# Maximum sum such that no two elements are adjacent | Set 2

Given an array of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array. So 3 2 7 10 should return 13 (sum of 3 and 10) or 3 2 5 10 7 should return 15 (sum of 3, 5 and 7).

Examples:

```Input :  arr[] = {3, 5, 3}
Output : 6
Explanation :
Selecting indexes 0 and 2 will maximise the sum
i.e 3+3 = 6

Input : arr[] = {2, 5, 2}
Output : 5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have already discussed the efficient approach of solving this problem in the previous article.

However, we can also solve this problem using Dynamic Programming approach.

Dynamic Programming Approach: Let’s decide the states of ‘dp’. Let dp[i] be the largest possible sum for the sub-array staring from index ‘i’ and ending at index ‘N-1’. Now, we have to find a recurrence relation between this state and a lower-order state.

In this case for an index ‘i’, we will have two choices.

```1) Choose the current index:
In this case, the relation will be dp[i] = arr[i] + dp[i+2]
2) Skip the current index:
Relation will be dp[i] = dp[i+1]
```

We will choose the path that maximizes our result.
Thus final relation will be:

```dp[i] = max(dp[i+2]+arr[i], dp[i+1])
```

Below is the implementation of the above approach:

 `// C++ program to implement above approach ` ` `  `#include ` `#define maxLen 10 ` `using` `namespace` `std; ` ` `  `// variable to store states of dp ` `int` `dp[maxLen]; ` ` `  `// variable to check if a given state ` `// has been solved ` `bool` `v[maxLen]; ` ` `  `// Function to find the maximum sum subsequence ` `// such that no two elements are adjacent ` `int` `maxSum(``int` `arr[], ``int` `i, ``int` `n) ` `{ ` `    ``// Base case ` `    ``if` `(i >= n) ` `        ``return` `0; ` ` `  `    ``// To check if a state has ` `    ``// been solved ` `    ``if` `(v[i]) ` `        ``return` `dp[i]; ` `    ``v[i] = 1; ` ` `  `    ``// Required recurrence relation ` `    ``dp[i] = max(maxSum(arr, i + 1, n), ` `                ``arr[i] + maxSum(arr, i + 2, n)); ` ` `  `    ``// Returning the value ` `    ``return` `dp[i]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 12, 9, 7, 33 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << maxSum(arr, 0, n); ` ` `  `    ``return` `0; ` `} `

 `// Java program to implement above approach  ` `class` `GFG ` `{ ` ` `  `static` `int` `maxLen = ``10``; ` ` `  `// variable to store states of dp  ` `static` `int` `dp[] = ``new` `int``[maxLen];  ` ` `  `// variable to check if a given state  ` `// has been solved  ` `static` `boolean` `v[] = ``new` `boolean``[maxLen];  ` ` `  `// Function to find the maximum sum subsequence  ` `// such that no two elements are adjacent  ` `static` `int` `maxSum(``int` `arr[], ``int` `i, ``int` `n)  ` `{  ` `    ``// Base case  ` `    ``if` `(i >= n)  ` `        ``return` `0``;  ` ` `  `    ``// To check if a state has  ` `    ``// been solved  ` `    ``if` `(v[i])  ` `        ``return` `dp[i];  ` `    ``v[i] = ``true``;  ` ` `  `    ``// Required recurrence relation  ` `    ``dp[i] = Math.max(maxSum(arr, i + ``1``, n),  ` `                ``arr[i] + maxSum(arr, i + ``2``, n));  ` ` `  `    ``// Returning the value  ` `    ``return` `dp[i];  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``int` `arr[] = { ``12``, ``9``, ``7``, ``33` `};  ` `    ``int` `n = arr.length;  ` `    ``System.out.println( maxSum(arr, ``0``, n));  ` `} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

 `# Python 3 program to implement above approach ` `maxLen ``=` `10` ` `  `# variable to store states of dp ` `dp ``=` `[``0` `for` `i ``in` `range``(maxLen)] ` ` `  `# variable to check if a given state  ` `# has been solved ` `v ``=` `[``0` `for` `i ``in` `range``(maxLen)] ` ` `  `# Function to find the maximum sum subsequence ` `# such that no two elements are adjacent ` `def` `maxSum(arr, i, n): ` `    ``# Base case ` `    ``if` `(i >``=` `n): ` `        ``return` `0` ` `  `    ``# To check if a state has ` `    ``# been solved ` `    ``if` `(v[i]): ` `        ``return` `dp[i] ` `    ``v[i] ``=` `1` ` `  `    ``# Required recurrence relation ` `    ``dp[i] ``=` `max``(maxSum(arr, i ``+` `1``, n), ` `            ``arr[i] ``+` `maxSum(arr, i ``+` `2``, n)) ` ` `  `    ``# Returning the value ` `    ``return` `dp[i] ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``12``, ``9``, ``7``, ``33``] ` ` `  `    ``n ``=` `len``(arr) ` `    ``print``(maxSum(arr, ``0``, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

 `// C# program to implement above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `maxLen = 10; ` ` `  `// variable to store states of dp  ` `static` `int``[] dp = ``new` `int``[maxLen];  ` ` `  `// variable to check if a given state  ` `// has been solved  ` `static` `bool``[] v = ``new` `bool``[maxLen];  ` ` `  `// Function to find the maximum sum subsequence  ` `// such that no two elements are adjacent  ` `static` `int` `maxSum(``int``[] arr, ``int` `i, ``int` `n)  ` `{  ` `    ``// Base case  ` `    ``if` `(i >= n)  ` `        ``return` `0;  ` ` `  `    ``// To check if a state has  ` `    ``// been solved  ` `    ``if` `(v[i])  ` `        ``return` `dp[i];  ` `    ``v[i] = ``true``;  ` ` `  `    ``// Required recurrence relation  ` `    ``dp[i] = Math.Max(maxSum(arr, i + 1, n),  ` `                ``arr[i] + maxSum(arr, i + 2, n));  ` ` `  `    ``// Returning the value  ` `    ``return` `dp[i];  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main() ` `{  ` `    ``int``[] arr = { 12, 9, 7, 33 };  ` `    ``int` `n = arr.Length;  ` `    ``Console.Write( maxSum(arr, 0, n));  ` `} ` `} ` ` `  `// This code is contributed by ChitraNayal `

 `= ``\$n``)  ` `        ``return` `0;  ` ` `  `    ``// To check if a state has  ` `    ``// been solved  ` `    ``if` `(``\$GLOBALS``[``'v'``][``\$i``])  ` `        ``return` `\$GLOBALS``[``'dp'``][``\$i``];  ` `         `  `    ``\$GLOBALS``[``'v'``][``\$i``] = 1;  ` ` `  `    ``// Required recurrence relation  ` `    ``\$GLOBALS``[``'dp'``][``\$i``] = max(maxSum(``\$arr``, ``\$i` `+ 1, ``\$n``),  ` `                ``\$arr``[``\$i``] + maxSum(``\$arr``, ``\$i` `+ 2, ``\$n``));  ` ` `  `    ``// Returning the value  ` `    ``return` `\$GLOBALS``[``'dp'``][``\$i``];  ` `}  ` ` `  `    ``// Driver code  ` `    ``\$arr` `= ``array``( 12, 9, 7, 33 );  ` ` `  `    ``\$n` `= ``count``(``\$arr``);  ` ` `  `    ``echo` `maxSum(``\$arr``, 0, ``\$n``);  ` ` `  `    ``// This code is contributed by AnkitRai01 ` `?> `

Output:
```45
```

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