Given an array arr[] of size N, the task is to find the maximum sum of the elements of the array after applying the given operation any number of times. In a single operation, choose an index 1 ? i < N and multiply both arr[i] and arr[i – 1] by -1.
Examples:
Input: arr[] = {-10, 5, -4}
Output: 19
Perform the operation for i = 1 and
the array becomes {10, -5, -4}
Perform the operation for i = 2 and
the array becomes {10, 5, 4}
10 + 5 + 4 = 19
Input: arr[] = {10, -4, -8, -11, 3}
Output: 30
Approach: This problem can be solved using dynamic programming. Since it is useless to choose and flip the same arr[i], we will consider flipping at most once for each element of the array in order from the left. Let dp[i][0] represents the maximum possible sum up to the ith index without flipping the ith index. dp[i][1] represents the maximum possible sum up to the ith index with flipping the ith index. So, dp(n, 0) is our required answer.
Base conditions:
- dp[0][0] = 0
- dp[0][1] = INT_MIN
Recurrence relation:
- dp[i + 1][0] = max(dp[i][0] + arr[i], dp[i][1] – arr[i])
- dp[i + 1][1] = max(dp[i][0] – arr[i], dp[i][1] + arr[i])
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int max_sum(int a[], int n)
{
vector<vector<int> > dp(n + 1,
vector<int>(2, 0));
dp[0][0] = 0, dp[0][1] = -999999;
for (int i = 0; i <= n - 1; i++) {
dp[i + 1][0] = max(dp[i][0] + a[i],
dp[i][1] - a[i]);
dp[i + 1][1] = max(dp[i][0] - a[i],
dp[i][1] + a[i]);
}
return dp[n][0];
}
int main()
{
int a[] = { -10, 5, -4 };
int n = sizeof(a) / sizeof(a[0]);
cout << max_sum(a, n);
return 0;
}
|
Java
class GFG
{
static int max_sum(int a[], int n)
{
int [][]dp = new int[n + 1][2];
dp[0][0] = 0; dp[0][1] = -999999;
for (int i = 0; i <= n - 1; i++)
{
dp[i + 1][0] = Math.max(dp[i][0] + a[i],
dp[i][1] - a[i]);
dp[i + 1][1] = Math.max(dp[i][0] - a[i],
dp[i][1] + a[i]);
}
return dp[n][0];
}
public static void main(String[] args)
{
int a[] = { -10, 5, -4 };
int n = a.length;
System.out.println(max_sum(a, n));
}
}
|
Python3
def max_sum(a, n):
dp = [[0 for i in range(2)] for j in range(n+1)]
dp[0][0] = 0; dp[0][1] = -999999;
for i in range(0, n):
dp[i + 1][0] = max(dp[i][0] + a[i],
dp[i][1] - a[i]);
dp[i + 1][1] = max(dp[i][0] - a[i],
dp[i][1] + a[i]);
return dp[n][0];
if __name__ == '__main__':
a = [-10, 5, -4 ];
n = len(a);
print(max_sum(a, n));
|
Javascript
<script>
function max_sum(a , n) {
var dp = Array(n + 1).fill().map(()=>Array(2).fill(0));
dp[0][0] = 0;
dp[0][1] = -999999;
for (i = 0; i <= n - 1; i++) {
dp[i + 1][0] = Math.max(dp[i][0] + a[i], dp[i][1] - a[i]);
dp[i + 1][1] = Math.max(dp[i][0] - a[i], dp[i][1] + a[i]);
}
return dp[n][0];
}
var a = [ -10, 5, -4 ];
var n = a.length;
document.write(max_sum(a, n));
</script>
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C#
using System;
class GFG
{
static int max_sum(int []a, int n)
{
int [,]dp = new int[n + 1, 2];
dp[0, 0] = 0; dp[0, 1] = -999999;
for (int i = 0; i <= n - 1; i++)
{
dp[i + 1, 0] = Math.Max(dp[i, 0] + a[i],
dp[i, 1] - a[i]);
dp[i + 1, 1] = Math.Max(dp[i, 0] - a[i],
dp[i, 1] + a[i]);
}
return dp[n, 0];
}
public static void Main(String[] args)
{
int []a = { -10, 5, -4 };
int n = a.Length;
Console.WriteLine(max_sum(a, n));
}
}
|
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(N), as we are using extra space for DP.
Efficient approach : Space Optimization O(1)
To optimize the space complexity in previous code we are only using the previous row of the dp table to compute the values of the current row. Therefore, we can maintain just two variables to keep track of the values of the previous row instead of using a 2D vector. This will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Initialize two variables prev0 and prev1 to 0 and -999999 respectively.
- Loop through the input array from index 0 to n-1.
- Compute two new variables curr0 and curr1 as follows.
- curr0 is the maximum of (prev0 + a[i]) and (prev1 – a[i])
- curr1 is the maximum of (prev0 – a[i]) and (prev1 + a[i])
- Update prev0 to curr0 and prev1 to curr1 for further iteration.
- At last return final answer stored in prev0.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int max_sum(int a[], int n)
{
int prev0 = 0, prev1 = -999999;
for (int i = 0; i <= n - 1; i++) {
int curr0 = max(prev0 + a[i], prev1 - a[i]);
int curr1 = max(prev0 - a[i], prev1 + a[i]);
prev0 = curr0;
prev1 = curr1;
}
return prev0;
}
int main()
{
int a[] = { -10, 5, -4 };
int n = sizeof(a) / sizeof(a[0]);
cout << max_sum(a, n);
return 0;
}
|
Java
import java.util.*;
class Main {
static int maxSum(int[] a, int n)
{
int prev0 = 0, prev1 = -999999;
for (int i = 0; i <= n - 1; i++) {
int curr0 = Math.max(prev0 + a[i], prev1 - a[i]);
int curr1 = Math.max(prev0 - a[i], prev1 + a[i]);
prev0 = curr0;
prev1 = curr1;
}
return prev0;
}
public static void main(String[] args)
{
int[] a = { -10, 5, -4 };
int n = a.length;
System.out.println(maxSum(a, n));
}
}
|
Python3
def max_sum(a, n):
prev0 = 0
prev1 = -999999
for i in range(n):
curr0 = max(prev0 + a[i], prev1 - a[i])
curr1 = max(prev0 - a[i], prev1 + a[i])
prev0 = curr0
prev1 = curr1
return prev0
a = [-10, 5, -4]
n = len(a)
print(max_sum(a, n))
|
C#
using System;
class MainClass
{
static int maxSum(int[] a, int n)
{
int prev0 = 0, prev1 = -999999;
for (int i = 0; i <= n - 1; i++)
{
int curr0 = Math.Max(prev0 + a[i], prev1 - a[i]);
int curr1 = Math.Max(prev0 - a[i], prev1 + a[i]);
prev0 = curr0;
prev1 = curr1;
}
return prev0;
}
public static void Main()
{
int[] a = { -10, 5, -4 };
int n = a.Length;
Console.WriteLine(maxSum(a, n));
}
}
|
Javascript
function max_sum(a, n) {
let prev0 = 0,
prev1 = -999999;
for (let i = 0; i <= n - 1; i++) {
let curr0 = Math.max(prev0 + a[i], prev1 - a[i]);
let curr1 = Math.max(prev0 - a[i], prev1 + a[i]);
prev0 = curr0;
prev1 = curr1;
}
return prev0;
}
let a = [-10, 5, -4];
let n = a.length;
console.log(max_sum(a, n));
|
Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time.
Auxiliary Space: O(1)