# Maximum sum in circular array such that no two elements are adjacent

• Difficulty Level : Medium
• Last Updated : 24 Jan, 2023

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

Naive Approach:

This problem is extension to the problem https://www.geeksforgeeks.org/find-maximum-possible-stolen-value-houses/ .In this problem the arr elements are arranged in circular fashion so the 1st and the last element are also adjacent. So we can not take first and last element together so we can take either of the element1 or last element.

So first we can take element 1 and exclude last element  and calculate the maximum sum subsequence where no two selected elements are adjacent and then also we can calculate another answer by excluding 1st element and including last element. And we will take maximum of two.

Implementation of recursion approach:

## C++

 // CPP program to find maximum sum in a circular array// such that no elements are adjacent in the sum.#includeusing namespace std; // calculate the maximum stolen valueint helper(int* arr, int n){    // base case    if (n < 0) {        return 0;    }     if (n == 0) {        return arr[0];    }     // if current element is pick then previous cannot be    // picked    int pick = arr[n] + helper(arr, n - 2);    // if current element is not picked then previous    // element is picked    int notPick = helper(arr, n - 1);     // return max of picked and not picked    return max(pick, notPick);} int maxLoot(int* arr, int n){    // case 1: Last element is exluded we have to calculate    // result for first n-1 houses.    int ans1 = helper(arr, n - 1);    // case 2: First element is excluded we have to    // calculate result 2 to n houses.    int ans2 = helper(arr + 1, n - 1);    return max(ans1, ans2);}// Driver to test above codeint main(){    int arr[] = { 1, 2, 3, 1 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << "Maximum loot possible : "         << maxLoot(arr, n - 1);    return 0;}// This code is contributed by Sanskar Bharadia

## 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 { // calculate the maximum stolen value    static int helper(int arr[],int l, int n){    // base case    if (n < 0) {        return 0;    }      if (n == 0) {        return arr[0];    }      // if current element is pick then previous cannot be    // picked    int pick = arr[n] + helper(arr,l, n - 2);    // if current element is not picked then previous    // element is picked    int notPick = helper(arr,l, n - 1);      // return max of picked and not picked    return Math.max(pick, notPick);}static int findMaxSum(int arr[], int n){    // case 1: Last element is exluded we have to calculate    // result for first n-1 houses.    int ans1 = helper(arr,0, n - 1);    // case 2: First element is excluded we have to    // calculate result 2 to n houses.    int ans2 = helper(arr,1, n - 1);    return Math.max(ans1, ans2);}// Driver Code    public static void main (String[] args) {                  int arr[] = { 1, 2, 3, 1 };        int n = arr.length;        System.out.println("Maximum loot possible : " + findMaxSum(arr, n));    }}//this article is contributed by aditya942003patil

## Python3

 # Python3 program to find maximum sum in a circular array# such that no elements are adjacent in the sum. # calculate the maximum stolen valuedef helper(arr, n):    # base case    if n < 0:        return 0    if n == 0:        return arr[0]           # if current element is pick then previous cannot be    # picked    pick = arr[n] + helper(arr, n - 2)         # if current element is not picked then previous    # element is picked    notPick = helper(arr, n - 1)     # return max of picked and not picked    return max(pick, notPick) def maxLoot(arr, n):    # case 1: Last element is exluded we have to calculate    # result for first n-1 houses.    ans1 = helper(arr, n - 1)         # case 2: First element is excluded we have to    # calculate result 2 to n houses.    ans2 = helper(arr[1:], n - 1)    return max(ans1, ans2) # Driver to test above codeif __name__ == "__main__":    arr = [1, 2, 3, 1]    n = len(arr)    print("Maximum loot possible : ", maxLoot(arr, n - 1))     # This code is contributed by lokeshpotta20

