LeetCode – Increasing Triplet Subsequence (Java)

 

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

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

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

Analysis

This problem can be converted to be finding if there is a sequence such that the_smallest_so_far < the_second_smallest_so_far < current. We use x, y and z to denote the 3 number respectively.

Java Solution

public boolean increasingTriplet(int[] nums) {
	int x = Integer.MAX_VALUE;
	int y = Integer.MAX_VALUE;
 
	for (int i = 0; i < nums.length; i++) {
		int z = nums[i];
 
		if (x >= z) {
			x = z;// update x to be a smaller value
		} else if (y >= z) {
			y = z; // update y to be a smaller value
		} else {
			return true;
		}
	}
 
	return false;
}

Related posts:

  1. LeetCode – Contains Duplicate II (Java)
  2. LeetCode – Minimum Size Subarray Sum (Java)
  3. LeetCode – Longest Increasing Subsequence (Java)
  4. LeetCode – Longest Increasing Path in a Matrix (Java)
Category >> Algorithms >> Interview  
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example: 
<pre><code> 
String foo = "bar";
</code></pre>

  1. Ankit Shah on 2016-3-24

    public class TrippletSubsequenceIncreasing {

    private static boolean isIncreasingTrippletSubsequence(int[] a) {
    int current = 0;
    int count = 1;
    for (int i = 1; i < a.length; i++) {
    if (a[i] == a[current] + 1) {
    count++;
    } else {
    count = 1;
    }
    current = current + 1;

    if (count == 3) {
    return true;
    }
    }

    return false;
    }

    public static void main(String[] args) {

    System.out.println(isIncreasingTrippletSubsequence(new int[] { 1, 2, 3, 4, 5 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 5, 4, 3, 2, 1 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 5, 6, 7, 8, 10 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 3, 4, 6, 7, 9 }));

    }

    }

  2. Ankit Shah on 2016-3-24


    public class TrippletSubsequenceIncreasing {

    private static boolean isIncreasingTrippletSubsequence(int[] a) {
    int count = 1;
    for (int i = 1; i a[i - 1]) {
    count++;
    } else {
    count = 1;
    }

    if (count == 3) {
    return true;
    }
    }
    return false;

    }

    public static void main(String[] args) {
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 1, 2, 3, 4, 5 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 5, 4, 3, 2, 1 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 5, 6, 7, 8, 10 }));
    System.out.println(isIncreasingTrippletSubsequence(new int[] { 3, 4, 6, 7, 9 }));

    }
    }

  3. Ashiq Imran on 2016-4-12

    You program will not work for [1,7,6,4,5] input?

  4. kash on 2016-5-14

    C++ solution

    bool increasing_triplet(int arr[], int n) {
    for (int i = 2; i arr[i-1] && arr[i-1] > arr[i-2])
    return true;
    }
    return false;
    }

  5. Samir Vasani on 2016-8-1

    public class IncreaseSubSequence {

    public static void main(String[] a){
    int[] num={7,8,1,4,5};
    int count=1;
    for(int i=0;i<num.length;i++){
    int j=i+1;
    if(j<=2 && i<2){
    if(num[i] < num[j]){
    count++;
    if(count == 3){
    break;
    }
    }else{
    count=0;
    break;
    }
    }
    }

    if(count == 3){
    System.out.println("true");
    }else{
    System.out.println("false");
    }
    }

  6. ryanlr on 2016-8-1

    They do not need to be consecutive.

  7. ryanlr on 2016-8-1

    This is not correct. For example, given the array {5,1,5,5,2,5,4}, it should return true. You code return false.

  8. ryanlr on 2016-8-1

    It works for this array.

  9. afj on 2016-8-12

    test case: [2,3,1,4] ?

Leave a comment

CAPTCHA

*