Merge branch 'DaleStudy:main' into main

Author: GotPrgmer

.github/workflows/integration.yaml

@@ -85,7 +85,7 @@ jobs:
       - name: Check filename rules
         if: ${{ steps.pr-labels.outputs.has_maintenance != 'true' }}
         run: |
-          files=$(git diff --name-only ${{ github.event.pull_request.base.sha }} ${{ github.sha }} | tr -d '"')
+          files=$(git diff --name-only ${{ github.event.pull_request.base.sha }} ${{ github.event.pull_request.head.sha }} | tr -d '"')
           pr_author="${{ github.event.pull_request.user.login }}"
           success=true
 

contains-duplicate/5YoonCheol.java

@@ -0,0 +1,16 @@
+class Solution {
+    /**
+     *   시간 복잡도: O(N)
+     *   공간 복잡도: O(N)
+     */
+    public boolean containsDuplicate(int[] nums) {
+        Set<Integer> set = new HashSet<>();
+
+        for (int num : nums) {
+            if (set.contains(num)) return true;
+            set.add(num);
+        }
+
+        return false;
+    }
+}

contains-duplicate/JisooPyo.kt

@@ -0,0 +1,38 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+
+/**
+ * Leetcode
+ * 217. Contains Duplicate
+ * Easy
+ */
+class ContainsDuplicate {
+    /**
+     * Runtime: 17 ms(Beats: 80.99 %)
+     * Time Complexity: O(n)
+     *   - 배열 순회
+     *
+     * Memory: 50.63 MB(Beats: 70.32 %)
+     * Space Complexity: O(n)
+     *   - HashSet에 최악의 경우 배열 원소 모두 저장
+     */
+    fun containsDuplicate(nums: IntArray): Boolean {
+        val set = hashSetOf<Int>()
+        for (i in nums) {
+            if (set.contains(i)) {
+                return true
+            }
+            set.add(i)
+        }
+        return false
+    }
+
+    @Test
+    fun test() {
+        containsDuplicate(intArrayOf(1, 2, 3, 1)) shouldBe true
+        containsDuplicate(intArrayOf(1, 2, 3, 4)) shouldBe false
+        containsDuplicate(intArrayOf(1, 1, 1, 3, 3, 4, 3, 2, 4, 2)) shouldBe true
+    }
+}

contains-duplicate/dusunax.py

@@ -0,0 +1,23 @@
+'''
+# Leetcode 217. Contains Duplicate
+
+use set to store distinct elements 🗂️
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to convert it to a set.
+
+### SC is O(n):
+- creating a set to store the distinct elements of the list.
+'''
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        return len(nums) != len(set(nums))
+

contains-duplicate/jeldo.py

@@ -0,0 +1,4 @@
+class Solution:
+    # O(n)
+    def containsDuplicate(self, nums: list[int]) -> bool:
+        return len(nums) != len(set(nums))  # O(n)

house-robber/5YoonCheol.java

@@ -0,0 +1,19 @@
+class Solution {
+    public int rob(int[] nums) {
+        //배열 길이 0이면 털 수 있는 집이 없음.
+        if (nums.length == 0) return 0;
+        //배열 길이가 1이면 한 집만 털 수 있음.
+        if (nums.length == 1) return nums[0];
+
+        //동적 계획법으로 풀이
+        int[] dp = new int[nums.length];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+
+        //배열 크기가 2이상일 경우 최대 금액의 범위 확장
+        for (int i = 2; i < nums.length; i++) {
+            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
+        }
+        return dp[nums.length - 1];
+    }
+}

