aboutsummaryrefslogtreecommitdiff
path: root/max_value_ordered_triplet1.c
diff options
context:
space:
mode:
Diffstat (limited to 'max_value_ordered_triplet1.c')
-rw-r--r--max_value_ordered_triplet1.c28
1 files changed, 28 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
+}