Skip to content

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 * 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?

Solutions

Solution 1

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution:
    def increasingTriplet(self, nums: List[int]) -> bool:
        mi, mid = inf, inf
        for num in nums:
            if num > mid:
                return True
            if num <= mi:
                mi = num
            else:
                mid = num
        return False
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
    public boolean increasingTriplet(int[] nums) {
        int n = nums.length;
        int[] lmi = new int[n];
        int[] rmx = new int[n];
        lmi[0] = Integer.MAX_VALUE;
        rmx[n - 1] = Integer.MIN_VALUE;
        for (int i = 1; i < n; ++i) {
            lmi[i] = Math.min(lmi[i - 1], nums[i - 1]);
        }
        for (int i = n - 2; i >= 0; --i) {
            rmx[i] = Math.max(rmx[i + 1], nums[i + 1]);
        }
        for (int i = 0; i < n; ++i) {
            if (lmi[i] < nums[i] && nums[i] < rmx[i]) {
                return true;
            }
        }
        return false;
    }
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
public:
    bool increasingTriplet(vector<int>& nums) {
        int mi = INT_MAX, mid = INT_MAX;
        for (int num : nums) {
            if (num > mid) return true;
            if (num <= mi)
                mi = num;
            else
                mid = num;
        }
        return false;
    }
};
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
func increasingTriplet(nums []int) bool {
    min, mid := math.MaxInt32, math.MaxInt32
    for _, num := range nums {
        if num > mid {
            return true
        }
        if num <= min {
            min = num
        } else {
            mid = num
        }
    }
    return false
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
function increasingTriplet(nums: number[]): boolean {
    let n = nums.length;
    if (n < 3) return false;
    let min = nums[0],
        mid = Number.MAX_SAFE_INTEGER;
    for (let num of nums) {
        if (num <= min) {
            min = num;
        } else if (num <= mid) {
            mid = num;
        } else if (num > mid) {
            return true;
        }
    }
    return false;
}
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
impl Solution {
    pub fn increasing_triplet(nums: Vec<i32>) -> bool {
        let n = nums.len();
        if n < 3 {
            return false;
        }
        let mut min = i32::MAX;
        let mut mid = i32