Maximize the sum of array by multiplying prefix of array with -1
Last Updated :
21 Aug, 2022
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++
#include<bits/stdc++.h>
using namespace std;
void maxSum(int *a, int n)
{
vector<int> l;
int s = 0;
for (int i = 0; i < n; i++)
{
s += abs(a[i]);
if (a[i] >= 0)
continue;
if (i == 0)
l.push_back(i + 1);
else
{
l.push_back(i + 1);
l.push_back(i);
}
}
cout << s << endl;
for (int i = 0; i < l.size(); i++)
cout << l[i] << " ";
}
int main()
{
int n = 4;
int a[] = {1, -2, -3, 4};
maxSum(a, n);
}
|
Java
import java.util.*;
class GFG
{
static void maxSum(int []a, int n)
{
Vector<Integer> l = new Vector<Integer>();
int s = 0;
for (int i = 0; i < n; i++)
{
s += Math.abs(a[i]);
if (a[i] >= 0)
continue;
if (i == 0)
l.add(i + 1);
else
{
l.add(i + 1);
l.add(i);
}
}
System.out.println(s);
for (int i = 0; i < l.size(); i++)
System.out.print(l.get(i) + " ");
}
public static void main(String[] args)
{
int n = 4;
int a[] = {1, -2, -3, 4};
maxSum(a, n);
}
}
|
Python3
def maxSum(arr, n):
s = 0
l = []
for i in range(len(a)):
s += abs(a[i])
if (a[i] >= 0):
continue
if (i == 0):
l.append(i + 1)
else:
l.append(i + 1)
l.append(i)
print(s)
print(*l)
n = 4
a = [1, -2, -3, 4]
maxSum(a, n)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void maxSum(int []a, int n)
{
List<int> l = new List<int>();
int s = 0;
for (int i = 0; i < n; i++)
{
s += Math.Abs(a[i]);
if (a[i] >= 0)
continue;
if (i == 0)
l.Add(i + 1);
else
{
l.Add(i + 1);
l.Add(i);
}
}
Console.WriteLine(s);
for (int i = 0; i < l.Count; i++)
Console.Write(l[i] + " ");
}
public static void Main(String[] args)
{
int n = 4;
int []a = {1, -2, -3, 4};
maxSum(a, n);
}
}
|
PHP
<?php
function maxSum($a, $n)
{
$s = 0;
$l = array();
for ($i = 0; $i < count($a); $i++)
{
$s += abs($a[$i]);
if ($a[$i] >= 0)
continue;
if ($i == 0)
array_push($l, $i + 1);
else
{
array_push($l, $i + 1);
array_push($l, $i);
}
}
echo $s . "\n";
for($i = 0; $i < count($l); $i++)
echo $l[$i] . " ";
}
$n = 4;
$a = array(1, -2, -3, 4);
maxSum($a, $n);
?>
|
Javascript
<script>
function maxSum(a, n)
{
let l = [];
let s = 0;
for (let i = 0; i < n; i++)
{
s += Math.abs(a[i]);
if (a[i] >= 0)
continue;
if (i == 0)
l.push(i + 1);
else
{
l.push(i + 1);
l.push(i);
}
}
document.write(s + "<br/>");
for (let i = 0; i < l.length; i++)
document.write(l[i] + " ");
}
let n = 4;
let a = [1, -2, -3, 4];
maxSum(a, n);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(n)
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