Leetcode Contains Duplicate III problem solution YASH PAL, 31 July 202420 January 2026 In this Leetcode Contains Duplicate III problem solution, we have given an integer array nums and two integers k and t, return true if there are two distinct indices i and j in the array such that abs(nums[i] – nums[j]) <= t and abs(i – j) <= k. Leetcode Contains Duplicate III problem solution in Python.class Solution: def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool: if t == 0 and len(nums) == len(set(nums)): return False for i in range(len(nums)-1): for j in range(i+1, min(i+1+k, len(nums))): if abs(nums[i] - nums[j]) <= t: return True return False Contains Duplicate III problem solution in Java.class Solution { public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) { TreeMap<Long,Integer> treemap=new TreeMap<>(); for(int i=0;i<nums.length;i++){ if(treemap.size() > k) treemap.remove((long)nums[i - k - 1]); Long ceil = treemap.ceilingKey((long)nums[i]); Long floor = treemap.floorKey((long)nums[i]); if(ceil != null && ceil <= (long)nums[i] + t) return true; if(floor != null && floor >= (long)nums[i] - t) return true; treemap.put((long)nums[i],i); } return false; } } Problem solution in C++.class Solution { public: bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) { set<long> s; for(int i=0;i<nums.size();i++){ if(i>k) s.erase(nums[i-k-1]); auto pos=s.lower_bound(long(nums[i])-long(t)); if(pos!=s.end() && *pos-nums[i]<=t) return true; s.insert(nums[i]); } return false; } }; coding problems solutions Leetcode Problems Solutions Leetcode