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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

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;
    }
};

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