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[환미니니] Week4 문제풀이 #425

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Sep 8, 2024
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32 changes: 32 additions & 0 deletions longest-consecutive-sequence/hwanmini.js
Original file line number Diff line number Diff line change
@@ -0,0 +1,32 @@
// 시간 복잡도 O(n log n)
// 공간 복잡도 O(n)

/**
* @param {number[]} nums
* @return {number}
*/
var longestConsecutive = function(nums) {
if (nums.length === 0) return []

let maxSequenceLength = -Infinity


const setNums = [...new Set(nums)].toSorted((a,b) => a - b)
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엇 그런데 문제 조건에서 O(n)에 동작하는 코드를 작성해야 한다고 했는데, 해당 코드는 시간복잡도가 nlogn이 되는 것 같습니다!

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앗 그냥 생각나는데로 풀어버렸네요 ㅜ
정렬을 안쓰고 풀어봐야겠어요 ~


let count = 0;
for (let i = 0 ; i < setNums.length; i++) {
if (setNums[i]+1 === setNums[i+1]) {
count += 1
} else {
count += 1
maxSequenceLength = Math.max(maxSequenceLength, count)
count = 0;
}
}

return maxSequenceLength
};


console.log(longestConsecutive([100,4,200,1,3,2]))
console.log(longestConsecutive([0,3,7,2,5,8,4,6,0,1]))
19 changes: 19 additions & 0 deletions missing-number/hwanmini.js
Original file line number Diff line number Diff line change
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// 시간복잡도 O(n log n)
// 공간복잡도 O(n)

/**
* @param {number[]} nums
* @return {number}
*/
var missingNumber = function(nums) {
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js에서 sort는 내부적으로 추가적인 메모리를 사용해서 공간복잡도가 O(1)은 아닐 것 같습니다!

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엇 그렇네요 정렬할 때 공간복잡도가 On 사용됩니다!
감사합니다 : )

nums.sort((a,b) => a - b);

for (let i = 0 ; i <= nums.length; i++) {
if (i !== nums[i]) return i
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숫자가 0부터 시작하니까 이렇게 비교해서 풀어도 좋겠네요! 배워갑니다 :)

}

};

console.log(missingNumber([3, 0, 1]))
console.log(missingNumber([0, 1]))
console.log(missingNumber([9,6,4,2,3,5,7,0,1]))
29 changes: 29 additions & 0 deletions valid-palindrome/hwanmini.js
Original file line number Diff line number Diff line change
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// 시간복잡도: O(n)
// 공간복잡도: O(n)

/**
* @param {string} s
* @return {boolean}
*/
var isPalindrome = function(s) {
const strs = s.replace(/[^a-z0-9]/gi, '').toLowerCase();

let leftIdx = 0;
let rightIdx = strs.length - 1

while (leftIdx <= rightIdx) {
if (strs[leftIdx] !== strs[rightIdx]) return false

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오 양쪽에서 인덱스를 증가시키거나 감소시키면서 하나씩 비교하니 풀이가 굉장히 깔끔하네요!

leftIdx++
rightIdx--
}


return true
};

const s = "A man, a plan, a canal: Panama"


console.log(isPalindrome(s))

69 changes: 69 additions & 0 deletions word-search/hwanmini.js
Original file line number Diff line number Diff line change
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// m은 board의 행 수, n은 board의 열 수, 4(상하좌우), l(word)

// 시간복잡도: O(m * n * 4L)
// 공간복잡도: O(m * n + L)
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const d_row = [1, -1, 0, 0]
const d_col = [0, 0, -1, 1]


const isMoveBoard = (new_row, new_col, board) => {
return new_row >= 0 && new_row < board.length && new_col >= 0 && new_col < board[0].length
}


/**
* @param {character[][]} board
* @param {string} word
* @return {boolean}
*/
var exist = function(board, word) {
let result = false

const visited = Array.from({length: board.length} , () => Array.from({length: board[0].length}).fill(false))

const dfs = (strs, row, col, count) => {

if (strs[count] !== word[count]) return
if (strs.length > word.length) return

if (strs === word) {
result = true
return
}

for (let i = 0 ; i < d_row.length; i++) {
const new_row = row + d_row[i]
const new_col = col + d_col[i]

if (isMoveBoard(new_row, new_col, board) && visited[new_row][new_col] !== true){
visited[new_row][new_col] = true
dfs(strs+board[new_row][new_col], new_row, new_col, count+1)
visited[new_row][new_col] = false
}
}



}


for (let row = 0 ; row < board.length ; row++) {
for (let col = 0 ; col < board[0].length ; col++) {
visited[row][col] = true
dfs(board[row][col], row, col , 0)
visited[row][col] = false
}
}


return result
};

console.log(exist([["A","B","C","E"],
["S","F","C","S"],
["A","D","E","E"]], "ABCCED"))
console.log(exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "SEE"))
console.log(exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCB"))