Merge pull request #708 from jungsiroo/main

[jungsiroo] Week2

Author: SamTheKorean

3sum/jungsiroo.py

@@ -0,0 +1,41 @@
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        # Naive Solution
+        # Tc : O(n^3) / Sc : O(nC_3)
+        
+        n = len(nums)
+        """
+        for i in range(n-2):
+            for j in range(i+1,n-1):
+                for k in range(j+1, n):
+                    if nums[i]+nums[j]+nums[k] == 0:
+                        ret.add(tuple(sorted([nums[i], nums[j], nums[k]])))
+
+        ret = [x for x in ret]
+        return ret 
+        """
+
+        # Better Solution
+        # two-sum question with fixed num (traversal in for loop)
+        # Tc : O(n^2) / Sc : O(n)
+        ret = []
+        nums.sort()
+
+        for i in range(n):
+            if i>0 and nums[i] == nums[i-1]:
+                continue
+            j, k = i+1, n-1
+
+            while j < k:
+                sum_ = nums[i] + nums[j] + nums[k]
+                if sum_ < 0 : j += 1
+                elif sum_ > 0 : k -= 1
+                else:
+                    ret.append([nums[i], nums[j], nums[k]])
+                    j += 1
+                
+                    while nums[j] == nums[j-1] and j<k:
+                        j += 1
+            
+        return ret
+

climbing-stairs/jungsiroo.py

@@ -0,0 +1,18 @@
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        """
+        dynamic programming
+        dp[0] = 1
+        dp[1] = 1
+        dp[2] = dp[0] + dp[1] = 2
+        dp[3] = dp[1] + dp[2] = 3
+        ...
+
+        Tc = O(n) / Sc = O(n)
+        """
+
+        dp = [1 for _ in range(n+1)]
+        for i in range(2, n+1):
+            dp[i] = dp[i-2] + dp[i-1]
+        return dp[n]
+        

decode-ways/jungsiroo.py

@@ -0,0 +1,26 @@
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        """
+        dp 이용 : 가능한 경우의 수를 구해놓고 그 뒤에 붙을 수 있는가?
+        ex) 1234
+            (1,2) -> (1,2,3)
+            (12) -> (12,3)
+            (1) -> (1,23)
+		
+		Tc : O(n) / Sc : O(n)
+        """
+        if s[0] == '0': return 0
+
+        n = len(s)
+        dp = [0]*(n+1)
+        dp[0] = dp[1] = 1
+
+        for i in range(2, n+1):
+            one = int(s[i-1])
+            two = int(s[i-2:i])
+
+            if 1<=one<=9: dp[i] += dp[i-1]
+            if 10<=two<=26 : dp[i] += dp[i-2]
+
+        return dp[n]
+        

valid-anagram/jungsiroo.py

@@ -0,0 +1,30 @@
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        # sort and compare
+        # Tc : O(nlogn) / Sc : O(n)(Because of python timsort)
+        """
+        s = sorted(s)
+        t = sorted(t)
+        for s_char, t_char in zip(s, t):
+            if s_char != t_char:
+                return False
+        return True
+        """
+
+        # dictionary to count letters
+        # Tc : O(n) / Sc : O(n)
+        letters_cnt = dict()
+        INF = int(1e6)
+        for s_char in s:
+            letters_cnt[s_char] = letters_cnt.get(s_char,0)+1
+        
+        for t_char in t:
+            letters_cnt[t_char] = letters_cnt.get(t_char,INF)-1
+            if letters_cnt[t_char] < 0:
+                return False
+
+        return sum(letters_cnt.values()) == 0
+