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

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

However, we can also solve this problem using the *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, the final relation will be:

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

Below is the implementation of the above approach:

## C++

`// C++ program to implement above approach` `#include <bits/stdc++.h>` `#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

`// 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` |

## Python3

`# 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#

`// 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` |

## PHP

`<?php` `// PHP program to implement above approach` `$maxLen` `= 10;` `// variable to store states of dp` `$dp` `= ` `array_fill` `(0, ` `$GLOBALS` `[` `'maxLen'` `], 0);` `// variable to check if a given state` `// has been solved` `$v` `= ` `array_fill` `(0, ` `$GLOBALS` `[` `'maxLen'` `], 0);` `// Function to find the maximum sum subsequence` `// such that no two elements are adjacent` `function` `maxSum(` `$arr` `, ` `$i` `, ` `$n` `)` `{` ` ` `// Base case` ` ` `if` `(` `$i` `>= ` `$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` `?>` |

## Javascript

`<script>` `// Javascript program to implement above approach` `var` `maxLen = 10;` `// variable to store states of dp` `var` `dp = Array(maxLen);` `// variable to check if a given state` `// has been solved` `var` `v = Array(maxLen);` `// Function to find the maximum sum subsequence` `// such that no two elements are adjacent` `function` `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] = Math.max(maxSum(arr, i + 1, n),` ` ` `arr[i] + maxSum(arr, i + 2, n));` ` ` `// Returning the value` ` ` `return` `dp[i];` `}` `// Driver code` `var` `arr = [12, 9, 7, 33 ];` `var` `n = arr.length;` `document.write( maxSum(arr, 0, n));` `</script>` |

**Output:**

45

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