0900 - 0999
0918 - Maximum Sum Circular Subarray (Medium)
Problem Link
https://leetcode.com/problems/maximum-sum-circular-subarray/
Problem Statement
Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray ofnums.
A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].
A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.
Example 1:
Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.Example 2:
Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.Example 3:
Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.Constraints:
n == nums.length1 <= n <= 3 * 10^4-3 * 10^4 <= nums[i] <= 3 * 10^4
Approach 1: Kadane's Algorithm
class Solution {
public:
// kadane's algo
int kadane(vector<int>& nums) {
int local = nums[0], global = nums[0];
for (int i = 1; i < nums.size(); i++) {
local = max(nums[i], local + nums[i]);
global = max(global, local);
}
return global;
}
// case 1: max subarray sum in [0 .. n - 1]
// i.e. kadane's algo
// case 2. circular subarray in [0 .. | n - 1 .. | .. 2 * n - 1]
// i.e. total sum - min subarray sum in [0 .. n - 1]
int maxSubarraySumCircular(vector<int>& nums) {
// use kadane's algo to find out max sub array sum (case 1)
int maxSubArraySum = kadane(nums);
// handle cases like [-3,-2,-3]
if (maxSubArraySum < 0) return maxSubArraySum;
// calculate the total sum
int totalSum = accumulate(nums.begin(), nums.end(), 0);
// in order to use the same kadane function, we flip the sign
for (auto &x : nums) x *= -1;
// use kadane's algo to find out min sub array sum
int minSubArraySum = kadane(nums) * -1;
// compare case 1 & case 2, take the max
return max(maxSubArraySum, totalSum - minSubArraySum);
}
};