Merge pull request #2101 from hongseoupyun/main

[hongseoupyun] WEEK 03 Solutions

Author: TonyKim9401

climbing-stairs/hongseoupyun.py

@@ -20,7 +20,7 @@ def climbStairs(self, n: int) -> int:
 
         prev1 = 1
         prev2 = 2
-        for i in range(n,n+1):
+        for i in range(3,n+1):
             current = prev2 + prev1
             prev1 = prev2
             prev2 = current

combination-sum/hongseoupyun.py

@@ -0,0 +1,95 @@
+# // Input: candidates = [2,3,5], target = 8
+# // Output: [[2,2,2,2],[2,3,3],[3,5]]
+
+# // DFS/backtracking should be used - Find possilbe combinations till the end and backtrack to previous step to find other combinations
+# // For example, if we have candidates [2,3,5] and target 8
+# // Start with empty combination [], target 8
+# // Add 2 -> [2], remaining 6
+# // Add 2 -> [2,2], remaining 4
+# // Add 2 -> [2,2,2], remaining 2
+# // Add 2 -> [2,2,2,2], remaining 0 (found a valid combination)
+# // Backtrack to [2,2,2,3], remaining -1 (remaining is negative, backtrack)
+# // Backtrack to [2,2,2,5], remaining -3 (remaining is negative, backtrack)
+# // Backtrack to [2,2,3], remaining 1 (dead end, backtrack)
+# // Backtrack to [2,2,5], remaining -3 (remaining is negative, backtrack)
+# // Backtrack to [2,3], remaining 3
+# // Add 3 -> [2,3,3], remaining 0 (found a valid combination)
+# // so on ..
+
+# // When you dfs, always start from current index and allow reusing the same element or next index only to avoid duplicates / to get unique combination
+# // Hence, if we starts from 2, next dfs calls can start from index of 2 or index of 3, but not index of 5 directly
+# // Moreover, if we start from 3, next dfs calls can start from index of 3 or index of 5, but not index of 2 directly
+
+# backtrack(8, [], 0)
+# │
+# ├─ i = 0 pick 2 → backtrack(6, [2], 0)
+# │   │
+# │   ├─ i = 0 pick 2 → backtrack(4, [2,2], 0)
+# │   │   │
+# │   │   ├─ i = 0 pick 2 → backtrack(2, [2,2,2], 0)
+# │   │   │   │
+# │   │   │   ├─ i = 0 pick 2 → backtrack(0, [2,2,2,2], 0) 🟢 success → return
+# │   │   │   │
+# │   │   │   ├─ i = 1 pick 3 → backtrack(-1, [2,2,2,3], 1) 🔴 → return
+# │   │   │   │
+# │   │   │   └─ i = 2 pick 5 → backtrack(-3, [2,2,2,5], 2) 🔴 → return
+# │   │   │
+# │   │   └─ loop end → pop → combination = [2,2]
+# │   │
+# │   ├─ i = 1 pick 3 → backtrack(1, [2,2,3], 1)
+# │   │   │
+# │   │   ├─ i = 1 pick 3 → backtrack(0, [2,2,3,3], 1) 🟢 success → return
+# │   │   │
+# │   │   └─ i = 2 pick 5 → backtrack(-4, [2,2,3,5], 2) 🔴 → return
+# │   │
+# │   ├─ loop end → pop → [2,2]
+# │   │
+# │   └─ i = 2 pick 5 → backtrack(-1, [2,2,5], 2) 🔴 → return
+# │
+# ├─ loop end → pop → [2]
+# │
+# ├─ i = 1 pick 3 → backtrack(3, [2,3], 1)
+# │   │
+# │   ├─ i = 1 pick 3 → backtrack(0, [2,3,3], 1) 🟢 success → return
+# │   │
+# │   └─ i = 2 pick 5 → backtrack(-2, [2,3,5], 2) 🔴 → return
+# │
+# ├─ loop end → pop → [2]
+# │
+# ├─ i = 2 pick 5 → backtrack(1, [2,5], 2)
+# │       │
+# │       └─ i = 2 pick 5 → backtrack(-4, [2,5,5], 2) 🔴 → return
+# │
+# └─ loop end → pop → []
+
+
+
+
+from typing import List
+
+class Solution:
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        result: List[List[int]] = []
+
+        def backtrack(remaining: int, combination: List[int], start_index: int):
+            # base cases
+            if remaining == 0:
+                # append a COPY of combination
+                result.append(combination[:])
+                return
+            if remaining < 0:
+                return
+
+            # try candidates from start_index onward
+            for i in range(start_index, len(candidates)):
+                current_number = candidates[i]
+                # choose
+                combination.append(current_number)
+                # explore (i, not i+1, because we can reuse same element)
+                backtrack(remaining - current_number, combination, i)
+                # undo (backtrack)
+                combination.pop()
+
+        # initial call
+        backtrack(target, [], 0)
+        return result

