Day 15 of 31-Day May LeetCode Challenge

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array. (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

Also, a subarray may only include each element of the fixed buffer A at most once. (Formally, for a subarray C[i], C[i+1], ..., C[j], there does not exist i <= k1, k2 <= j with k1 % A.length = k2 % A.length.)

class Solution {
public:
    int kadane(vector<int> arr, int n){
        int dp[n];
        dp[0] = arr[0];
        for(int i=1;i<n;++i){
            dp[i] = max(arr[i], dp[i-1]+arr[i]);
        }
        int max_sum = INT_MIN;
        for(int i=0;i<n;++i){
            max_sum = max(max_sum, dp[i]);
        }
        return max_sum;
    }
    int maxSubarraySumCircular(vector<int>& A) {
        
        //method : using kadane's algorithm
        //O(n) time & O(1) space
        int n = A.size();
        
        //STEP 1 :  find maximum subarray sum in normal array
        int max_kadane = kadane(A, n);
        
        
        //STEP 2 : find maximum subarray sum in circular array
        //maximum subarray sum in circular array = overall_sum - minimum subarray sum in normal array
        //BUT instead of modifying kadane's() to find minimum subarray sum in normal array , INSTEAD
        //we invert all elemets and then apply kadane() on array & ADD it to sum(instead of substracting)
        
        int overall_sum = 0;
        for(int i=0;i<n;++i) overall_sum += A[i];
        
        for(int i=0;i<n;++i) A[i] *= -1;
        int max_wrap = overall_sum + kadane(A, n);
        
        //absolute values are considered to handle a corner case i.e when all array elements are negative
        //in this case step 2 will always give 0 which may or may not be correct
        return (abs(max_wrap)>abs(max_kadane))?max_wrap:max_kadane;
        
    }
};