220. Contains Duplicate III

Problem Link

Description

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You are given an integer array nums and two integers indexDiff and valueDiff.

Find a pair of indices (i, j) such that:

• i != j,
• abs(i - j) <= indexDiff.
• abs(nums[i] - nums[j]) <= valueDiff, and

Return true if such pair exists or false otherwise.

 

Example 1:

Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output: true
Explanation: We can choose (i, j) = (0, 3).
We satisfy the three conditions:
i != j --> 0 != 3
abs(i - j) <= indexDiff --> abs(0 - 3) <= 3
abs(nums[i] - nums[j]) <= valueDiff --> abs(1 - 1) <= 0

Example 2:

Input: nums = [1,5,9,1,5,9], indexDiff = 2, valueDiff = 3
Output: false
Explanation: After trying all the possible pairs (i, j), we cannot satisfy the three conditions, so we return false.

 

Constraints:

• 2 <= nums.length <= 105
• -109 <= nums[i] <= 109
• 1 <= indexDiff <= nums.length
• 0 <= valueDiff <= 109

Solution

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Python3

class Solution:
    def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, w: int) -> bool:
        N = len(nums)
        mp = {}
 
        for i, x in enumerate(nums):
            bucket = x // (w + 1)
 
            if bucket in mp: return True
            if bucket + 1 in mp and mp[bucket + 1] - x <= w: return True
            if bucket - 1 in mp and x - mp[bucket - 1] <= w: return True
 
            mp[bucket] = x
 
            if i >= k:
                del mp[(nums[i - k]) // (w + 1)]
            
        return False