Leetcode Increasing Triplet Subsequence problem solution YASH PAL, 31 July 2024 In this Leetcode Increasing Triplet Subsequence problem solution we have 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. Problem solution in Python. class Solution: def increasingTriplet(self, nums: List[int]) -> bool: if len(nums) < 3: return False seq = [None] * 3 for num in nums: for j in range(3): if seq[j] is None or num <= seq[j]: seq[j] = num break if seq[2] is not None: return True return False Problem solution in Java. class Solution { public boolean increasingTriplet(int[] A) { int n = A.length; int[] min = new int[n]; int[] max = new int[n]; min[0] = A[0]; max[n-1] = A[n-1]; for(int i=1,j;i<n;i++){ j = n-i-1; min[i] = Math.min(min[i-1], A[i]); max[j] = Math.max(max[j+1], A[j]); } for(int i=1;i<n-1;i++){ int l = min[i-1]; int r = max[i+1]; if(l< A[i] && A[i]<r) return true; } return false; } } Problem solution in C++. class Solution { public: bool increasingTriplet(vector<int>& nums) { if(nums.size() < 3) return false; int minOne = INT_MAX; int minTwo = INT_MAX; for(int i = 0; i<nums.size(); i++) { if(nums[i] < minOne) minOne = nums[i]; if(nums[i] > minOne) minTwo = min(nums[i],minTwo); if(nums[i] > minTwo) return true; } return false; } }; Problem solution in C. bool increasingTriplet(int* nums, int numsSize) { if (numsSize < 3) return false; int l = nums[0], m = 0x7fffffff; for (int i = 1; i < numsSize; i++) { int a = nums[i]; if (a <= l) l = a; else if (a < m) m = a; else if (a > m) return true; } return false; } coding problems