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

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

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; 
} 

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));  
          
    } 
} 

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 

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 

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 
?> 

Output:

Time Complexity: O(N^2)

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

C++

Java

Python 3

C#

PHP

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