334. Increasing Triplet Subsequence

Posted 2020-07-20 蜃利的阴影下

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.

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

Hide Similar Problems

 (M) Longest Increasing Subsequence

 

Soluion1:

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;
        for (int n : nums) {
            if (n <= small)
                small = n; 
            else if (n <= big) 
                big = n;
            else 
                return true;
        }
        return false;
    }
}

 

 

 

Solution2:

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int min = Integer.MAX_VALUE;
        int mid = Integer.MAX_VALUE;
        
        int altMin = Integer.MAX_VALUE;
        
        for(int i =0; i<nums.length; ++i)
        {
            if(nums[i]<min)
            {
                if(mid == Integer.MAX_VALUE) //If mid val hasn‘t been set
                    min = nums[i];   
                else if(altMin < nums[i])    //If in (alterMin, Min), reset Min & Mid
                {
                    min = altMin;
                    mid = nums[i];
                    altMin = Integer.MAX_VALUE;
                }
                else if(nums[i] < altMin)     //update the alternative min, which is less than min.
                    altMin = nums[i];
                
            }
            else if(nums[i]<mid)
            {
                if(nums[i] > altMin)        //If in (alterMin, Mid), reset Min & Mid
                {
                    min = altMin;
                    mid = nums[i];
                    altMin = Integer.MAX_VALUE;
                }
                else if(nums[i] > min)
                    mid = nums[i];          // Reset mid if alterMin is not set.
            }
            else if(nums[i]>mid)
            {
                return true;
            }
        }
        return false;
    }
}