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-rw-r--r-- | max_value_ordered_triplet1.c | 28 | ||||
-rw-r--r-- | max_value_ordered_triplet1.py | 32 |
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])) |