Given an integer array nums
, return true
if there exists a triple of indices (i, j, k)
such that i < j < k
and nums[i] < nums[j] < nums[k]
. If no such indices exists, return false
.
Example 1:
Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1]
Output: false
Explanation: No triplet exists.
Example 3:
Input: nums = [2,1,5,0,4,6]
Output: true
Explanation: The triplet (3, 4, 5) is valid because nums[3] == 0 < nums[4] == 4 < nums[5] == 6.
Constraints:
-
1 <= nums.length <= 5 * 105
-
-231 <= nums[i] <= 231 - 1
Follow up: Could you implement a solution that runs in O(n)
time complexity and O(1)
space complexity?SOLUTION:
class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
n = len(nums)
minval, maxval = min(nums), max(nums)
for i in range(1, n - 1):
if nums[i] > minval and nums[i] < maxval:
if min(nums[:i]) < nums[i] and nums[i] < max(nums[i+1:]):
return True
return False
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