Maximize difference between sum of even and odd-indexed elements of a subsequence | Set 2
Last Updated :
23 Jan, 2023
Given an array arr[] consisting of N positive integers, the task is to find the maximum value of the difference between the sum of elements at even and odd indices for any subsequence of the array.
Note: The value of N is always greater than 1.
Examples:
Input: arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 }
Output: 15
Explanation:
Considering the subsequences { 3, 1, 5, 1, 9 } from the array
Sum of even-indexed array elements = 3 + 5 + 9 = 17
Sum of odd-indexed array elements = is 1 + 1 = 2
Therefore, the difference between the sum of even and odd-indexed elements present in the subsequence = (17 – 2) = 15, which is the maximum possible.
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 6
Naive Approach: The simple approach to solve the given problem is to generate all possible subsequences of the given array and for each subsequence maximize the difference between the sum of elements at even and odd indices. After checking for all subsequences, print the maximum value obtained.
Time Complexity: O(N*2N)
Auxiliary Space: O(1)
Local Maxima Approach: The approach to solve this problem using Local Maxima and Local Minima is discussed in this post. In this article, we have discussed the dynamic programming approach.
Efficient Approach: The above approach can also be optimized by using Dynamic Programming as the above approach has Optimal Substructure and Overlapping Subproblems. Follow the steps below to solve the given problem:
- Initialize two arrays, dp1[] and dp2[] of size N, and initialize with values -1 such that dp1[i] stores the maximum sum from a subsequence of the odd length till the ith index and dp2[i] stores the maximum sum from a subsequence of an even length till the ith index.
- Update the value of dp1[0] as arr[0] and dp2[0] as 0.
- Iterate over the range [1, N] using the variable i and perform the following steps:
- Update the value of dp1[i] as dp1[i] = max(dp1[i – 1], dp2[i – 1] + arr[i]), as the ith element will be appended at an even index in the subsequence. Hence, the array element arr[i] is added.
- Update the value of dp2[i] as dp2[i] = max(dp2[i – 1], dp1[i – 1] – arr[i]) as the ith element will be appended at an odd index in the subsequence. Hence, the array element arr[i] is subtracted.
- After performing the above steps, print the maximum of (dp1[N – 1], dp2[N – 1]) as the result.
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
using namespace std;
void findMaximumPeakSum( int arr[], int n)
{
int dp1[n], dp2[n];
for ( int i = 0; i < n; i++) {
dp1[i] = -1;
dp2[i] = -1;
}
dp2[0] = 0;
dp1[0] = arr[0];
for ( int i = 1; i < n; i++) {
dp1[i] = max(dp1[i - 1],
dp2[i - 1] + arr[i]);
dp2[i] = max(dp2[i - 1],
dp1[i - 1] - arr[i]);
}
int ans = max(dp1[n - 1], dp2[n - 1]);
cout << ans;
}
int main()
{
int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
int N = sizeof (arr) / sizeof (arr[0]);
findMaximumPeakSum(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void findMaximumPeakSum( int arr[], int n)
{
int []dp1 = new int [n];
int []dp2 = new int [n];
for ( int i = 0 ; i < n; i++) {
dp1[i] = - 1 ;
dp2[i] = - 1 ;
}
dp2[ 0 ] = 0 ;
dp1[ 0 ] = arr[ 0 ];
for ( int i = 1 ; i < n; i++) {
dp1[i] = Math.max(dp1[i - 1 ],
dp2[i - 1 ] + arr[i]);
dp2[i] = Math.max(dp2[i - 1 ],
dp1[i - 1 ] - arr[i]);
}
int ans = Math.max(dp1[n - 1 ], dp2[n - 1 ]);
System.out.print(ans);
}
public static void main(String[] args)
{
int arr[] = { 3 , 2 , 1 , 4 , 5 , 2 , 1 , 7 , 8 , 9 };
int N = arr.length;
findMaximumPeakSum(arr, N);
}
}
|
Python3
def findMaximumPeakSum(arr, n):
dp1 = [ 0 ] * n
dp2 = [ 0 ] * n
for i in range (n):
dp1[i] = - 1
dp2[i] = - 1
dp2[ 0 ] = 0
dp1[ 0 ] = arr[ 0 ]
for i in range ( 1 , n):
dp1[i] = max (dp1[i - 1 ],
dp2[i - 1 ] + arr[i])
