Update sum sum += Math.abs(arr[i]); // Update SmValue SmValue = Math.min(SmValue, Math.abs(arr[i])); } // Update MaxAltSum MaxAltSum = sum; // If cntNeg is // an odd number if (cntNeg % 2 == 1) { // Update MaxAltSum MaxAltSum -= 2 * SmValue; } return MaxAltSum;}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 1, -2, -4, 5 }; int N = arr.length; System.out.print(findMaxSumByAlternatingSign( arr, N));}}// This code is contributed by jana_sayantan |
Python3
# Python3 program to implement# the above approach# Function to find the maximum sum by# alternating the signs of adjacent# elements of the arraydef findMaxSumByAlternatingSign(arr, N): # Stores count of negative # elements in the array cntNeg = 0 # Stores maximum sum by alternating # the signs of adjacent elements MaxAltSum = 0 # Stores smallest absolute # value of array elements SmValue = 0 # Stores sum of absolute # value of array elements sum = 0 # Traverse the array for i in range(N): # If arr[i] is # a negative number if (arr[i] < 0): # Update cntNeg cntNeg += 1 # Update sum sum += abs(arr[i]) # Update SmValue SmValue = min(SmValue, abs(arr[i])) # Update MaxAltSum MaxAltSum = sum # If cntNeg is # an odd number if (cntNeg & 1): # Update MaxAltSum MaxAltSum -= 2 * SmValue return MaxAltSum# Driver Codeif __name__ == '__main__': arr = [ 1, 1, -2, -4, 5 ] N = len(arr) print(findMaxSumByAlternatingSign(arr, N))# This code is contributed by SURENDRA_GANGWAR |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to find the maximum sum by alternating// the signs of adjacent elements of the arraystatic int findMaxSumByAlternatingSign(int []arr, int N){ // Stores count of negative // elements in the array int cntNeg = 0; // Stores maximum sum by alternating // the signs of adjacent elements int MaxAltSum = 0; // Stores smallest absolute // value of array elements int SmValue = 0; // Stores sum of absolute // value of array elements int sum = 0; // Traverse the array for(int i = 0; i < N; i++) { // If arr[i] is // a negative number if (arr[i] < 0) { // Update cntNeg cntNeg += 1; } // Update sum sum += Math.Abs(arr[i]); // Update SmValue SmValue = Math.Min(SmValue, Math.Abs(arr[i])); } // Update MaxAltSum MaxAltSum = sum; // If cntNeg is // an odd number if (cntNeg % 2 == 1) { // Update MaxAltSum MaxAltSum -= 2 * SmValue; } return MaxAltSum;}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 1, -2, -4, 5 }; int N = arr.Length; Console.Write(findMaxSumByAlternatingSign( arr, N));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the maximum sum by alternating// the signs of adjacent elements of the arrayfunction findMaxSumByAlternatingSign(arr, N){ // Stores count of negative // elements in the array let cntNeg = 0; // Stores maximum sum by alternating // the signs of adjacent elements let MaxAltSum = 0; // Stores smallest absolute // value of array elements let SmValue = 0; // Stores sum of absolute // value of array elements let sum = 0; // Traverse the array for(let i = 0; i < N; i++) { // If arr[i] is // a negative number if (arr[i] < 0) { // Update cntNeg cntNeg += 1; } // Update sum sum += Math.abs(arr[i]); // Update SmValue SmValue = Math.min(SmValue, Math.abs(arr[i])); } // Update MaxAltSum MaxAltSum = sum; // If cntNeg is // an odd number if (cntNeg % 2 == 1) { // Update MaxAltSum MaxAltSum -= 2 * SmValue; } return MaxAltSum;} // Driver Code let arr = [ 1, 1, -2, -4, 5 ]; let N = arr.length; document.write(findMaxSumByAlternatingSign( arr, N));// This code is contributed by souravghosh0416.</script> |
Output:
13
Time Complexity: O(N)
Auxiliary Space:O(1)
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