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
Naive Approach:
This problem is extension to the problem https://www.w3wiki.net/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. #include<iostream> using namespace std; // calculate the maximum stolen value int 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 excluded 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 code int 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 excluded 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 value def 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 excluded 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 code if __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 system using 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 excluded 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 value function 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 excluded 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. |
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 <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++) { // 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 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++) { // 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 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): # 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 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++) { // 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 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 ++) { // bottom-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 ?> |
Javascript
<script> // JavaScript 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) { let dp = new Array(n); let 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) { let dp = new Array(n); let 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++) { // bottom-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) { let t = Math.max(maxSum1(arr, n), maxSum2(arr, n)); return t; } // Driver Code let arr = [1, 2, 3, 1 ]; let n = arr.length; document.write(findMaxSum(arr, n)); // This code is contributed by mohit kumar 29. </script> |
4
Time Complexity: O(N^2)
Auxiliary Space: O(N)
Space optimization approach: As we can see the dp[i] is calculated from previous values so we can just store those previous values in variables.
Implementation:
C++14
// C++ program to find the maximum stolen value #include <iostream> using namespace std; // calculate the maximum stolen value 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 = 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 code int 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 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; 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 value def 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 code if __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 value function 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 houses let 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 |
19
Complexity Analysis:
- Time Complexity: O(N).
- Space Complexity: O(1).
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