leetcode
December 22, 2023

Viiisit [LeetCode] - 219. Contains Duplicate II

#php#javascript#ruby

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Problem - Contains Duplicate II

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.

翻譯蒟蒻

找出在陣列中兩個相同的元素，且其索引值差值不超過 k。

Solution - JavaScript

Solution 1:

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/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var containsNearbyDuplicate = function(nums, k) {
  for (let i = 0; i < nums.length; i++) {
    for (let j = i + 1; j <= i + k && j < nums.length; j++) {
      if (nums[i] == nums[j]) {
        return true;
      }
    }
  }
  return false
};

Solution 2:

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/**
 * @param {number[]} nums
 * @param {number} k
 * @return {boolean}
 */
var containsNearbyDuplicate = function(nums, k) {
  let map = new Map();

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

Solution - Ruby

Solution 1:
(Time Limit Exceeded)

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def contains_nearby_duplicate(nums, k)
  for i in 0...nums.length
    for j in i+1...nums.length
      if nums[i] == nums[j] && (j - i) <= k
        return true
      end
    end
  end
  return false
end

Solution 2:

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def contains_nearby_duplicate(nums, k)
    hash = {}
    nums.each_with_index do |num, index|
        if hash[num] && (index - hash[num]).abs <= k
            return true
        else
            hash[num] = index
        end
    end
    false
end

Solution - PHP

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function containsNearbyDuplicate($nums, $k) {
  $map = [];

  for ($i = 0; $i < count($nums); $i++) {
    if (isset($map[$nums[$i]]) && $i - $map[$nums[$i]] <= $k) {
      return true;
    }
    $map[$nums[$i]] = $i;
  }

  return false;
}

LeetCode 傳送門 - Contains Duplicate II