Merge pull request #925 from jungsiroo/main

[jungsiroo] Week 7

Author: TonyKim9401

longest-substring-without-repeating-characters/jungsiroo.py

@@ -0,0 +1,31 @@
+class Solution:
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        """
+        딕셔너리를 사용하여 이전에 나왔던 인덱스를 기억
+        중복되지 않았다면 answer를 저장해 나감
+        중복이 발생하고 begin과 end사이라면 중복 발생 원인 다음을 begin으로 세팅 후 다시 계산
+
+        Time Complexity : O(n)
+        Space Complexity : O(n)
+
+        """ 
+        answer, begin, end = 0, 0, 0
+        hash_map = dict()
+
+        while end < len(s):
+            ch = s[end]
+            hash_map[ch] = hash_map.get(ch, -1)
+
+            if hash_map[ch] == -1: #한 번도 나오지 않았다.
+                hash_map[ch] = end
+                answer = max(answer, end-begin+1)
+            else:
+                if begin<=hash_map[ch]<end: #중복이 생겼다.
+                    begin = hash_map[ch]+1
+                else:
+                    answer = max(answer, end-begin+1)
+                hash_map[ch] = end
+            end += 1
+        
+        return answer
+

number-of-islands/jungsiroo.py

@@ -0,0 +1,50 @@
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+		# TC : O(n*m)
+		# SC : O(n*m)
+
+        m, n = len(grid), len(grid[0])
+        dx = [-1,1,0,0]
+        dy = [0,0,-1,1]
+
+        def in_range(r, c):
+            if r<0 or c<0 or r>=m or c>=n:
+                return False
+            return True
+
+        # DFS Way
+
+        def dfs(r, c):
+            grid[r][c] = 'x'
+
+            for i in range(4):
+                nr, nc = r+dx[i], c+dy[i]
+                if not in_range(nr, nc) or grid[nr][nc] != '1':
+                    continue
+                dfs(nr, nc)
+        
+        # BFS Way
+        from collections import deque
+        def bfs(r, c):
+            grid[r][c] = 'x'
+            queue = deque()
+            queue.append([r, c])
+
+            while queue:
+                cr, cc = queue.popleft()
+                for i in range(4):
+                    nr, nc = cr+dx[i], cc+dy[i]
+                    
+                    if not in_range(nr, nc) or grid[nr][nc] != '1':
+                        continue
+                    grid[nr][nc] = 'x'
+                    queue.append([nr, nc])
+
+        ret = 0
+        for i in range(m):
+            for j in range(n):
+                if grid[i][j] == '1':
+                    bfs(i, j)
+                    ret += 1
+        return ret
+

reverse-linked-list/jungsiroo.py

@@ -0,0 +1,29 @@
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        # iterative way
+
+        """
+        prev, curr = None, head
+        while curr:
+            tmp_nxt = curr.next
+            
+            curr.next = prev
+            prev, curr = curr, tmp_nxt
+
+        return prev
+        """
+
+        # Recursive Way
+        if head is None or head.next is None:
+            return head
+
+        new_head = self.reverseList(head.next)
+        head.next.next = head # reversing pointer
+        head.next = None
+        return new_head
+
+        # 둘 다 시간복잡도 O(n)
+        # 하지만 재귀의 경우 콜스택에 따른 공간복잡도 O(n)을 소요
+        # iterative 방식은 O(1)
+                
+            

set-matrix-zeroes/jungsiroo.py

@@ -0,0 +1,43 @@
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+
+        """
+        # Naive change : save rows and cols
+        # SC : O(m+n)
+        row_zero, col_zero = set(), set()
+        
+        rows, cols = len(matrix), len(matrix[0])
+        for i in range(rows):
+            for j in range(cols):
+                if matrix[i][j] == 0:
+                    row_zero.add(i)
+                    col_zero.add(j)
+        
+        for i in range(rows):
+            for j in range(cols):
+                if i in row_zero or j in col_zero:
+                    matrix[i][j] = 0
+        """
+
+        # Constant Space Complexity using bitmasking
+		# 0인 구간을 toggle 시켜놓고 확인하는 방법
+        def is_on(number, k):
+            return (number & (1<<k)) != 0
+        
+        row_zero, col_zero = 0, 0
+        
+        rows, cols = len(matrix), len(matrix[0])
+        for i in range(rows):
+            for j in range(cols):
+                if matrix[i][j] == 0:
+                    row_zero |= (1 << i)
+                    col_zero |= (1 << j)
+        
+        for i in range(rows):
+            for j in range(cols):
+                if is_on(row_zero, i) or is_on(col_zero, j):
+                    matrix[i][j] = 0
+        

unique-paths/jungsiroo.py

@@ -0,0 +1,15 @@
+from math import comb 
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # mathematic way
+        # 중학교 때 배운 최단거리 조합으로 구하기 문제
+        # 문제 자체가 오른쪽, 아래밖에 못가기에 최단거리가 될 수 밖에 없음
+
+        """
+        TC : O(max(m, n))
+        SC : O(1)
+        """
+
+        return comb(m+n-2, m-1)
+