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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
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tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /max_sum_circular_subarr.cpp
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+// 918. Maximum Sum Circular Subarray
+#include "leetcode.h"
+
+/*
+ * given: circular int array 'nums' of length 'n', return the max possible
+ * sum of non-empty subarray of 'nums'
+ *
+ * a circular array means the end of the array connects to the beginning.
+ * formally, the previous element of 'nums[i]' is 'nums[(i -/+ 1) % 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
+ */
+
+class Solution {
+public:
+ int maxSubarraySumCircular(vector<int> &nums) {
+ int sum = 0;
+ for (int i : nums)
+ sum += i;
+ int s1 = helper(nums.begin(), nums.end());
+ for (int &i : nums)
+ i *= -1;
+ int s2 = sum + helper(nums.begin(), nums.end() - 1);
+ int s3 = sum + helper(nums.begin() + 1, nums.end());
+ return max(s1, max(s2, s3));
+ }
+
+private:
+ int helper(vector<int>::iterator p, vector<int>::iterator q) {
+ int res = INT_MIN, n = 0;
+ for (auto i = p; i != q; i++) {
+ n += *i;
+ res = max(res, n);
+ if (n < 0)
+ n = 0;
+ }
+ return res;
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> nums1 = {-2}, nums2 = {5, -3, 5}, nums3 = {-3, -2, -3};
+ cout << obj.maxSubarraySumCircular(nums1) << endl; // expect: 3
+ cout << obj.maxSubarraySumCircular(nums2) << endl; // expect: 10
+ cout << obj.maxSubarraySumCircular(nums3) << endl; // expect: -2
+}