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 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 starting 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 10using namespace std;// variable to store states of dpint dp[maxLen];// variable to check if a given state// has been solvedbool v[maxLen];// Function to find the maximum sum subsequence// such that no two elements are adjacentint 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 codeint 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 approachclass GFG{static int maxLen = 10;// variable to store states of dpstatic int dp[] = new int[maxLen];// variable to check if a given state// has been solvedstatic boolean v[] = new boolean[maxLen];// Function to find the maximum sum subsequence// such that no two elements are adjacentstatic 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 codepublic 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 approachmaxLen = 10# variable to store states of dpdp = [0 for i in range(maxLen)]# variable to check if a given state# has been solvedv = [0 for i in range(maxLen)]# Function to find the maximum sum subsequence# such that no two elements are adjacentdef 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 codeif __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 approachusing System;class GFG{static int maxLen = 10;// variable to store states of dpstatic int[] dp = new int[maxLen];// variable to check if a given state// has been solvedstatic bool[] v = new bool[maxLen];// Function to find the maximum sum subsequence// such that no two elements are adjacentstatic 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 codepublic 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, $GLOBALS['maxLen'], 0);// Function to find the maximum sum subsequence// such that no two elements are adjacentfunction maxSum($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 approachvar maxLen = 10;// variable to store states of dpvar dp = Array(maxLen);// variable to check if a given state// has been solvedvar v = Array(maxLen);// Function to find the maximum sum subsequence// such that no two elements are adjacentfunction 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 codevar arr = [12, 9, 7, 33 ];var n = arr.length;document.write( maxSum(arr, 0, n));</script> |
Output:
45
Time Complexity : O(n)
Auxiliary Space : O(10)
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