## C#

 // Include namespace systemusing System; public class GFG{  // calculate the maximum stolen value  public static int helper(int[] arr, int l, int n)  {    // base case    if (n < 0)    {      return 0;    }    if (n == 0)    {      return arr[0];    }     // if current element is pick then previous cannot be    // picked    var pick = arr[n] + GFG.helper(arr, l, n - 2);     // if current element is not picked then previous    // element is picked    var notPick = GFG.helper(arr, l, n - 1);     // return max of picked and not picked    return Math.Max(pick,notPick);  }  public static int findMaxSum(int[] arr, int n)  {     // case 1: Last element is exluded we have to calculate    // result for first n-1 houses.    var ans1 = GFG.helper(arr, 0, n - 1);     // case 2: First element is excluded we have to    // calculate result 2 to n houses.    var ans2 = GFG.helper(arr, 1, n - 1);    return Math.Max(ans1,ans2);  }   // Driver Code  public static void Main(String[] args)  {    int[] arr = {1, 2, 3, 1};    var n = arr.Length;    Console.WriteLine("Maximum loot possible : " + GFG.findMaxSum(arr, n).ToString());  }} // This code is contributed by aadityaburujwale.

## Javascript

 // calculate the maximum stolen valuefunction helper(arr, l, n){    // base case    if (n < 0)    {        return 0;    }    if (n == 0)    {        return arr[0];    }         // if current element is pick then previous cannot be    // picked    var pick = arr[n] + helper(arr, l, n - 2);         // if current element is not picked then previous    // element is picked    var notPick = helper(arr, l, n - 1);         // return max of picked and not picked    return Math.max(pick,notPick);}function findMaxSum(arr, n){    // case 1: Last element is exluded we have to calculate    // result for first n-1 houses.    var ans1 = helper(arr, 0, n - 1);         // case 2: First element is excluded we have to    // calculate result 2 to n houses.    var ans2 = helper(arr, 1, n - 1);    return Math.max(ans1,ans2);} // Driver Code    var arr = [1, 2, 3, 1];    var n = arr.length;    console.log("Maximum loot possible : " + findMaxSum(arr, n));         // This code is contributed by sourabhdalal0001.

Output

Maximum loot possible : 4

Complexity Analysis:

• Time Complexity:  O(2n).
• Auxiliary Space: O(N)

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 using namespace std; // Function to calculate the sum// from 0th position to(n-2)th positionint 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 positionint 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++) {         // bottom-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 Codeint 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 positionstatic 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 positionstatic 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++) {         // bottom-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 positiondef 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 positiondef 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):         # bottom-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 Codeif __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 positionstatic 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 positionstatic 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++)    {         // bottom-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 Codestatic 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



## Javascript



Output

4

Time Complexity: O(N^2)

Space optimization approach: As we can see the dp[i] is calculated form previous values so we can just store those previous values in variables.

Implementation:

## C++14

 // C++ program to find the maximum stolen value#include using namespace std; // calculate the maximum stolen valueint findMaxSum(int arr[], int n){    if (n == 0)        return 0;     int value1 = arr[0];    if (n == 1)        return value1;     int value2 = 0;     // case 1: check for 1 to n-1 houses.    // contains maximum stolen value at the end    int max_val1, max_val2;     // Fill remaining positions    for (int i = 1; i < n - 1; i++) {        int first = value1;        int second = arr[i];        if (i > 1) {            second += value2;        }        int curr = max(first, second);         value2 = value1;        value1 = curr;    }    //  case 2: check for 2 to n houses.    max_val1 = value1;    value1 = arr[1];    value2 = 0;    for (int i = 2; i < n; i++) {        int first = value1;        int second = arr[i];        if (i > 1) {            second += value2;        }        int curr = max(first, second);         value2 = value1;        value1 = curr;    }    max_val2 = value1;    return max(max_val1, max_val2);} // Driver to test above codeint main(){    // Value of houses    int arr[] = { 6, 7, 1, 3, 8, 2, 4 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << findMaxSum(arr, n);    return 0;}

## Java

 // Java program to find the maximum stolen value  public class GFG {  // calculate the maximum stolen valuestatic int findMaxSum(int[] arr, int n){    if (n == 0)        return 0;      int value1 = arr[0];    if (n == 1)        return value1;      int value2 = 0;      // case 1: check for 1 to n-1 houses.    // contains maximum stolen value at the end    int max_val1, max_val2;      // Fill remaining positions    for (int i = 1; i < n - 1; i++) {        int first = value1;        int second = arr[i];        if (i > 1) {            second += value2;        }        int curr = Math.max(first, second);          value2 = value1;        value1 = curr;    }    //  case 2: check for 2 to n houses.    max_val1 = value1;    value1 = arr[1];    value2 = 0;    for (int i = 2; i < n; i++) {        int first = value1;        int second = arr[i];        if (i > 1) {            second += value2;        }        int curr = Math.max(first, second);          value2 = value1;        value1 = curr;    }    max_val2 = value1;    return Math.max(max_val1, max_val2);}      // Driver Code    public static void main (String[] args)    {        // Value of houses        int arr[] = { 6, 7, 1, 3, 8, 2, 4 };        int n = arr.length;        System.out.println(findMaxSum(arr, n));    }}// This code is contributed by aditya942003patil

