[LeetCode 217] Contians Duplicate

Contains Duplicate I

• Question: 217. Contains Duplicate

HashSet

• add(): add an element into hashset, if it has the element return false, otherwise return true.

Code

public class Solution {
    public boolean containsDuplicate(int[] nums) {
        HashSet<Integer> hs = new HashSet<Integer>();
        for (int num: nums) {
            // if the set already contains the element, the add() will return false
            if (!hs.add(num)) {
                return true;
            }
        }
        return false;
    }
}

Contains Duplicate II

• Question: 219. Contains Duplicate II

Explanation

• use sliding window
• keep a window with size k using HashSet
• move forward one element, we deleted the last one and check if the window has an element as same as the new one.

Code

public class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        HashSet<Integer> hs = new HashSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (!hs.add(nums[i])) {
                return true;
            }
            if (i >= k) {
                hs.remove(nums[i - k]);
            }
        }
        return false;
    }
}

Contains Duplicate III

• Question: 220. Contains Duplicate III My Submissions Question

TreeSet

• floor(num): return the greatest element less than or equal to num.
• ceiling(num): return the least element greater than or equal to num.

Explanation

Code

public class Solution {
    public boolean containsNearbyAlmostDuplicate(int[] nums, int k, int t) {
        TreeSet<Integer> window = new TreeSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            Integer ceiling = window.ceiling(nums[i]);
            Integer floor = window.floor(nums[i]);
            if (ceiling != null && nums[i] >= ceiling - t) {
                return true;
            if (floor != null && nums[i] <= floor + t) {
                return true;
            window.add(nums[i]);
            if (i >= k) {
                window.remove(nums[i - k]);
        }
        return false;
    }
}