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Leetcode: Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should:
Return true if there exists i, j, k 
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.

Examples:
Given [1, 2, 3, 4, 5],
return true.

Given [5, 4, 3, 2, 1],
return false.

Naive Solution: use DP, Time O(N^2), Space O(N)

  dp[i] represents the length of longest increasing subsequence till i including element i in nums array. dp[i] is initialized to be 1.

  dp[i] = max(dp[i], dp[j]+1), where j is an index before i

 1 public class Solution {
 2     public boolean increasingTriplet(int[] nums) {
 3         int[] dp = new int[nums.length];
 4         for (int i=0; i<nums.length; i++) {
 5             dp[i] = 1;
 6             for (int j=0; j<i; j++) {
 7                 if (nums[j] < nums[i]) {
 8                     dp[i] = Math.max(dp[i], dp[j]+1);
 9                 }
10                 if (dp[i] == 3) return true;
11             }
12         }
13         return false;
14     }
15 }

Better Solution: keep two values. Once find a number bigger than both, while both values have been updated, return true.

small: is the minimum value ever seen untill now

big: the smallest value that has something before it that is even smaller. That ‘something before it that is even smaller‘ does not have to be the current min value.

Example:
3,2,1,4,0,5

When you see 5, min value is 0, and the smallest second value is 4, which is not after the current min value.

 

Leetcode: Increasing Triplet Subsequence