aboutsummaryrefslogtreecommitdiff
path: root/other/max_sum_3_noverlap_subarr.cpp
diff options
context:
space:
mode:
Diffstat (limited to 'other/max_sum_3_noverlap_subarr.cpp')
-rw-r--r--other/max_sum_3_noverlap_subarr.cpp57
1 files changed, 0 insertions, 57 deletions
diff --git a/other/max_sum_3_noverlap_subarr.cpp b/other/max_sum_3_noverlap_subarr.cpp
deleted file mode 100644
index c326ad3..0000000
--- a/other/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");
-}