Increasing Triplet Subsequence

 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.



Brute Force : Failed for some testcase

class Solution {
    public boolean increasingTriplet(int[] nums) {

        int n=nums.length;
        int i=0;
        int j=i+1;
        int k=j+1;

        while(i <n && j<n && k<n)
        {

        if(nums[i]<nums[j] &&i<j)
        {
       
        if(nums[j]<nums[k] && j <k)
        {
         return true;
        }
        else
        {
         i++;
         j++;
         k++;
        }
        }
        else
        {
            i++;
            j++;
            k++;
        }
        }
        return false;
       
    }
}



GreedyApproach


class Solution {
    public boolean increasingTriplet(int[] nums) {
        int first = Integer.MAX_VALUE;
        int second = Integer.MAX_VALUE;

        for (int num : nums) {
            if (num <= first) {
                first = num;         // new smallest
            } else if (num <= second) {
                second = num;        // new second smallest
            } else {
                return true;         // num > first and num > second
            }
        }
        return false;
    }
}


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