## Python3

 # C++ program to find the maximum stolen value # calculate the maximum stolen valuedef findMaxSum(arr, n):     if n == 0 :        return 0     value1 = arr[0]    if n == 1 :        return value1     value2 = 0     # case 1: check for 1 to n-1 houses.    # contains maximum stolen value at the end    max_val1, max_val2 = 0,0     # Fill remaining positions    for i in range(1,n) :        first = value1        second = arr[i]        if i > 1 :            second += value2        curr = max(first, second)         value2 = value1        value1 = curr             #  case 2: check for 2 to n houses.    max_val1 = value1    value1 = arr[1]    value2 = 0    for i in range(2,n) :        first = value1        second = arr[i]        if i > 1 :            second += value2        curr = max(first, second)         value2 = value1        value1 = curr             max_val2 = value1    return max(max_val1, max_val2) # Driver to test above codeif __name__ == "__main__":         # Value of houses    arr = [ 6, 7, 1, 3, 8, 2, 4 ]    n = len(arr)    print(findMaxSum(arr, n)) # This code is contributed by aditya942003patil

## C#

 // C# program to find the maximum stolen value using System;using System.Collections;using System.Collections.Generic; public class GFG {   // calculate the maximum stolen value  static int findMaxSum(int[] arr, int n)  {    if (n == 0)      return 0;     int value1 = arr[0];    if (n == 1)      return value1;     int value2 = 0;     // case 1: check for 1 to n-1 houses.    // contains maximum stolen value at the end    int max_val1, max_val2;     // Fill remaining positions    for (int i = 1; i < n - 1; i++) {      int first = value1;      int second = arr[i];      if (i > 1) {        second += value2;      }      int curr = Math.Max(first, second);       value2 = value1;      value1 = curr;    }    //  case 2: check for 2 to n houses.    max_val1 = value1;    value1 = arr[1];    value2 = 0;    for (int i = 2; i < n; i++) {      int first = value1;      int second = arr[i];      if (i > 1) {        second += value2;      }      int curr = Math.Max(first, second);       value2 = value1;      value1 = curr;    }    max_val2 = value1;    return Math.Max(max_val1, max_val2);  }   // Driver Code  public static void Main(string[] args)  {    // Value of houses    int[] arr = { 6, 7, 1, 3, 8, 2, 4 };    int n = arr.Length;    Console.WriteLine(findMaxSum(arr, n));  }}// This code is contributed by karandeep1234.

## Javascript

 // JS program to find the maximum stolen value  // calculate the maximum stolen valuefunction findMaxSum(arr, n) {    if (n == 0)        return 0;     let value1 = arr[0];    if (n == 1)        return value1;     let value2 = 0;     // case 1: check for 1 to n-1 houses.    // contains maximum stolen value at the end    let max_val1, max_val2;     // Fill remaining positions    for (let i = 1; i < n - 1; i++) {        let first = value1;        let second = arr[i];        if (i > 1) {            second += value2;        }        let curr = Math.max(first, second);         value2 = value1;        value1 = curr;    }    // case 2: check for 2 to n houses.    max_val1 = value1;    value1 = arr[1];    value2 = 0;    for (let i = 2; i < n; i++) {        let first = value1;        let second = arr[i];        if (i > 1) {            second += value2;        }        let curr = Math.max(first, second);         value2 = value1;        value1 = curr;    }    max_val2 = value1;    return Math.max(max_val1, max_val2);} // Driver to test above code  // Value of houseslet arr = [6, 7, 1, 3, 8, 2, 4];let n = arr.length;console.log(findMaxSum(arr, n));return 0; // This code is contributed by adityamaharshi21

Output

19

Complexity Analysis:

• Time Complexity:  O(N).
• Space Complexity: O(1)

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