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[gmlwls96] Week4 #807

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Jan 5, 2025
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25 changes: 25 additions & 0 deletions coin-change/gmlwls96.kt
Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
class Solution {
// 알고리즘 : dp
/** 풀이
* dp배열에 최소한의 동전의 개수를 저장.
* dp[i] = min(dp[i - 동전값], dp[i]) 중 더 작은값이 최소 동전의 개수.
* */
// 시간 : O(coins.len*amount), 공간 : O(amount)
fun coinChange(coins: IntArray, amount: Int): Int {
val initValue = Int.MAX_VALUE / 2
val dp = IntArray(amount + 1) { initValue }
dp[0] = 0
for (i in 1..amount) {
coins.forEach { c ->
if (c <= i) {
dp[i] = min(dp[i - c] + 1, dp[i])
}
}
}
return if (dp[amount] == initValue) {
-1
} else {
dp[amount]
}
}
}
Comment on lines +8 to +25
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아직 이 문제를 풀지 못했는데, 파이썬에서 sys.maxsize // 2를 초기값으로 설정하면, 오버플로우를 방지할 수 있다는 아이디어를 얻었어요. 문제를 풀 때 이 방식을 사용해봐야겠습니다. 이번주도 고생하셨습니다!

50 changes: 50 additions & 0 deletions merge-two-sorted-lists/gmlwls96.kt
Original file line number Diff line number Diff line change
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/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
// 시간, 공간 : o(n+m),
fun mergeTwoLists(list1: ListNode?, list2: ListNode?): ListNode? {
var currentList1: ListNode? = list1
var currentList2: ListNode? = list2
val answerList = LinkedList<Int>()
while (currentList1 != null || currentList2 != null) {
when {
currentList1 == null -> {
answerList.offer(currentList2!!.`val`)
currentList2 = currentList2.next
}
currentList2 == null -> {
answerList.offer(currentList1.`val`)
currentList1 = currentList1.next
}
Comment on lines +18 to +25
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@DaleSeo DaleSeo Dec 31, 2024

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여기서 왜 currentList2 상대로만 non-null assertion이 필요한 건가요? 🤔

currentList1.`val` <= currentList2.`val` -> {
answerList.offer(currentList1.`val`)
currentList1 = currentList1.next
}
currentList2.`val` < currentList1.`val` -> {
answerList.offer(currentList2.`val`)
currentList2 = currentList2.next
}
}
}
var answer: ListNode? = null
var currentAnswer: ListNode? = null
while (answerList.isNotEmpty()) {
val num = answerList.poll()
if (answer == null) {
answer = ListNode(num)
currentAnswer = answer
} else {
currentAnswer?.next = ListNode(num)
currentAnswer = currentAnswer?.next
}
}
return answer
}
}
13 changes: 13 additions & 0 deletions missing-number/gmlwls96.kt
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이 풀이만 복잡도 표기를 누락하신 것 같네요. (의도하신 게 아니라면요)

Original file line number Diff line number Diff line change
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class Solution {
fun missingNumber(nums: IntArray): Int {
nums.sort()
var num = nums[0]
for (i in 1 until nums.size) {
if (nums[i] != (num + 1)) {
return num + 1
}
num = nums[i]
}
return num + 1
}
}
Comment on lines +1 to +13
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저는 nums[i]를 i와 비교하는 방식을 사용했는데, @gmlwls96 님처럼 이전 값과 현재 값의 차이를 비교하는 방식도 있다는 걸 배웠어요. 이 방식으로도 문제를 풀어봐야겠습니다 😊

28 changes: 28 additions & 0 deletions palindromic-substrings/gmlwls96.kt
Original file line number Diff line number Diff line change
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class Solution {
// 알고리즘 : brute-force
/** 풀이
* 1. 모든 substring을 전부 뽑아낸다.
* 2. 해당 substring이 palindrome인지 체크한다.
*/
// 시간 : O(n^3)
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깔끔한 풀이지만 실제 면접에서는 시간 복잡도에 대해서 챌린지를 받으실 수 있으실 것 같아요. 아직 마감까지 시간이 많으니 더 효율적인 알고리즘도 고민해보시면 좋을 것 같습니다.

fun countSubstrings(s: String): Int {
var count = 0
for (len in 1..s.length) { // 길이
for (i in 0..(s.length - len)) { // i : sub string start index.
if (checkPalindrome(s.substring(i, i + len))) {
count++
}
}
}
return count
}

private fun checkPalindrome(subStr: String): Boolean {
return if (subStr.length == 1) {
true
} else {
val reverse = subStr.reversed()
subStr == reverse
}
}
}
67 changes: 67 additions & 0 deletions word-search/gmlwls96.kt
Original file line number Diff line number Diff line change
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class Solution {

// 풀이 : dfs
// 시간 :O(m * n * 4^w), 공간 :O(m * n + w)
val movePos = arrayOf(
intArrayOf(-1, 0),
intArrayOf(0, -1),
intArrayOf(1, 0),
intArrayOf(0, 1)
)

fun exist(board: Array<CharArray>, word: String): Boolean {
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@DaleSeo DaleSeo Dec 31, 2024

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만약에 board[["a"]]로 주어지고, word"a"로 주어진다면 어떨까요?

for (y in board.indices) {
for (x in board[y].indices) {
if (existDfs(
board,
Array(board.size) { BooleanArray(board[it].size) },
word,
"",
y,
x
)
) {
return true
}
}
}
return false
}

private fun existDfs(
board: Array<CharArray>,
visit: Array<BooleanArray>,
findWord: String,
currentWord: String,
y: Int,
x: Int
): Boolean {
if (findWord == currentWord) return true
val findChar = findWord[currentWord.length]
if (board[y][x] == findChar) {
val newWord = currentWord + board[y][x]
visit[y][x] = true
for (pos in movePos) {
val newY = y + pos[0]
val newX = x + pos[1]
if (newY >= 0 && newX >= 0
&& newY < board.size
&& newX < board[newY].size
&& !visit[newY][newX]
&& existDfs(
board = board,
visit = visit,
findWord = findWord,
currentWord = newWord,
y = newY,
x = newX
)
) {
return true
}
}
visit[y][x] = false
}
return false
}
}
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