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Diffstat (limited to 'max_sum_3_noverlap_subarr.cpp')
-rw-r--r-- | max_sum_3_noverlap_subarr.cpp | 57 |
1 files changed, 0 insertions, 57 deletions
diff --git a/max_sum_3_noverlap_subarr.cpp b/max_sum_3_noverlap_subarr.cpp deleted file mode 100644 index c326ad3..0000000 --- a/max_sum_3_noverlap_subarr.cpp +++ /dev/null @@ -1,57 +0,0 @@ -// 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"); -} |