220. Contains Duplicate III
Description
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
contains-duplicate-iii.py
class Solution:
def containsNearbyAlmostDuplicate(self, nums: List[int], k: int, t: int) -> bool:
if t < 0: return False
d = {}
w = t + 1
for i,n in enumerate(nums):
m = n // w
if m in d:
return True
if m - 1 in d and abs(n - d[m - 1]) < w:
return True
if m + 1 in d and abs(n - d[m + 1]) < w:
return True
d[m] = n
if i >= k: del d[nums[i - k] // w]
return False