334. Increasing Triplet Subsequence
Description
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 * 10^52^31 <= nums[i] <= 2^31 - 1
My Solution
// Failed: Time Limit Exceeded
/**
* @param {number[]} nums
* @return {boolean}
*/
var increasingTriplet = function (nums) {
for (let i = 0; i < nums.length - 2; i++) {
let first = nums[i]
for (let j = i + 1; j < nums.length - 1; j++) {
let second
if (first < nums[j]) {
second = nums[j]
} else {
continue
}
for (let k = j + 1; k < nums.length; k++) {
if (second < nums[k]) {
return true
} else {
continue
}
}
}
}
return false
}
Solution from others
var increasingTriplet = function (nums) {
let first = Infinity
let second = Infinity
for (let i = 0; i < nums.length; i++) {
if (nums[i] <= first) {
first = nums[i]
} else if (nums[i] <= second) {
second = nums[i]
} else {
return true
}
}
return false
}