Given an array of elements ‘arr’, the task is to maximize the sum of the elements of this array after performing the following operation:
You can take any prefix of ‘arr’ and multiply each element of the prefix with ‘-1’.
In the first line, print the maximized sum than in the next line, print the index upto which the sequence of prefixes were chosen.
Examples:
Input: arr = {1, -2, -3, 4}
Output: 10
2 1 3 2
Flip the prefix till 2nd element then the sequence is -1 2 -3 4
Flip the prefix till 1st element then the sequence is 1 2 -3 4
Flip the prefix till 3rd element then the sequence is -1 -2 3 4
Flip the prefix till 2nd element then the sequence is 1 2 3 4
And, the final maximised sum is 10
Input: arr = {1, 2, 3, 4}
Output: 10
As, all the elements are already positive.
Approach: The max sum will always be 
as all the numbers of the array can be changed from negative to positive with the given operation.
- Traverse the array from left to right, if the element at index ‘i’ is negative then choose ‘i’ as the ending index of the prefix array and multiply each element by ‘-1’.
- Due to the operation in the previous step, all the elements in the array before index ‘i’ must be negative. So, take the prefix array ending at index ‘i-1’ and apply the same operation again to change all the elements to positive.
- Repeat the above steps until the complete array has been traversed and print the sum of all the elements along with all the ending indices of the chosen prefix arrays in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include<bits/stdc++.h>using namespace std;void maxSum(int *a, int n){ vector<int> l; // To store sum int s = 0; // To store ending indices // of the chosen prefix array vect for (int i = 0; i < n; i++) { // Adding the absolute // value of a[i] s += abs(a[i]); if (a[i] >= 0) continue; // If i == 0 then there is no index // to be flipped in (i-1) position if (i == 0) l.push_back(i + 1); else { l.push_back(i + 1); l.push_back(i); } } // print the maximized sum cout << s << endl; // print the ending indices // of the chosen prefix arrays for (int i = 0; i < l.size(); i++) cout << l[i] << " ";}// Driver Code int main(){ int n = 4; int a[] = {1, -2, -3, 4}; maxSum(a, n);} // This code is contributed by// Surendra_Gangwar |
Java
// Java implementation of the approachimport java.util.*;class GFG { static void maxSum(int []a, int n){ Vector<Integer> l = new Vector<Integer>(); // To store sum int s = 0; // To store ending indices // of the chosen prefix array vect for (int i = 0; i < n; i++) { // Adding the absolute // value of a[i] s += Math.abs(a[i]); if (a[i] >= 0) continue; // If i == 0 then there is no index // to be flipped in (i-1) position if (i == 0) l.add(i + 1); else { l.add(i + 1); l.add(i); } } // print the maximised sum System.out.println(s); // print the ending indices // of the chosen prefix arrays for (int i = 0; i < l.size(); i++) System.out.print(l.get(i) + " ");}// Driver Code public static void main(String[] args) { int n = 4; int a[] = {1, -2, -3, 4}; maxSum(a, n);}}// This code is contributed by 29AjayKumar |
Python3
# Python implementation of the approachdef maxSum(arr, n): # To store sum s = 0 # To store ending indices # of the chosen prefix arrays l = [] for i in range(len(a)): # Adding the absolute # value of a[i] s += abs(a[i]) if (a[i] >= 0): continue # If i == 0 then there is # no index to be flipped # in (i-1) position if (i == 0): l.append(i + 1) else: l.append(i + 1) l.append(i) # print the # maximised sum print(s) # print the ending indices # of the chosen prefix arrays print(*l) n = 4a = [1, -2, -3, 4]maxSum(a, n) |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG { static void maxSum(int []a, int n){ List<int> l = new List<int>(); // To store sum int s = 0; // To store ending indices // of the chosen prefix array vect for (int i = 0; i < n; i++) { // Adding the absolute // value of a[i] s += Math.Abs(a[i]); if (a[i] >= 0) continue; // If i == 0 then there is no index // to be flipped in (i-1) position if (i == 0) l.Add(i + 1); else { l.Add(i + 1); l.Add(i); } } // print the maximised sum Console.WriteLine(s); // print the ending indices // of the chosen prefix arrays for (int i = 0; i < l.Count; i++) Console.Write(l[i] + " ");}// Driver Code public static void Main(String[] args) { int n = 4; int []a = {1, -2, -3, 4}; maxSum(a, n);}}// This code is contributed by PrinciRaj1992 |
PHP
<?php// PHP implementation of the approachfunction maxSum($a, $n){ // To store sum $s = 0; // To store ending indices // of the chosen prefix arrays $l = array(); for ($i = 0; $i < count($a); $i++) { // Adding the absolute // value of a[i] $s += abs($a[$i]); if ($a[$i] >= 0) continue; // If i == 0 then there is // no index to be flipped // in (i-1) position if ($i == 0) array_push($l, $i + 1); else { array_push($l, $i + 1); array_push($l, $i); } } // print the // maximised sum echo $s . "\n"; // print the ending indices // of the chosen prefix arrays for($i = 0; $i < count($l); $i++) echo $l[$i] . " ";}// Driver Code$n = 4;$a = array(1, -2, -3, 4);maxSum($a, $n);// This code is contributed by mits?> |
Javascript
<script>// Javascript implementation of the above approach function maxSum(a, n){ let l = []; // To store sum let s = 0; // To store ending indices // of the chosen prefix array vect for (let i = 0; i < n; i++) { // Adding the absolute // value of a[i] s += Math.abs(a[i]); if (a[i] >= 0) continue; // If i == 0 then there is no index // to be flipped in (i-1) position if (i == 0) l.push(i + 1); else { l.push(i + 1); l.push(i); } } // print the maximised sum document.write(s + "<br/>"); // print the ending indices // of the chosen prefix arrays for (let i = 0; i < l.length; i++) document.write(l[i] + " ");}// driver code let n = 4; let a = [1, -2, -3, 4]; maxSum(a, n); </script> |
Output:
10 2 1 3 2
Time Complexity: O(n)
Auxiliary Space: O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
