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authorJack Sangdahl <[email protected]>2024-12-27 19:04:58 -0700
committerJack Sangdahl <[email protected]>2024-12-27 19:04:58 -0700
commitbe2dd2d507ad02c2ebfb3e371cfda1d4612de271 (patch)
tree6ec2021ad3c81294a4e99416cbf0eafc5f3140ea
parenta52ccc2d81430bae1862eebf40055815e1ff1c73 (diff)
downloadleetcode-be2dd2d507ad02c2ebfb3e371cfda1d4612de271.tar.gz
leetcode-be2dd2d507ad02c2ebfb3e371cfda1d4612de271.zip
689. maximum sum of 3 non-overlapping subarrays
-rw-r--r--max_sum_3_noverlap_subarr.cpp57
-rw-r--r--max_sum_3_noverlap_subarr.py53
2 files changed, 110 insertions, 0 deletions
diff --git a/max_sum_3_noverlap_subarr.cpp b/max_sum_3_noverlap_subarr.cpp
new file mode 100644
index 0000000..c326ad3
--- /dev/null
+++ b/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");
+}
diff --git a/max_sum_3_noverlap_subarr.py b/max_sum_3_noverlap_subarr.py
new file mode 100644
index 0000000..8240c46
--- /dev/null
+++ b/max_sum_3_noverlap_subarr.py
@@ -0,0 +1,53 @@
+# 689. Maximum Sum of 3 Non-Overlapping Subarrays
+
+"""
+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(object):
+ def max_sumOfThreeSubarrays(self, nums, k):
+ """
+ :type nums: List[int]
+ :type k: int
+ :rtype: List[int]
+ """
+ n = len(nums)
+ max_sum = 0
+ sum = [0]
+ left = [0] * n
+ right = [n - k] * n
+ ans = [0, 0, 0]
+ for i in nums:
+ sum.append(sum[-1] + i)
+ tot = sum[k] - sum[0]
+ for i in range(k, n):
+ 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]
+ tot = sum[n] - sum[n - k]
+ for i in range(n - k - 1, -1, -1):
+ if sum[i + k] - sum[i] > tot:
+ right[i] = i
+ tot = sum[i + k] - sum[i]
+ else:
+ right[i] = right[i + 1]
+ for i in range(k, n - 2 * k + 1):
+ l, r = left[i - 1], right[i + k]
+ 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
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.max_sumOfThreeSubarrays(nums=[1, 2, 1, 2, 6, 7, 5, 1], k=2))
+ print(obj.max_sumOfThreeSubarrays(nums=[1, 2, 1, 2, 1, 2, 1, 2, 1], k=2))