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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /max_sum_circular_subarr.cpp | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'max_sum_circular_subarr.cpp')
-rw-r--r-- | max_sum_circular_subarr.cpp | 49 |
1 files changed, 49 insertions, 0 deletions
diff --git a/max_sum_circular_subarr.cpp b/max_sum_circular_subarr.cpp new file mode 100644 index 0000000..a86eafa --- /dev/null +++ b/max_sum_circular_subarr.cpp @@ -0,0 +1,49 @@ +// 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 +} |