Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most 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
Related Topics:
Sort, Ordered Map
Similar Questions:
// OJ: https://leetcode.com/problems/contains-duplicate-iii/
// Author: github.com/lzl124631x
// Time: O(NlogK)
// Space: O(K)
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& A, int k, int t) {
if (t < 0) return false;
multiset<long> s;
for (int i = 0; i < A.size(); ++i) {
if (i > k) s.erase(s.find(A[i - k - 1]));
if (s.lower_bound((long)A[i] - t) != s.upper_bound((long)A[i] + t)) return true;
s.insert(A[i]);
}
return false;
}
};
// OJ: https://leetcode.com/problems/contains-duplicate-iii/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(K)
// Ref: https://leetcode.com/problems/contains-duplicate-iii/discuss/61645/AC-O(N)-solution-in-Java-using-buckets-with-explanation
class Solution {
public:
bool containsNearbyAlmostDuplicate(vector<int>& A, int k, int t) {
if (A.size() < 2 || k < 1 || t < 0) return false;
unordered_map<long, long> m;
for (int i = 0; i < A.size(); ++i) {
if (i > k) m.erase(((long)A[i - k - 1] - INT_MIN) / ((long)t + 1));
long n = (long)A[i] - INT_MIN, id = n / ((long)t + 1);
if (m.count(id) || (m.count(id - 1) && A[i] - m[id - 1] <= t) || (m.count(id + 1) && m[id + 1] - A[i] <= t)) return true;
m[id] = A[i];
}
return false;
}
};