house-robber/JisooPyo.kt

@@ -0,0 +1,74 @@
+package leetcode_study
+
+import io.kotest.matchers.shouldBe
+import org.junit.jupiter.api.Test
+import kotlin.math.max
+
+/**
+ * Leetcode
+ * 198. House Robber
+ * Medium
+ *
+ * 사용된 알고리즘: Dynamic Programming
+ *
+ * i번째 집에서 얻을 수 있는 최대 금액은 다음 두 가지 중 큰 값입니다.
+ *   - (i-2)번째 집까지의 최대 금액 + 현재 집의 금액
+ *   - (i-1)번째 집까지의 최대 금액
+ */
+class HouseRobber {
+    /**
+     * Runtime: 0 ms(Beats: 100.00 %)
+     * Time Complexity: O(n)
+     *
+     * Memory: 34.65 MB(Beats: 40.50 %)
+     * Space Complexity: O(n)
+     */
+    fun rob(nums: IntArray): Int {
+        if (nums.size == 1) {
+            return nums[0]
+        }
+        if (nums.size == 2) {
+            return max(nums[0], nums[1])
+        }
+        val dp = IntArray(nums.size)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for (i in 2 until nums.size) {
+            dp[i] = max(dp[i - 2] + nums[i], dp[i - 1])
+        }
+        return dp[nums.size - 1]
+    }
+
+    /**
+     * 공간 복잡도를 개선
+     * Runtime: 0 ms(Beats: 100.00 %)
+     * Time Complexity: O(n)
+     *
+     * Memory: 34.95 MB(Beats: 36.98 %)
+     * Space Complexity: O(1)
+     */
+    fun rob2(nums: IntArray): Int {
+        if (nums.size == 1) return nums[0]
+        if (nums.size == 2) return max(nums[0], nums[1])
+
+        var twoBack = nums[0]
+        var oneBack = max(nums[0], nums[1])
+        var current = oneBack
+
+        for (i in 2 until nums.size) {
+            current = max(twoBack + nums[i], oneBack)
+            twoBack = oneBack
+            oneBack = current
+        }
+
+        return current
+    }
+
+    @Test
+    fun test() {
+        rob(intArrayOf(1, 2, 3, 1)) shouldBe 4
+        rob(intArrayOf(2, 7, 9, 3, 1)) shouldBe 12
+        rob2(intArrayOf(1, 2, 3, 1)) shouldBe 4
+        rob2(intArrayOf(2, 7, 9, 3, 1)) shouldBe 12
+    }
+}

house-robber/dusunax.py

@@ -0,0 +1,44 @@
+'''
+# Leetcode 198. House Robber
+
+use **dynamic programming** to solve this problem. (bottom-up approach) 🧩
+
+choose bottom-up approach for less space complexity.
+
+## DP relation
+
+```
+dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+```
+
+- **dp[i - 1]:** skip and take the value from the previous house
+- **dp[i - 2]:** rob the current house, add its value to the maximum money from two houses before
+
+## Time and Space Complexity
+
+```
+TC: O(n)
+SC: O(n)
+```
+
+### TC is O(n):
+- iterating through the list just once to calculate the maximum money. = O(n)
+
+### SC is O(n):
+- using a list to store the maximum money at each house. = O(n)
+
+'''
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 1:
+            return nums[0]
+
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+
+        for i in range(2, len(nums)):
+            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])
+            
+        return dp[-1]

house-robber/jeldo.py

@@ -0,0 +1,9 @@
+class Solution:
+    # O(n)
+    def rob(self, nums: list[int]) -> int:
+        if len(nums) <= 2:
+            return max(nums)
+        nums[2] += nums[0]
+        for i in range(3, len(nums)):
+            nums[i] += max(nums[i-3], nums[i-2])
+        return max(nums[-1], nums[-2])

longest-consecutive-sequence/5YoonCheol.java

@@ -0,0 +1,34 @@
+import java.util.*;
+
+class Solution {
+    public int longestConsecutive(int[] nums) {
+        if (nums == null || nums.length == 0) return 0;
+
+        //모든 요소 HashSet 삽입
+        HashSet<Integer> set = new HashSet<>();
+        for (int num : nums) {
+            set.add(num);
+        }
+
+        int longest = 0;
+
+        // 시작지점 체크
+        for (int num : nums) {
+            //배열 요소보다 1 작은 수 가 없는 경우 새로운 시작 지점이 됨
+            if (!set.contains(num - 1)) {
+                int start = num;
+                int currentLength = 1;
+
+                // 1씩 증가시키면서 연속된 수의 개수 탐색
+                while (set.contains(start + 1)) {
+                    start++;
+                    currentLength++;
+                }
+                // 기존 longest와 현재 연속된 수를 비교
+                longest = Math.max(longest, currentLength);
+            }
+        }
+
+        return longest;
+    }
+}