Maximum sum in circular array such that no two elements are adjacent
Given a circular array containing of positive integers value. The task is to find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array.
Examples:
Input: circular arr = {1, 2, 3, 1}
Output : 4
subsequence will be(1, 3), hence 1 + 3 = 4
Input: circular arr = {1, 2, 3, 4, 5, 1}
Output: 9
subsequence will be(1, 3, 5), hence 1 + 3 + 5 = 9
Approach The problem can be solved using DP. An approach has already been discussed in this post, but it for an array. We can treat the circular subarray a two arrays one from (0th to n-2-th) and (1st to n-1-th) index, and use the approach used in the previous post. The maximum sum returned by both will be the answer.
Below is the implementation of the above approach.
C++
// CPP program to find maximum sum in a circular array // such that no elements are adjacent in the sum. #include <bits/stdc++.h> using namespace std; // Function to calculate the sum // from 0th position to(n-2)th position int maxSum1(int arr[], int n) { int dp[n]; int maxi = 0; for (int i = 0; i < n - 1; i++) { // copy the element of original array to dp[] dp[i] = arr[i]; // find the maximum element in the array if (maxi < arr[i]) maxi = arr[i]; } // start from 2nd to n-1th pos for (int i = 2; i < n - 1; i++) { // traverse for all pairs // bottom-up approach for (int j = 0; j < i - 1; j++) { // dp-condition if (dp[i] < dp[j] + arr[i]) { dp[i] = dp[j] + arr[i]; // find maximum sum if (maxi < dp[i]) maxi = dp[i]; } } } // return the maximum return maxi; } // Function to find the maximum sum // from 1st position to n-1-th position int maxSum2(int arr[], int n) { int dp[n]; int maxi = 0; for (int i = 1; i < n; i++) { dp[i] = arr[i]; if (maxi < arr[i]) maxi = arr[i]; } // Traverse from third to n-th pos for (int i = 3; i < n; i++) { // bootom-up approach for (int j = 1; j < i - 1; j++) { // dp condition if (dp[i] < arr[i] + dp[j]) { dp[i] = arr[i] + dp[j]; // find max sum if (maxi < dp[i]) maxi = dp[i]; } } } // return max return maxi; } int findMaxSum(int arr[], int n) { return max(maxSum1(arr, n), maxSum2(arr, n)); } // Driver Code int main() { int arr[] = { 1, 2, 3, 1 }; int n = sizeof(arr)/sizeof(arr[0]); cout << findMaxSum(arr, n); return 0; } |
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Java
// Java program to find maximum sum in a circular array // such that no elements are adjacent in the sum. import java.io.*; class GFG { // Function to calculate the sum // from 0th position to(n-2)th position static int maxSum1(int arr[], int n) { int dp[]=new int[n]; int maxi = 0; for (int i = 0; i < n - 1; i++) { // copy the element of original array to dp[] dp[i] = arr[i]; // find the maximum element in the array if (maxi < arr[i]) maxi = arr[i]; } // start from 2nd to n-1th pos for (int i = 2; i < n - 1; i++) { // traverse for all pairs // bottom-up approach for (int j = 0; j < i - 1; j++) { // dp-condition if (dp[i] < dp[j] + arr[i]) { dp[i] = dp[j] + arr[i]; // find maximum sum if (maxi < dp[i]) maxi = dp[i]; } } } // return the maximum return maxi; } // Function to find the maximum sum // from 1st position to n-1-th position static int maxSum2(int arr[], int n) { int dp[]=new int[n]; int maxi = 0; for (int i = 1; i < n; i++) { dp[i] = arr[i]; if (maxi < arr[i]) maxi = arr[i]; } // Traverse from third to n-th pos for (int i = 3; i < n; i++) { // bootom-up approach for (int j = 1; j < i - 1; j++) { // dp condition if (dp[i] < arr[i] + dp[j]) { dp[i] = arr[i] + dp[j]; // find max sum if (maxi < dp[i]) maxi = dp[i]; } } } // return max return maxi; } static int findMaxSum(int arr[], int n) { int t=Math.max(maxSum1(arr, n), maxSum2(arr, n)); return t; } // Driver Code public static void main (String[] args) { int arr[] = { 1, 2, 3, 1 }; int n = arr.length; System.out.println(findMaxSum(arr, n)); } } |
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Python 3
