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-rw-r--r--max_value_ordered_triplet1.c28
-rw-r--r--max_value_ordered_triplet1.py32
2 files changed, 60 insertions, 0 deletions
diff --git a/max_value_ordered_triplet1.c b/max_value_ordered_triplet1.c
new file mode 100644
index 0000000..9a62e4a
--- /dev/null
+++ b/max_value_ordered_triplet1.c
@@ -0,0 +1,28 @@
+// 2873. Maximum Value of an Ordered Triplet I
+#include "leetcode.h"
+
+/*
+ * you are given a 0-indexed integer array 'nums'. return the maximum value over
+ * all triplets of indices '(i, j, k)' such that 'i < j < k'. if all such
+ * triplets have a negative value, return 0. the value of a triplet of indices
+ * '(i, j, k)' is equal to '(nums[i] - nums[j]) * nums[k]'
+ */
+
+long long maximumTripletValue(int *nums, int numsSize) {
+ if (numsSize < 3)
+ return 0;
+ long long ans = 0, curr = nums[0], diff = 0;
+ for (int i = 1; i < numsSize; i++) {
+ ans = fmax(ans, (long long)diff * nums[i]);
+ diff = fmax(diff, curr - nums[i]);
+ curr = fmax(curr, nums[i]);
+ }
+ return ans;
+}
+
+int main() {
+ int n1[] = {12, 6, 1, 2, 7}, n2[] = {1, 10, 3, 4, 19}, n3[] = {1, 2, 3};
+ printf("%lld\n", maximumTripletValue(n1, ARRAY_SIZE(n1))); // expect: 77
+ printf("%lld\n", maximumTripletValue(n2, ARRAY_SIZE(n2))); // expect: 133
+ printf("%lld\n", maximumTripletValue(n3, ARRAY_SIZE(n3))); // expect: 0
+}
diff --git a/max_value_ordered_triplet1.py b/max_value_ordered_triplet1.py
new file mode 100644
index 0000000..e91c340
--- /dev/null
+++ b/max_value_ordered_triplet1.py
@@ -0,0 +1,32 @@
+# 2873. Maximum Value of an Ordered Triplet I
+
+"""
+you are given a 0-indexed integer array 'nums'. return the maximum value over
+all triplets of indices '(i, j, k)' such that 'i < j < k'. if all such
+triplets have a negative value, return 0. the value of a triplet of indices
+'(i, j, k)' is equal to '(nums[i] - nums[j]) nums[k]'
+"""
+
+
+class Solution(object):
+ def maximumTripletValue(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ n = len(nums)
+ if n < 3:
+ return 0
+ ans, curr, diff = 0, nums[0], 0
+ for i in range(1, n):
+ ans = max(ans, diff * nums[i])
+ diff = max(diff, curr - nums[i])
+ curr = max(curr, nums[i])
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.maximumTripletValue(nums=[12, 6, 1, 2, 7]))
+ print(obj.maximumTripletValue(nums=[1, 10, 3, 4, 19]))
+ print(obj.maximumTripletValue(nums=[1, 2, 3]))