dp2[i] = max (dp2[i - 1 ],
dp1[i - 1 ] - arr[i])
ans = max (dp1[n - 1 ], dp2[n - 1 ])
print (ans)
arr = [ 3 , 2 , 1 , 4 , 5 , 2 , 1 , 7 , 8 , 9 ]
N = len (arr)
findMaximumPeakSum(arr, N)
|
C#
using System;
public class GFG{
static void findMaximumPeakSum( int []arr, int n)
{
int []dp1 = new int [n];
int []dp2 = new int [n];
for ( int i = 0; i < n; i++) {
dp1[i] = -1;
dp2[i] = -1;
}
dp2[0] = 0;
dp1[0] = arr[0];
for ( int i = 1; i < n; i++) {
dp1[i] = Math.Max(dp1[i - 1],
dp2[i - 1] + arr[i]);
dp2[i] = Math.Max(dp2[i - 1],
dp1[i - 1] - arr[i]);
}
int ans = Math.Max(dp1[n - 1], dp2[n - 1]);
Console.Write(ans);
}
public static void Main(String[] args)
{
int []arr = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
int N = arr.Length;
findMaximumPeakSum(arr, N);
}
}
|
Javascript
<script>
function findMaximumPeakSum(arr, n) {
let dp1 = new Array(n);
let dp2 = new Array(n);
for (let i = 0; i < n; i++) {
dp1[i] = -1;
dp2[i] = -1;
}
dp2[0] = 0;
dp1[0] = arr[0];
for (let i = 1; i < n; i++) {
dp1[i] = Math.max(dp1[i - 1],
dp2[i - 1] + arr[i]);
dp2[i] = Math.max(dp2[i - 1],
dp1[i - 1] - arr[i]);
}
let ans = Math.max(dp1[n - 1], dp2[n - 1]);
document.write(ans);
}
let arr = [3, 2, 1, 4, 5, 2, 1, 7, 8, 9];
let N = arr.length;
findMaximumPeakSum(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Optimized Approach: The above approach can be further be optimized by using two variables, odd and even, instead of two arrays, dp1[] and dp2[] to maintain the maximum difference between the sum of elements at even and odd indices. For every index i, only the maximum sums of even and odd length subsequences of the previous index are required to calculate the current states are needed.
Below is the implementation of the above approach:
C++14
#include <bits/stdc++.h>
using namespace std;
void findMaximumPeakSum( int arr[], int n)
{
int even = 0;
int odd = arr[0];
for ( int i = 1; i < n; i++) {
int temp = odd;
odd = max(odd, even + arr[i]);
even = max(even, temp - arr[i]);
}
int ans = max(odd, even);
cout << ans;
}
int main()
{
int arr[] = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
int N = sizeof (arr) / sizeof (arr[0]);
findMaximumPeakSum(arr, N);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void findMaximumPeakSum( int arr[], int n)
{
int even = 0 ;
int odd = arr[ 0 ];
for ( int i = 1 ; i < n; i++) {
int temp = odd;
odd = Math.max(odd, even + arr[i]);
even = Math.max(even, temp - arr[i]);
}
int ans = Math.max(odd, even);
System.out.print(ans);
}
public static void main(String[] args) {
int arr[] = { 3 , 2 , 1 , 4 , 5 , 2 , 1 , 7 , 8 , 9 };
int N = arr.length;
findMaximumPeakSum(arr, N);
}
}
|
Python3
def findMaximumPeakSum(arr, n):
even = 0
odd = arr[ 0 ]
for i in range ( 1 , n):
temp = odd
odd = max (odd, even + arr[i])
even = max (even, temp - arr[i])
ans = max (odd, even)
print (ans)
if __name__ = = "__main__" :
arr = [ 3 , 2 , 1 , 4 , 5 , 2 , 1 , 7 , 8 , 9 ]
N = len (arr)
findMaximumPeakSum(arr, N)
|
C#
using System;
class GFG {
static void findMaximumPeakSum( int [] arr, int n)
{
int even = 0;
int odd = arr[0];
for ( int i = 1; i < n; i++) {
int temp = odd;
odd = Math.Max(odd, even + arr[i]);
even = Math.Max(even, temp - arr[i]);
}
int ans = Math.Max(odd, even);
Console.Write(ans);
}
public static void Main () {
int [] arr = { 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 };
int N = arr.Length;
findMaximumPeakSum(arr, N);
}
}
|
Javascript
<script>
function findMaximumPeakSum(arr , n) {
var even = 0;
var odd = arr[0];
for (i = 1; i < n; i++) {
var temp = odd;
odd = Math.max(odd, even + arr[i]);
even = Math.max(even, temp - arr[i]);
}
var ans = Math.max(odd, even);
document.write(ans);
}
var arr = [ 3, 2, 1, 4, 5, 2, 1, 7, 8, 9 ];
var N = arr.length;
findMaximumPeakSum(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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