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Home Code Solutions

Increasing Triplet Subsequence – LeetCode Solution

Increasing Triplet Subsequence - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 6, 2022
Reading Time: 2 mins read
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Leetcode All Problems Solutions

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Table of Contents

  •  Increasing Triplet Subsequence – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++  Increasing Triplet Subsequence LeetCode Solution
  • Java  Increasing Triplet Subsequence LeetCode Solution
  • Python 3  Increasing Triplet Subsequence LeetCode Solution
    • Leave a comment below
      • Related posts:

 Increasing Triplet Subsequence – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

 

Constraints:

  • 1 <= nums.length <= 5 * 105
  • -231 <= nums[i] <= 231 - 1

C++  Increasing Triplet Subsequence LeetCode Solution


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bool increasingTriplet(vector<int>& nums) {
    int c1 = INT_MAX, c2 = INT_MAX;
    for (int x : nums) {
        if (x <= c1) {
            c1 = x;           // c1 is min seen so far (it's a candidate for 1st element)
        } else if (x <= c2) { // here when x > c1, i.e. x might be either c2 or c3
            c2 = x;           // x is better than the current c2, store it
        } else {              // here when we have/had c1 < c2 already and x > c2
            return true;      // the increasing subsequence of 3 elements exists
        }
    }
    return false;
}

Java  Increasing Triplet Subsequence LeetCode Solution


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 public boolean increasingTriplet(int[] nums) {
        // start with two largest values, as soon as we find a number bigger than both, while both have been updated, return true.
        int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;
        for (int n : nums) {
            if (n <= small) { small = n; } // update small if n is smaller than both
            else if (n <= big) { big = n; } // update big only if greater than small but smaller than big
            else return true; // return if you find a number bigger than both
        }
        return false;
    }

 



Python 3  Increasing Triplet Subsequence LeetCode Solution


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def increasingTriplet(nums):
    first = second = float('inf')
    for n in nums:
        if n <= first:
            first = n
        elif n <= second:
            second = n
        else:
            return True
    return False



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Tags: Cc++14full solutionGoIncreasing Triplet Subsequencejavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionpypy 3Python 2python 3rubyrustSolution
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