Given n numbers (both +ve and -ve), arranged in a circle, find the maximum sum of consecutive numbers.
Examples:
Input: a[] = {8, -8, 9, -9, 10, -11, 12}
Output: 22 (12 + 8 - 8 + 9 - 9 + 10)
Input: a[] = {10, -3, -4, 7, 6, 5, -4, -1}
Output: 23 (7 + 6 + 5 - 4 -1 + 10)
Input: a[] = {-1, 40, -14, 7, 6, 5, -4, -1}
Output: 52 (7 + 6 + 5 - 4 - 1 - 1 + 40)
Method 1 There can be two cases for the maximum sum:
- Case 1: The elements that contribute to the maximum sum are arranged such that no wrapping is there. Examples: {-10, 2, -1, 5}, {-2, 4, -1, 4, -1}. In this case, Kadane’s algorithm will produce the result.
- Case 2: The elements which contribute to the maximum sum are arranged such that wrapping is there. Examples: {10, -12, 11}, {12, -5, 4, -8, 11}. In this case, we change wrapping to non-wrapping. Let us see how. Wrapping of contributing elements implies non-wrapping of non-contributing elements, so find out the sum of non-contributing elements and subtract this sum from the total sum. To find out the sum of non-contributions, invert the sign of each element and then run Kadane’s algorithm.
Our array is like a ring and we have to eliminate the maximum continuous negative that implies maximum continuous positive in the inverted arrays. Finally, we compare the sum obtained in both cases and return the maximum of the two sums.
The following are implementations of the above method.
C++
// C++ program for maximum contiguous circular sum problem #include <bits/stdc++.h> using namespace std; // Standard Kadane's algorithm to // find maximum subarray sum int kadane(int a[], int n); // The function returns maximum // circular contiguous sum in a[] int maxCircularSum(int a[], int n) { // Case 1: get the maximum sum using standard kadane' // s algorithm int max_kadane = kadane(a, n); // if maximum sum using standard kadane' is less than 0 if(max_kadane < 0) return max_kadane; // Case 2: Now find the maximum sum that includes // corner elements. int max_wrap = 0, i; for (i = 0; i < n; i++) { max_wrap += a[i]; // Calculate array-sum a[i] = -a[i]; // invert the array (change sign) } // max sum with corner elements will be: // array-sum - (-max subarray sum of inverted array) max_wrap = max_wrap + kadane(a, n); // The maximum circular sum will be maximum of two sums return (max_wrap > max_kadane) ? max_wrap : max_kadane; } // Standard Kadane's algorithm to find maximum subarray sum // See https:// www.neveropen.co.uk/archives/576 for details int kadane(int a[], int n) { int max_so_far = 0, max_ending_here = 0; int i; for (i = 0; i < n; i++) { max_ending_here = max_ending_here + a[i]; if (max_so_far < max_ending_here) max_so_far = max_ending_here; if (max_ending_here < 0) max_ending_here = 0; } return max_so_far; } /* Driver program to test maxCircularSum() */int main() { int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 }; int n = sizeof(a) / sizeof(a[0]); cout << "Maximum circular sum is " << maxCircularSum(a, n) << endl; return 0; } // This is code is contributed by rathbhupendra |
Output:
Maximum circular sum is 31
Complexity Analysis:
- Time Complexity: O(n), where n is the number of elements in the input array.
As only linear traversal of the array is needed. - Auxiliary Space: O(1).
As no extra space is required.
Note that the above algorithm doesn’t work if all numbers are negative, e.g., {-1, -2, -3}. It returns 0 in this case. This case can be handled by adding a pre-check to see if all the numbers are negative before running the above algorithm.
Method 2
Approach: In this method, modify Kadane’s algorithm to find a minimum contiguous subarray sum and the maximum contiguous subarray sum, then check for the maximum value between the max_value and the value left after subtracting min_value from the total sum.
Algorithm
- We will calculate the total sum of the given array.
- We will declare the variable curr_max, max_so_far, curr_min, min_so_far as the first value of the array.
- Now we will use Kadane’s Algorithm to find the maximum subarray sum and minimum subarray sum.
- Check for all the values in the array:-
- If min_so_far is equaled to sum, i.e. all values are negative, then we return max_so_far.
- Else, we will calculate the maximum value of max_so_far and (sum – min_so_far) and return it.
The implementation of the above method is given below.
C++
// C++ program for maximum contiguous circular sum problem #include <bits/stdc++.h> using namespace std; // The function returns maximum // circular contiguous sum in a[] int maxCircularSum(int a[], int n) { // Corner Case if (n == 1) return a[0]; // Initialize sum variable which store total sum of the array. int sum = 0; for (int i = 0; i < n; i++) { sum += a[i]; } // Initialize every variable with first value of array. int curr_max = a[0], max_so_far = a[0], curr_min = a[0], min_so_far = a[0]; // Concept of Kadane's Algorithm for (int i = 1; i < n; i++) { // Kadane's Algorithm to find Maximum subarray sum. curr_max = max(curr_max + a[i], a[i]); max_so_far = max(max_so_far, curr_max); // Kadane's Algorithm to find Minimum subarray sum. curr_min = min(curr_min + a[i], a[i]); min_so_far = min(min_so_far, curr_min); } if (min_so_far == sum) return max_so_far; // returning the maximum value return max(max_so_far, sum - min_so_far); } /* Driver program to test maxCircularSum() */int main() { int a[] = { 11, 10, -20, 5, -3, -5, 8, -13, 10 }; int n = sizeof(a) / sizeof(a[0]); cout << "Maximum circular sum is " << maxCircularSum(a, n) << endl; return 0; } |
Output:
Maximum circular sum is 31
Complexity Analysis:
- Time Complexity: O(n), where n is the number of elements in the input array.
As only linear traversal of the array is needed. - Auxiliary Space: O(1).
As no extra space is required.
Please refer complete article on Maximum circular subarray sum for more details!
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