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

Solutions

Solution 1: Sliding Window + Ordered Set

We maintain a sliding window of size $k$, and the elements in the window are kept in order.

We traverse the array nums. For each element $nums[i]$, we look for the first element in the ordered set that is greater than or equal to $nums[i] - t$. If the element exists, and this element is less than or equal to $nums[i] + t$, it means we have found a pair of elements that meet the conditions, and we return true. Otherwise, we insert $nums[i]$ into the ordered set, and if the size of the ordered set exceeds $k$, we need to remove the earliest added element from the ordered set.

The time complexity is $O(n \times \log k)$, where $n$ is the length of the array nums. For each element, we need $O(\log k)$ time to find the element in the ordered set, and there are $n$ elements in total, so the total time complexity is $O(n \times \log k)$.

Python3JavaC++GoTypeScriptC#

from sortedcontainers import SortedSet


class Solution:
    def containsNearbyAlmostDuplicate(
        self, nums: List[int], indexDiff: int, valueDiff: int
    ) -> bool:
        s = SortedSet()
        for i, v in enumerate(nums):
            j = s.bisect_left(v - valueDiff)
            if j < len(s) and s[j] <= v + valueDiff:
                return True
            s.add(v)
            if i >= indexDiff:
                s.remove(nums[i - indexDiff])
        return False

class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int indexDiff, int valueDiff) {
        TreeSet<Long> ts = new TreeSet<>();
        for (int i = 0; i < nums.length; ++i) {
            Long x = ts.ceiling((long) nums[i] - (long) valueDiff);
            if (x != null && x <= (long) nums[i] + (long) valueDiff) {
                return true;
            }
            ts.add((long) nums[i]);
            if (i >= indexDiff) {
                ts.remove((long) nums[i - indexDiff]);
            }
        }
        return false;
    }
}

class Solution {
public:
    bool containsNearbyAlmostDuplicate(vector<int>& nums, int indexDiff, int valueDiff) {
        set<long> s;
        for (int i = 0; i < nums.size(); ++i) {
            auto it = s.lower_bound((long) nums[i] - valueDiff);
            if (it != s.end() && *it <= (long) nums[i] + valueDiff) return true;
            s.insert((long) nums[i]);
            if (i >= indexDiff) s.erase((long) nums[i - indexDiff]);
        }
        return false;
    }
};

func containsNearbyAlmostDuplicate(nums []int, k int, t int) bool {
    n := len(nums)
    left, right := 0, 0
    rbt := redblacktree.NewWithIntComparator()
    for right < n {
        cur := nums[right]
        right++
        if p, ok := rbt.Floor(cur); ok && cur-p.Key.(int) <= t {
            return true
        }
        if p, ok := rbt.Ceiling(cur); ok && p.Key.(int)-cur <= t {
            return true
        }
        rbt.Put(cur, struct{}{})
        if right-left > k {
            rbt.Remove(nums[left])
            left++
        }
    }
    return false
}