combination-sum/hongseoupyun.ts

@@ -0,0 +1,100 @@
+// Input: candidates = [2,3,5], target = 8
+// Output: [[2,2,2,2],[2,3,3],[3,5]]
+
+// DFS/backtracking should be used - Find possilbe combinations till the end and backtrack to previous step to find other combinations
+// For example, if we have candidates [2,3,5] and target 8
+// Start with empty combination [], target 8
+// Add 2 -> [2], remaining 6
+// Add 2 -> [2,2], remaining 4
+// Add 2 -> [2,2,2], remaining 2
+// Add 2 -> [2,2,2,2], remaining 0 (found a valid combination)
+// Backtrack to [2,2,2,3], remaining -1 (remaining is negative, backtrack)
+// Backtrack to [2,2,2,5], remaining -3 (remaining is negative, backtrack)
+// Backtrack to [2,2,3], remaining 1 (dead end, backtrack)
+// Backtrack to [2,2,5], remaining -3 (remaining is negative, backtrack)
+// Backtrack to [2,3], remaining 3
+// Add 3 -> [2,3,3], remaining 0 (found a valid combination)
+// so on ..
+
+// When you dfs, always start from current index and allow reusing the same element or next index only to avoid duplicates / to get unique combination
+// Hence, if we starts from 2, next dfs calls can start from index of 2 or index of 3, but not index of 5 directly
+// Moreover, if we start from 3, next dfs calls can start from index of 3 or index of 5, but not index of 2 directly
+
+/*
+backtrack(8, [], 0)
+│
+├─ i = 0 pick 2 → backtrack(6, [2], 0)
+│   │
+│   ├─ i = 0 pick 2 → backtrack(4, [2,2], 0)
+│   │   │
+│   │   ├─ i = 0 pick 2 → backtrack(2, [2,2,2], 0)
+│   │   │   │
+│   │   │   ├─ i = 0 pick 2 → backtrack(0, [2,2,2,2], 0) 🟢 success → return
+│   │   │   │
+│   │   │   ├─ i = 1 pick 3 → backtrack(-1, [2,2,2,3], 1) 🔴 → return
+│   │   │   │
+│   │   │   └─ i = 2 pick 5 → backtrack(-3, [2,2,2,5], 2) 🔴 → return
+│   │   │
+│   │   └─ loop end → pop → combination = [2,2]
+│   │
+│   ├─ i = 1 pick 3 → backtrack(1, [2,2,3], 1)
+│   │   │
+│   │   ├─ i = 1 pick 3 → backtrack(0, [2,2,3,3], 1) 🟢 success → return
+│   │   │
+│   │   └─ i = 2 pick 5 → backtrack(-4, [2,2,3,5], 2) 🔴 → return
+│   │
+│   ├─ loop end → pop → [2,2]
+│   │
+│   └─ i = 2 pick 5 → backtrack(-1, [2,2,5], 2) 🔴 → return
+│
+├─ loop end → pop → [2]
+│
+├─ i = 1 pick 3 → backtrack(3, [2,3], 1)
+│   │
+│   ├─ i = 1 pick 3 → backtrack(0, [2,3,3], 1) 🟢 success → return
+│   │
+│   └─ i = 2 pick 5 → backtrack(-2, [2,3,5], 2) 🔴 → return
+│
+├─ loop end → pop → [2]
+│
+├─ i = 2 pick 5 → backtrack(1, [2,5], 2)
+│       │
+│       └─ i = 2 pick 5 → backtrack(-4, [2,5,5], 2) 🔴 → return
+│
+└─ loop end → pop → []
+*/
+
+
+function combinationSum(candidates: number[], target: number): number[][] {
+    const result: number[][] = []
+
+    function backtrack(remaining: number, combination: number[], startindex: number) {
+        // Remaining is target - sum of combination
+        // Base Case1: If remaining is 0, we found a valid combination -> add that combinations to result
+        if (remaining === 0) {
+            result.push([...combination]);
+            return;
+        }
+
+        // Base Case2: If remaining is negative, no valid combination -> backtrack
+        if (remaining < 0) {
+            return;
+        }
+
+        // Explore further by trying each candidate starting from startindex
+        // candidates = [2,3,5], target = 8
+        for (let i = startindex; i < candidates.length; i++) {
+            // Choose the current number
+            const currentNumber = candidates[i];
+            // Add the current number to the combination
+            combination.push(currentNumber);
+
+            // Explore further with updated remaining and current combination
+            backtrack(remaining - currentNumber, combination, i);
+            // Backtrack: remove the last added candidate to try next candidate
+            combination.pop();
+        }
+    }
+        backtrack(target, [], 0);
+        return result;
+};

