Hello guys👋

In this article you will learn how to solve the Contains Duplicate II Question on Leetcode and I will solve the question using a very easy method.

Stay Calm and Let's get in✨

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

Solution

This has a similar approach to the Contains Duplicate we solved earlier.

• Set up a map to store our values by index.

  const map = new Map();
  

• Loop through the nums array and set the values of the array in the map by index.

  for (let i = 0; i < nums.length; i++) {
    map.set(nums[i], i);
  }
  

• We need to meet two conditions according to the questions and the conditions are:

• nums[i] == nums[j].

• abs(i-j) <= k.

• So for nums[i] == nums[j], that means it appears more than once or it is a duplicate. We will need to check the map if it has occurred before.

  for (let i = 0; i < nums.length; i++) {
    if (map.has(nums[i])) {
    }
  
    map.set(nums[i], i);
  }
  

• If it has occurred before we will get the index of the duplicate and follow the second condition which is abs(i-j) <= k.

  for (let i = 0; i < nums.length; i++) {
    if (map.has(nums[i])) {
      const j = map.get(nums[i]);
  
      if (Math.abs(i - j) <= k) {
      }
    }
  
    map.set(nums[i], i);
  }
  

• If this second condition is met then we will return true.

  for (let i = 0; i < nums.length; i++) {
    if (map.has(nums[i])) {
      const j = map.get(nums[i]);
  
      if (Math.abs(i - j) <= k) {
        return true;
      }
    }
  
    map.set(nums[i], i);
  }
  

• We are done but we will need to return false if none of this condition is met.

  return false;
  

  The Compilation of the code above is found below:

  for (let i = 0; i < nums.length; i++) {
    if (map.has(nums[i])) {
      const j = map.get(nums[i]);
  
      if (Math.abs(i - j) <= k) {
        return true;
      }
    }
  
    map.set(nums[i], i);
  }
  
  return false;
  

• Time Complexity: O(n)
• Space Complexity: O(n)

Conclusion

I hope you have learnt a lot from this article and your contribution is also highly welcomed because we get better by learning from others, you can reach out to me on Twitter here . Watch out for more helpful contents on Data Structures and Algorithm from me.

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