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# Contains Duplicate III LeetCode Solution

## Problem – Contains Duplicate III

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`.

Example 1:

``````Input: nums = [1,2,3,1], k = 3, t = 0
Output: true``````

Example 2:

``````Input: nums = [1,0,1,1], k = 1, t = 2
Output: true``````

Example 3:

``````Input: nums = [1,5,9,1,5,9], k = 2, t = 3
Output: false``````

Constraints:

• `1 <= nums.length <= 2 * 104`
• `-231 <= nums[i] <= 231 - 1`
• `0 <= k <= 104`
• `0 <= t <= 231 - 1`

### Contains Duplicate III LeetCode Solution in Python

``````def containsNearbyAlmostDuplicate(self, nums, k, t):
if t < 0: return False
n = len(nums)
d = {}
w = t + 1
for i in xrange(n):
m = nums[i] / w
if m in d:
return True
if m - 1 in d and abs(nums[i] - d[m - 1]) < w:
return True
if m + 1 in d and abs(nums[i] - d[m + 1]) < w:
return True
d[m] = nums[i]
if i >= k: del d[nums[i - k] / w]
return False
``````

### Contains Duplicate III LeetCode Solution in Java

``````private long getID(long i, long w) {
return i < 0 ? (i + 1) / w - 1 : i / w;
}

public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (t < 0) return false;
Map<Long, Long> d = new HashMap<>();
long w = (long)t + 1;
for (int i = 0; i < nums.length; ++i) {
long m = getID(nums[i], w);
if (d.containsKey(m))
return true;
if (d.containsKey(m - 1) && Math.abs(nums[i] - d.get(m - 1)) < w)
return true;
if (d.containsKey(m + 1) && Math.abs(nums[i] - d.get(m + 1)) < w)
return true;
d.put(m, (long)nums[i]);
if (i >= k) d.remove(getID(nums[i - k], w));
}
return false;
}
``````

### Contains Duplicate III LeetCode Solution in C++

``````class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {
int n = nums.size();

if(n == 0 || k < 0  || t < 0) return false;

unordered_map<int,int> buckets;

for(int i=0; i<n; ++i) {
int bucket = nums[i] / ((long)t + 1);

// For negative numbers, we need to decrement bucket by 1
// to ensure floor division.
// For example, -1/2 = 0 but -1 should be put in Bucket[-1].
// Therefore, decrement by 1.
if(nums[i] < 0) --bucket;

if(buckets.find(bucket) != buckets.end()) return true;
else {
buckets[bucket] = nums[i];
if(buckets.find(bucket-1) != buckets.end() && (long) nums[i] - buckets[bucket-1] <= t) return true;
if(buckets.find(bucket+1) != buckets.end() && (long) buckets[bucket+1] - nums[i] <= t) return true;

if(buckets.size() > k) {
int key_to_remove = nums[i-k] / ((long)t + 1);

if(nums[i-k] < 0) --key_to_remove;

buckets.erase(key_to_remove);
}
}
}

return false;
}
};
``````
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