valid-palindrome/hongseoupyun.py

@@ -0,0 +1,44 @@
+
+import re
+
+class Solution:
+# Palindrome is a word, phrase, number, or other sequences of characters that reads the same forward and backward (ignoring non-alphanumeric characters - letters and number).
+# the solution should first filter out non-alphanumeric characters and convert the string to lowercase.
+# Then, it should check if the cleaned string is equal to its reverse.
+
+# Time complexity = O(n), as it checks, replaces, converts to lowercase, and reverses all characters in the string
+# Space complexity = O(n), as it creates a new string with a maximum length equal to the original string
+    def isPalindrome(self, s: str) -> bool:
+        filtered_string = re.sub(r'[^a-zA-Z0-9]','',s).lower()
+        reversed_string = filtered_string[::-1]
+        return reversed_string == filtered_string
+
+
+# Using Two-pointer; It searches from given string directly so the Space complexity is O(n) - better method
+# By using two index, searchs and compares from start and end of the string like this -><- 
+# Do this while start index is less than end index
+# Skip non-alphanumeric characters
+# If both pointer are in characters, then convert to lowercase and compare
+# If not equal, return false
+# If all characters are equal, return true
+
+# Time complexity = O(n), as it checks and converts to lowercase all characters in the string
+# Space complexity = O(1), as it uses only a constant amount of extra space
+    def isPalindromeTwoPointer(self, s: str) -> bool:
+        left = 0
+        right = len(s) - 1
+
+        while left < right:
+            if not s[left].isalnum():
+                left+=1
+                continue
+            if not s[right].isalnum():
+                right-=1
+                continue
+            if s[left].lower() != s[right].lower():
+                return False
+            
+            left+=1
+            right-=1
+
+        return True

valid-palindrome/hongseoupyun.ts

@@ -0,0 +1,56 @@
+// # Palindrome is a word, phrase, number, or other sequences of characters that reads the same forward and backward (ignoring non-alphanumeric characters - letters and number).
+// # the solution should first filter out non-alphanumeric characters and convert the string to lowercase.
+// # Then, it should check if the cleaned string is equal to its reverse.
+
+
+// Time complexity = O(n), as it checks, replaces, converts to lowercase, and reverses all characters in the string
+// Space complexity = O(n), as it creates a new string with a maximum length equal to the original string
+function isPalindrome(s: string): boolean {
+    let filteredString: string = s.replace(/[^a-zA-Z0-9]/g, '').toLowerCase();
+    let reversed: string = filteredString.split("").reverse().join("")
+
+    if (filteredString === reversed) {
+        return true
+    } else {
+        return false
+    }
+};
+
+//Palindrom example: "A man, a plan, a canal: Panama"
+
+// Using Two-pointer; It searches from given string directly so the Space complexity is O(n) - better method
+// By using two index, searchs and compares from start and end of the string like this-><- 
+// Do this while start index is less than end index
+// Skip non-alphanumeric characters
+// If both pointer are in characters, then convert to lowercase and compare
+// If not equal, return false
+// If all characters are equal, return true
+
+// Time complexity = O(n), as it checks and converts to lowercase all characters in the string
+// Space complexity = O(1), as it uses only a constant amount of extra space
+function isPalindrome2(s: string): boolean {
+
+    let left: number = 0;
+    let right: number = s.length - 1;
+
+    let isAlphaNum = (str: string): boolean => {
+        return /^[a-zA-Z0-9]$/.test(str)
+    }
+
+    while (left < right) {
+        if (!isAlphaNum(s[left])) {
+            left++
+            continue
+        }
+        if (!isAlphaNum(s[right])) {
+            right--
+            continue
+        }
+        if (s[left].toLowerCase() !== s[right].toLowerCase()) {
+            return false
+        }
+        left++
+        right--
+    }
+     return true
+};