# Python 3 program to find maximum sum # in a circular array such that no # elements are adjacent in the sum. # Function to calculate the sum from # 0th position to(n-2)th position def maxSum1(arr, n): dp = [0] * n maxi = 0 for i in range(n - 1): # copy the element of original # array to dp[] dp[i] = arr[i] # find the maximum element in the array if (maxi < arr[i]): maxi = arr[i] # start from 2nd to n-1th pos for i in range(2, n - 1): # traverse for all pairs bottom-up # approach for j in range(i - 1) : # dp-condition if (dp[i] < dp[j] + arr[i]): dp[i] = dp[j] + arr[i] # find maximum sum if (maxi < dp[i]): maxi = dp[i] # return the maximum return maxi # Function to find the maximum sum # from 1st position to n-1-th position def maxSum2(arr, n): dp = [0] * n maxi = 0 for i in range(1, n): dp[i] = arr[i] if (maxi < arr[i]): maxi = arr[i] # Traverse from third to n-th pos for i in range(3, n): # bootom-up approach for j in range(1, i - 1) : # dp condition if (dp[i] < arr[i] + dp[j]): dp[i] = arr[i] + dp[j] # find max sum if (maxi < dp[i]): maxi = dp[i] # return max return maxi def findMaxSum(arr, n): return max(maxSum1(arr, n), maxSum2(arr, n)) # Driver Code if __name__ == "__main__": arr = [ 1, 2, 3, 1 ] n = len(arr) print(findMaxSum(arr, n)) # This code is contributed by ita_c |
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C#
// C# program to find maximum sum // in a circular array such that // no elements are adjacent in the sum. using System; class GFG { // Function to calculate the sum // from 0th position to(n-2)th position static int maxSum1(int []arr, int n) { int []dp = new int[n]; int maxi = 0; for (int i = 0; i < n - 1; i++) { // copy the element of original // array to dp[] dp[i] = arr[i]; // find the maximum element // in the array if (maxi < arr[i]) maxi = arr[i]; } // start from 2nd to n-1th pos for (int i = 2; i < n - 1; i++) { // traverse for all pairs // bottom-up approach for (int j = 0; j < i - 1; j++) { // dp-condition if (dp[i] < dp[j] + arr[i]) { dp[i] = dp[j] + arr[i]; // find maximum sum if (maxi < dp[i]) maxi = dp[i]; } } } // return the maximum return maxi; } // Function to find the maximum sum // from 1st position to n-1-th position static int maxSum2(int []arr, int n) { int []dp = new int[n]; int maxi = 0; for (int i = 1; i < n; i++) { dp[i] = arr[i]; if (maxi < arr[i]) maxi = arr[i]; } // Traverse from third to n-th pos for (int i = 3; i < n; i++) { // bootom-up approach for (int j = 1; j < i - 1; j++) { // dp condition if (dp[i] < arr[i] + dp[j]) { dp[i] = arr[i] + dp[j]; // find max sum if (maxi < dp[i]) maxi = dp[i]; } } } // return max return maxi; } static int findMaxSum(int []arr, int n) { int t = Math.Max(maxSum1(arr, n), maxSum2(arr, n)); return t; } // Driver Code static public void Main () { int []arr = { 1, 2, 3, 1 }; int n = arr.Length; Console.WriteLine(findMaxSum(arr, n)); } } // This code is contributed // by Sach_Code |
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PHP
<?php // PHP program to find maximum sum in // a circular array such that no // elements are adjacent in the sum. // Function to calculate the sum // from 0th position to(n-2)th position function maxSum1($arr, $n) { $dp[$n] = array(); $maxi = 0; for ($i = 0; $i < $n - 1; $i++) { // copy the element of original // array to dp[] $dp[$i] = $arr[$i]; // find the maximum element in the array if ($maxi < $arr[$i]) $maxi = $arr[$i]; } // start from 2nd to n-1th pos for ($i = 2; $i < $n - 1; $i++) { // traverse for all pairs // bottom-up approach for ( $j = 0; $j < $i - 1; $j++) { // dp-condition if ($dp[$i] < $dp[$j] + $arr[$i]) { $dp[$i] = $dp[$j] + $arr[$i]; // find maximum sum if ($maxi < $dp[$i]) $maxi = $dp[$i]; } } } // return the maximum return $maxi; } // Function to find the maximum sum // from 1st position to n-1-th position function maxSum2($arr, $n) { $dp[$n] = array(); $maxi = 0; for ($i = 1; $i < $n; $i++) { $dp[$i] = $arr[$i]; if ($maxi < $arr[$i]) $maxi = $arr[$i]; } // Traverse from third to n-th pos for ($i = 3; $i < $n; $i++) { // bootom-up approach for ($j = 1; $j < $i - 1; $j++) { // dp condition if ($dp[$i] < $arr[$i] + $dp[$j]) { $dp[$i] = $arr[$i] + $dp[$j]; // find max sum if ($maxi < $dp[$i]) $maxi = $dp[$i]; } } } // return max return $maxi; } function findMaxSum($arr, $n) { return max(maxSum1($arr, $n), maxSum2($arr, $n)); } // Driver Code $arr = array(1, 2, 3, 1 ); $n = sizeof($arr); echo findMaxSum($arr, $n); // This code is contributed // by Sach_Code ?> |
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Output:
4
Time Complexity: O(N^2)
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