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# Minimize sum of max sized Subsequence such that no two elements are adjacent

Given an array arr[] of positive numbers, the task is to find the minimum sum of a subsequence of maximum possible elements with the constraint that no two numbers in the sequence are adjacent.

Examples:

Input: arr[]={1, 7, 3, 2}
Output: 3
Explanation: Pick the subsequence {1, 2}.
The sum is 3 and no two elements are adjacent.
This is the least possible sum.

Input: arr[]={1, 8, 4, 10, 6}
Output: 11
Explanation: Pick the subsequence {1, 4, 6}.
The sum is 11 and no two elements are adjacent.
This is the least possible sum.

Input: arr[]={2, 0, 11, 2, 0, 2}
Output: 4
Explanation: Pick the subsequence {0, 2, 2}.
The sum is 4 and no two elements are adjacent.
This is the least possible sum.

Approach: To solve the problem follow the below observations:

There are two different cases:

• Array size is Odd – the only choice is to pick the even indexed elements(Zero – Based Indexing).
• Array size is Even – The subsequence size will be N/2. It can be in three possible ways
• All the even indexed elements.
• All the odd indexed elements.
• Even indexed elements till i and odd indexed elements starting from i+3
• The minimum among these 3 will be the answer.

Follow the steps to solve the problem:

• If the size of arr[] is Odd, the sum of all the even indexed elements is the answer.
• For Size of arr to Even, store the sum of all the even indices into a variable.
• Iterating from the end by adding the odd index elements and removing the even indexed elements from the end.
• Update the minimum answer in each index.
• After the iteration is over, return the minimum value as the answer.

Below is the implementation of the above approach.

## C++

```// C++ code for the above approach:

#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum sum
int FindMinSum(vector<int>& vec, int n)
{
int sum = 0;
for (int i = 0; i < n; i += 2) {

// Sum of all the even indices
sum += vec[i];
}
if (n % 2 != 0)
return sum;
int ans = sum;
for (int i = n - 1; i > 0; i -= 2) {

// Removing the even index and
sum = sum - vec[i - 1] + vec[i];

// Checking for the minimum sum
ans = min(sum, ans);
}

// Return the minimum sum
return ans;
}

// Driver Code
int main()
{

// Creating the array
vector<int> arr = { 2, 0, 11, 2, 0, 2 };
int N = vec.size();

// Function call
cout << FindMinSum(arr, N) << endl;

return 0;
}```

## C

```#include <stdio.h>

int FindMinSum(int arr[], int n)
{
int sum = 0;
for (int i = 0; i < n; i += 2) {
// sum of all the odd indexes
sum += arr[i];
}
if (n % 2 == 1)
return sum;
int ans = sum;
for (int i = n - 1; i > 0; i -= 2) {
// removing the odd index and addinig the even index
sum = sum - arr[i - 1] + arr[i];
// checking for the minimum sum
if (sum < ans)
ans = sum;
}

// Return the minimum sum
return ans;
}
// Driver Code
int main()
{
// Creating the array
int arr[] = { 2, 0, 11, 2, 0, 2 };
int n = *(&arr + 1) - arr;

// Function call

return 0;
}```

## Java

```import java.io.*;
import java.util.*;

public class GFG {

static int solve(int vec[], int n)
{
int sum = 0;
if (n % 2 != 0) {
for (int i = 0; i < n; i = i + 2) {
// sum of all the odd indexes
sum = sum + vec[i];
}
return sum;
}
else {
int ans = sum;
for (int i = n - 1; i > 0; i -= 2) {
// removing the odd index and addinig the
// even index
ans = ans - vec[i - 1] + vec[i];
// checking for the minimum sum
ans = Math.min(ans, sum);
}
// Return the minimum sum
return ans;
}
}
public static void main(String args[])
throws IOException
{
// Creating the array
int vec[] = { 2, 1, 2 };

int n = vec.length;

// Function call
System.out.println(solve(vec, n));
}
}```

## Python

```# Python code to implement the approach
def FindMinSum(list, n):
sum = 0
for i in range(0, n, 2):
# sum of all the odd indexes
sum = sum + list[i]
if n % 2 == 1:
return sum
ans = sum
for i in range(n-1, 0, -2):
# removing the odd index and addinig the even index
sum = sum - list[i-1] + list[i]
# checking for the minimum sum
ans = min(sum, ans)
# Return the minimum sum
return ans

# Driver Code
# Creating the list
list = [2, 0, 11, 2, 0, 2]
n = 6
# Function call
k = FindMinSum(list, n)
print(k)
```

## C#

```// C# code for the above approach:
using System;

public class GFG {
static int solve(int[] vec, int n)
{
int sum = 0;
if (n % 2 != 0) {
for (int i = 0; i < n; i = i + 2)
{

// sum of all the odd indexes
sum = sum + vec[i];
}
return sum;
}
else {
int ans = sum;
for (int i = n - 1; i > 0; i -= 2)
{

// removing the odd index and addinig the
// even index
ans = ans - vec[i - 1] + vec[i];

// checking for the minimum sum
ans = Math.Min(ans, sum);
}

// Return the minimum sum
return ans;
}
}

// Driver Code
static public void Main()
{
// Creating the array
int[] vec = { 2, 1, 2 };

int n = vec.Length;

// Function call
Console.WriteLine(solve(vec, n));
}
}

// This code is contributed by Rohit Pradhan```

## Javascript

```    <script>
// JavaScript code for the above approach:

// Function to find the minimum sum
const FindMinSum = (vec, n) => {
let sum = 0;
for (let i = 0; i < n; i += 2) {

// Sum of all the even indices
sum += vec[i];
}
if (n % 2 != 0)
return sum;
let ans = sum;
for (let i = n - 1; i > 0; i -= 2) {

// Removing the even index and
sum = sum - vec[i - 1] + vec[i];

// Checking for the minimum sum
ans = Math.min(sum, ans);
}

// Return the minimum sum
return ans;
}

// Driver Code

// Creating the array
let arr = [2, 0, 11, 2, 0, 2];
let N = arr.length;

// Function call
document.write(FindMinSum(arr, N));

// This code is contributed by rakeshsahni

</script>```
Output

`4`

Time Complexity: O(N)
Auxiliary Space: O(1)

My Personal Notes arrow_drop_up
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