Given an integer array nums and two integers k and t, return true if there are two distinct indicesi 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:
0 <= nums.length <= 2 * 104
-231 <= nums[i] <= 231 - 1
0 <= k <= 104
0 <= t <= 231 - 1
ac1: TreeSet
TreeSet is Black-Red Tree, self balance tree, sorted.
aclass Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
TreeSet<Long> set = new TreeSet<Long>();
for (int i = 0; i < nums.length; i++) {
long val = (long) nums[i];
Long smaller = set.floor(val);
if (smaller != null && val - smaller <= t) return true;
Long larger = set.ceiling(val);
if (larger != null && larger - val <= t) return true;
set.add(val);
if (set.size() > k) set.remove((long)nums[i-k]);
}
return false;
}
}
// O(nlogk)
ac2: Bucket
class Solution {
public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
if (t < 0) return false;
Map<Long, Long> buckets = new HashMap<Long, Long>();
long size = (long)t + 1;
for (int i = 0; i < nums.length; i++) {
long ival = (long) nums[i];
long label = getId(ival, size);
if (buckets.containsKey(label)) return true;
if (buckets.containsKey(label - 1) && Math.abs(buckets.get(label-1) - ival) <= t) return true;
if (buckets.containsKey(label + 1) && Math.abs(buckets.get(label+1) - ival) <= t) return true;
buckets.put(label, ival);
if (i >= k) {
buckets.remove(getId((long)nums[i-k], size));
}
}
return false;
}
private Long getId(long num, long size) {
return num < 0 ? num / size - 1: num / size;
// in Java, -3 / 5 == 0, but we need -3 / 5 == -1, so we minus 1.
}
}
// O(n)
// Why "Math.abs(buckets.get(label-1) - ival) <= t)"? Because 1 bucket can only has one element, if it has more than 1, "buckets.containsKey(label)" will return true earlier.