notes/OJ notes/pages/Leetcode Increasing-Triplet-Subsequence.md

Leetcode Increasing-Triplet-Subsequence

2022-09-04 14:23

Algorithms:

#algorithm #greedy

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty_medium

Additional tags:

#leetcode #CS_list_need_understanding

Revisions:

2022-09-21

Links:

• Link to problem
• Solution with explanation

---

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.

Examples

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

Thoughts

[!summary] This is a #greedy problem.

Use two variables: small and big to keep track of stuff.

For each element, there are three possibilities:

• the element is smaller than small, assign the element to small

small and big doesn't represent i and j, accurately, they get updated lazily. And because we don't need to show i, j, k, this will be just fine

• the element is greater than small:
  • the element is smaller than big, change big, this allows more possibilities
  • the element is bigger than big, found, exit loop.

Why not #stack ?

Similar to Leetcode Next-Greater-Element-I, the property of stack can be used for continuity related problem, but here we don't need this.

Solution

class Solution {
public:
  bool increasingTriplet(vector<int> &nums) {
    // greedy algorithm
    int small = INT_MAX, big = INT_MAX;

    for (int i = 0, size = nums.size(); i < size; i++) {
      if (nums[i] <= small) {
        // Use equal sign here, to avoid small and big being same.
        small = nums[i];
      } else if (nums[i] <= big) {
        // Use equal sign here, to avoid three consecutive
        // same numbers
        big = nums[i];
      } else {
        return true;
      }
    }

    return false;
  }
};