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334. Increasing Triplet Subsequence

DifficultyMedium

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: One of the valid triplet is (1, 4, 5), because nums[1] == 1 < 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

Thinking

Decide whether an increasing triplet of indices exists. Trying every middle index is \(O(n^2)\). We only need a smaller left value and a candidate second value.

Keep \(mi<mid\). A value above \(mid\) finishes the triplet; otherwise update \(mi\) if it is no larger, else update \(mid\). The old index of \(mid\) need not be stored: it already sits after a smaller \(mi\).

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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

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class Solution {
    public boolean increasingTriplet(int[] nums) {
        int min = Integer.MAX_VALUE, mid = Integer.MAX_VALUE;
        for (int num : nums) {
            if (num > mid) {
                return true;
            }
            if (num <= min) {
                min = num;
            } else {
                mid = num;
            }
        }
        return false;
    }
}

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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;
    }
};

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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
}

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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;
}

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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::MAX;
        for num in nums.into_iter() {
            if num <= min {
                min = num;
            } else if num <= mid {
                mid = num;
            } else {
                return true;
            }
        }
        false
    }
}

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