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Apr 29, 2024
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30 changes: 30 additions & 0 deletions best-time-to-buy-and-sell-stock/nhistory.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,30 @@
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function (prices) {
// Initiate left and right pointer (left: buy price / right: sell price)
// Make profit with initiative value 0
let l = 0;
let r = 1;
let profit = 0;
// Iterate to check profit = prices[r] - prices[l]
while (r < prices.length) {
// If profit is positive, compare profit with previous one
if (prices[r] > prices[l]) {
profit = Math.max(profit, prices[r] - prices[l]);
r++;
} else {
// If profit is negative, move forware left and right pointer
l = r;
r++;
}
}
return profit;
};

// TC: O(n)
// SC: O(3)

console.log(maxProfit([7, 1, 5, 3, 6, 4])); //5
console.log(maxProfit([7, 6, 4, 3, 1])); //0
24 changes: 24 additions & 0 deletions contains-duplicate/nhistoy.js
Original file line number Diff line number Diff line change
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/**
* @param {number[]} nums
* @return {boolean}
*/
var containsDuplicate = function (nums) {
// 1. Make hashmap that has previous value
let map = {};
// 2. Iterate nums to check nums[i] is duplicated or not
for (let i = 0; i < nums.length; i++) {
// 2.1 Check there is same value on the map
// 3. Return true when their is duplicated value
if (nums[i] in map) {
return true;
} else map[nums[i]] = i;
}
return false;
};

// TC: O(n)
// SC: O(n)

console.log(containsDuplicate([1, 2, 3, 1])); // true
console.log(containsDuplicate([1, 2, 3, 4])); // false
console.log(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])); // true
Comment on lines +19 to +24
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캬~ 복잡도 분석에 셀프 테스트까지? 멋지십니다! 👏👏👏

23 changes: 23 additions & 0 deletions two-sum/nhistory.js
Original file line number Diff line number Diff line change
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/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function (nums, target) {
// 1. Make a Hashmap to store { key(index) : value(target-num)}
let map = {};
// 2. Iterate to find value is equal to (target - nums[i])
// There is only one solution
for (let i = 0; i < nums.length; i++) {
const diff = target - nums[i];
// 3. If there is an index that has different value, return array
if (diff in map) return [map[diff], i];
map[nums[i]] = i;
}
};

// TC: O(n)
// SC: O(n)

console.log(twoSum([2, 7, 11, 15], 9));
console.log(twoSum([3, 2, 4], 6));
38 changes: 38 additions & 0 deletions valid-anagram/nhistory.js
Original file line number Diff line number Diff line change
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/**
* @param {string} s
* @param {string} t
* @return {boolean}
*/
var isAnagram = function (s, t) {
/*
// Split string and sort charaters
// Compare s and t after sorting
return s.split("").sort().join() === t.split("").sort().join()
// 92ms, 53.29MB
*/

// Made Hashmap and count number of each charater
let map = {};

// Check character length between s and t
if (s.length !== t.length) return false;
// Put a character and add or substract number
for (let i = 0; i < s.length; i++) {
// map[s[i]] = map[s[i]] ? map[s[i]] + 1 : 1;
map[s[i]] = (map[s[i]] ?? 0) + 1;
// map[t[i]] = map[t[i]] ? map[t[i]] - 1 : -1;
map[t[i]] = (map[t[i]] ?? 0) - 1;
}

for (let i = 0; i < s.length; i++) {
if (map[s[i]] !== 0) return false;
}

return true;
};

console.log(isAnagram("anagram", "nagaram"));
console.log(isAnagram("rat", "car"));

// TC: O(n)
// SC: O(n)
43 changes: 43 additions & 0 deletions valid-palindrome/nhistory.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,43 @@
/**
* @param {string} s
* @return {boolean}
*/
var isPalindrome = function (s) {
// 1. Create left and right pointer
let l = 0;
let r = s.length - 1;
// 2. Iterate while loop with l<r condition
while (l < r) {
// 3. Check character with left pointer is alphanumeric character or not
if (
(s[l] >= "a" && s[l] <= "z") ||
(s[l] >= "A" && s[l] <= "Z") ||
(s[l] >= "0" && s[l] <= "9")
) {
// 4. Check character with right pointer is alphanumeric character or not
if (
(s[r] >= "a" && s[r] <= "z") ||
(s[r] >= "A" && s[r] <= "Z") ||
(s[r] >= "0" && s[r] <= "9")
) {
// 5. Compare left and right pointer character
if (s[l].toLowerCase() !== s[r].toLowerCase()) {
return false;
} else {
l++;
r--;
}
// If not, go to next location for right pointer
} else r--;

// If not, go to next location for left pointer
} else l++;
Comment on lines +12 to +34
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@DaleSeo DaleSeo Apr 29, 2024

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@nhistory 님, 코드가 이미 머지가 되었지만 사소한 피드백을 드리고 싶어서 코멘트 드립니다.
이 3중 중첩 조건문에 주석이 없었다면 코드를 바로 이해하기가 쉽지 않을 수도 있었겠다는 생각이 들었습니다.
아무래도 가독성이 떨어지는 코드는 일반적으로 코딩 인터뷰에서 지원자에게 불리하게 작용하거든요.
어떻게 하면 이 부분에서 조건 중첩을 줄일 수 있을지 한번 고민해보시면 좋을 것 같습니다.

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@DaleSeo 귀한 피드백 감사드립니다!
중첩 조건문을 줄이는 방법으로 다음과 같이 수정해 보았습니다.

  // 2. Iterate while loop with l<r condition
  while (l < r) {
    // 3. Check left and right pointer has alphanumeric character
    while (l < r && !isAlphanumeric(s[l])) l++;
    while (l < r && !isAlphanumeric(s[r])) r--;

    // 4. Compare left and right pointer character
    if (s[l].toLowerCase() !== s[r].toLowerCase()) {
      return false;
    }
    l++;
    r--;
  }
  return true;
};

}
return true;
};

// TC: O(n)
// SC: O(n)

console.log(isPalindrome("A man, a plan, a canal: Panama")); //true
console.log(isPalindrome("race a car")); //false