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author | Jack Sangdahl <[email protected]> | 2025-03-10 00:36:13 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-03-10 00:36:13 -0600 |
commit | 7c60dee83b436e8e8a667bf9f191beced8de4fc3 (patch) | |
tree | eb7a24ee2067f92245c57bb118e85a1baf15df50 /other/max_sum_3_noverlap_subarr.cpp | |
parent | 5d7cdc215e4ba1ec4cbb24c084cccb5e5e641468 (diff) | |
download | leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.tar.gz leetcode-7c60dee83b436e8e8a667bf9f191beced8de4fc3.zip |
chore: move cpp and rust code to directory
Diffstat (limited to 'other/max_sum_3_noverlap_subarr.cpp')
-rw-r--r-- | other/max_sum_3_noverlap_subarr.cpp | 57 |
1 files changed, 57 insertions, 0 deletions
diff --git a/other/max_sum_3_noverlap_subarr.cpp b/other/max_sum_3_noverlap_subarr.cpp new file mode 100644 index 0000000..c326ad3 --- /dev/null +++ b/other/max_sum_3_noverlap_subarr.cpp @@ -0,0 +1,57 @@ +// 689. Maximum Sum of 3 Non-Overlapping Subarrays +#include "leetcode.h" + +/* + * given an integer array 'nums' and an integer 'k', find three non-overlapping + * subarrays of length 'k' with maximum sum and return them. return the result + * as a list of indices representing the starting position of each interval + * (0-indexed). if there are multiple answers, return the lexicographcially + * smallest one + */ + +class Solution { +public: + vector<int> maxSumOfThreeSubarrays(vector<int> &nums, int k) { + int n = nums.size(), max_sum = 0; + vector<int> sum = {0}, left(n, 0), right(n, n - k), ans(3, 0); + for (int i : nums) + sum.push_back(sum.back() + i); + for (int i = k, tot = sum[k] - sum[0]; i < n; i++) { + if (sum[i + 1] - sum[i + 1 - k] > tot) { + left[i] = i + 1 - k; + tot = sum[i + 1] - sum[i + 1 - k]; + } else + left[i] = left[i - 1]; + } + for (int i = n - k - 1, tot = sum[n] - sum[n - k]; i >= 0; i--) { + if (sum[i + k] - sum[i] >= tot) { + right[i] = i; + tot = sum[i + k] - sum[i]; + } else + right[i] = right[i + 1]; + } + for (int i = k; i <= n - 2 * k; i++) { + int l = left[i - 1], r = right[i + k]; + int tot = + (sum[i + k] - sum[i]) + (sum[l + k] - sum[l]) + (sum[r + k] - sum[r]); + if (tot > max_sum) { + max_sum = tot; + ans = {l, i, r}; + } + } + return ans; + } +}; + +int main() { + Solution obj; + vector<int> n1 = {1, 2, 1, 2, 6, 7, 5, 1}, n2 = {1, 2, 1, 2, 1, 2, 1, 2, 1}; + vector<int> msots1 = obj.maxSumOfThreeSubarrays(n1, 2); + vector<int> msots2 = obj.maxSumOfThreeSubarrays(n2, 2); + for (auto i : msots1) + printf("%d ", i); // expect: 0 3 5 + printf("\n"); + for (auto i : msots2) + printf("%d ", i); // expect: 0 2 4 + printf("\n"); +} |