Problem
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 * 10^5-2^31 <= nums[i] <= 2^31 - 1
Follow up: Could you implement a solution that runs in O(n) time complexity and O(1) space complexity?
Solution (Java)
class Solution {
public boolean increasingTriplet(int[] nums) {
int n = nums.length;
int i = 0;
int low = Integer.MAX_VALUE;
int high = Integer.MAX_VALUE;
while (i < n) {
if (low >= nums[i]) {
low = nums[i++];
} else if (high >= nums[i]) {
high = nums[i++];
} else {
return true;
}
}
return false;
}
}
Solution (Javascript)
/**
* @param {number[]} nums
* @return {boolean}
*/
var increasingTriplet = function(nums) {
let a = Infinity
let b = Infinity
for(let num of nums) {
if(num <= a) {
a = num
} else if(num <= b) {
b = num
} else {
return true
}
}
return false
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).