219. Contains Duplicate II 

Description

Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

Example 2:

Input: nums = [1,0,1,1], k = 1
Output: true

Example 3:

Input: nums = [1,2,3,1,2,3], k = 2
Output: false

Constraints:

1 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹
0 <= k <= 10⁵

Solutions

Solution 1: Hash Table

We use a hash table $\textit{d}$ to store the recently traversed numbers and their corresponding indices.

Traverse the array $\textit{nums}$. For the current element $\textit{nums}[i]$, if it exists in the hash table and the difference between the indices is no more than $k$, return $\text{true}$. Otherwise, add the current element to the hash table.

After traversing, return $\text{false}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
        d = {}
        for i, x in enumerate(nums):
            if x in d and i - d[x] <= k:
                return True
            d[x] = i
        return False

Java

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Map<Integer, Integer> d = new HashMap<>();
        for (int i = 0; i < nums.length; ++i) {
            if (i - d.getOrDefault(nums[i], -1000000) <= k) {
                return true;
            }
            d.put(nums[i], i);
        }
        return false;
    }
}

C++

class Solution {
public:
    bool containsNearbyDuplicate(vector<int>& nums, int k) {
        unordered_map<int, int> d;
        for (int i = 0; i < nums.size(); ++i) {
            if (d.count(nums[i]) && i - d[nums[i]] <= k) {
                return true;
            }
            d[nums[i]] = i;
        }
        return false;
    }
};

Go

func containsNearbyDuplicate(nums []int, k int) bool {
	d := map[int]int{}
	for i, x := range nums {
		if j, ok := d[x]; ok && i-j <= k {
			return true
		}
		d[x] = i
	}
	return false
}

TypeScript

function containsNearbyDuplicate(nums: number[], k: number): boolean {
    const d: Map<number, number> = new Map();
    for (let i = 0; i < nums.length; ++i) {
        if (d.has(nums[i]) && i - d.get(nums[i])! <= k) {
            return true;
        }
        d.set(nums[i], i);
    }
    return false;
}

JavaScript

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var containsNearbyDuplicate = function (nums, k) {
    const d = new Map();
    for (let i = 0; i < nums.length; ++i) {
        if (d.has(nums[i]) && i - d.get(nums[i]) <= k) {
            return true;
        }
        d.set(nums[i], i);
    }
    return false;
};

C#

public class Solution {
    public bool ContainsNearbyDuplicate(int[] nums, int k) {
        var d = new Dictionary<int, int>();
        for (int i = 0; i < nums.Length; ++i) {
            if (d.ContainsKey(nums[i]) && i - d[nums[i]] <= k) {
                return true;
            }
            d[nums[i]] = i;
        }
        return false;
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @param Integer $k
     * @return Boolean
     */
    function containsNearbyDuplicate($nums, $k) {
        $d = [];
        for ($i = 0; $i < count($nums); ++$i) {
            if (array_key_exists($nums[$i], $d) && $i - $d[$nums[$i]] <= $k) {
                return true;
            }
            $d[$nums[$i]] = $i;
        }
        return false;
    }
}