leetcode

commit d9ee70a4fdca0799186b7923f2d24fc41b281b17
parent 4b640063f1be1c56ae6aed07d3b50e345e518323
Author: Jack Sangdahl <jack@pngu.org>
Date:   Tue,  9 Jun 2026 20:53:17 -0600

3689. maximum total subarray value 1

Diffstat:
A
max_total_subarr_val1.c
 | 
30
++++++++++++++++++++++++++++++
A
max_total_subarr_val1.py
 | 
25
+++++++++++++++++++++++++

2 files changed, 55 insertions(+), 0 deletions(-)
diff --git a/max_total_subarr_val1.c b/max_total_subarr_val1.c
@@ -0,0 +1,30 @@
+// 3689. Maximum Total Subarray Value I
+#include "leetcode.h"
+
+/*
+ * you are given an integer array 'nums' of length 'n' and an integer 'k'. you
+ * need to choose exactly 'k' non-empty subarrays 'nums[l ... r]' of 'nums'.
+ * subarrays may overlap, and the exact same subarray can be chosen more than
+ * once. the value of a subarray 'nums[l ... r]' is defined as 'max(nums[l ...
+ * r]) - min(nums[l ... r])'. the total value is the sum of the values of all
+ * chosen subarrays. return the maximum possible total value you can achieve.
+ */
+
+long long maxTotalValue(int *nums, int numsSize, int k) {
+  int min = nums[0], max = nums[0];
+  for (int i = 0; i < numsSize; i++) {
+    min = fmin(min, nums[i]);
+    max = fmax(max, nums[i]);
+  }
+  return (long long)(max - min) * k;
+}
+
+int main() {
+  int n1[] = {1, 3, 2}, n2[] = {4, 2, 5, 1};
+  long long r1 = maxTotalValue(n1, ARRAY_SIZE(n1), 2);
+  long long r2 = maxTotalValue(n2, ARRAY_SIZE(n2), 3);
+  printf("%lld\n", r1);
+  assert(r1 == 4);
+  printf("%lld\n", r2);
+  assert(r2 == 12);
+}
diff --git a/max_total_subarr_val1.py b/max_total_subarr_val1.py
@@ -0,0 +1,25 @@
+# 3689. Maximum Total Subarray Value I
+
+"""
+you are given an integer array 'nums' of length 'n' and an integer 'k'. you
+need to choose exactly 'k' non-empty subarrays 'nums[l ... r]' of 'nums'.
+subarrays may overlap, and the exact same subarray can be chosen more than
+once. the value of a subarray 'nums[l ... r]' is defined as 'max(nums[l ...
+r]) - min(nums[l ... r])'. the total value is the sum of the values of all
+chosen subarrays. return the maximum possible total value you can achieve.
+"""
+
+
+class Solution:
+    def maxTotalValue(self, nums: List[int], k: int) -> int:
+        mn = mx = nums[0]
+        for i in nums:
+            mn = min(mn, i)
+            mx = max(mx, i)
+        return (mx - mn) * k
+
+
+if __name__ == "__main__":
+    obj = Solution()
+    print(obj.maxTotalValue(nums=[1, 3, 2], k=2))
+    print(obj.maxTotalValue(nums=[4, 2, 5, 1], k=3))