Merge pull request #1718 from ivanpenaloza/august02

adding updats

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/round_7/k_free_subsets.py

@@ -0,0 +1,111 @@
+from typing import List, Union, Collection, Mapping, Optional
+from collections import defaultdict
+
+class Solution:
+    def countTheNumOfKFreeSubsets(self, nums: List[int], k: int) -> int:
+        """
+        Count k-Free subsets using dynamic programming.
+        
+        Approach:
+        1. Group elements by (num % k) to find independent groups
+        2. Within each group, sort and build chains where elements differ by k
+        3. For each chain, use House Robber DP to count valid subsets
+        4. Multiply results across all independent chains
+        
+        Time: O(n log n), Space: O(n)
+        """
+        # Group numbers by their remainder when divided by k
+        groups = defaultdict(list)
+        for num in nums:
+            groups[num % k].append(num)
+        
+        res = 1
+        
+        # Process each group independently
+        for group in groups.values():
+            group.sort()
+            
+            # Build chains within this group
+            i = 0
+            while i < len(group):
+                chain = [group[i]]
+                j = i + 1
+                
+                # Build chain where each element is exactly k more than previous
+                while j < len(group) and group[j] == chain[-1] + k:
+                    chain.append(group[j])
+                    j += 1
+                
+                # House Robber DP for this chain
+                m = len(chain)
+                if m == 1:
+                    chain_res = 2  # {} or {chain[0]}
+                else:
+                    take = 1  # Take first element
+                    skip = 1  # Skip first element
+                    
+                    for idx in range(1, m):
+                        new_take = skip  # Can only take current if we skipped previous
+                        new_skip = take + skip  # Can skip current regardless
+                        take, skip = new_take, new_skip
+                    
+                    chain_res = take + skip
+                
+                res *= chain_res
+                i = j
+        
+        return res
+
+
+
+    
+
+'''
+Detailed Algorithm Explanation
+Part 1: Why Group by num % k?
+Two numbers can have a difference of exactly k only if they have the same remainder when divided by k.
+
+Mathematical proof:
+
+If a - b = k, then a = b + k
+Therefore: a % k = (b + k) % k = b % k
+Example: nums = [2, 3, 5, 8], k = 5
+
+num | num % 5 | group
+----|---------|-------
+2   |    2    | Group A
+3   |    3    | Group B  
+5   |    0    | Group C
+8   |    3    | Group B
+
+
+Why this matters: Elements from different groups can never differ by k, so they're independent. We can combine any subset from Group A with any subset from Group B.
+
+Part 2: Building Chains
+Within each group, we sort and find chains where consecutive elements differ by exactly k.
+
+Example with Group B: [3, 8]
+
+Sorted: [3, 8]
+Check: 8 - 3 = 5 ✓
+Chain: 3 → 8
+
+Another example: nums = [1, 6, 11, 21], k = 5 (all have remainder 1)
+
+Sorted: [1, 6, 11, 21]
+Check: 6-1=5 ✓, 11-6=5 ✓, 21-11=10 ✗
+Chains: [1 → 6 → 11], [21]
+
+
+Part 3: House Robber DP - The Core Logic
+For a chain like [3 → 8], we can't pick both 3 and 8 (they differ by k). This is the House Robber problem: count all subsets where we don't pick adjacent elements.
+
+DP State Variables
+take = number of valid subsets that INCLUDE the current element
+skip = number of valid subsets that EXCLUDE the current element
+
+DP Transitions
+new_take = skip      # To take current, we MUST have skipped previous
+new_skip = take + skip  # To skip current, we can take or skip previous
+
+'''    

src/my_project/interviews/amazon_high_frequency_23/round_7/longest_common_subsequence.py

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+from typing import List, Union, Collection, Mapping, Optional
+
+class Solution:
+    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
+
+        # Make a grid of 0's with len(text2) + 1 columns 
+        # and len(text1) + 1 rows.
+        len_1 = len(text1)
+        len_2 = len(text2)
+        dp_grid = [[0]*(len_2+1) for _ in range(len_1+1)]
+
+        # Iterate up each column, starting from the last one. 
+        for j in reversed(range(len_2)):
+            for i in reversed(range(len_1)):
+                if text1[i] == text2[j]:
+                    dp_grid[i][j] = dp_grid[i+1][j+1] + 1
+                else:
+                    dp_grid[i][j] = max(dp_grid[i+1][j], dp_grid[i][j+1])
+
+        # The original problem's answer is in dp_grid[0][0]. Return it.  
+        return dp_grid[0][0]

src/my_project/interviews/google_top_exercises/round_1/28_alien_dictionary.py

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+from typing import List
+from collections import deque, defaultdict
+
+class Solution:
+    def alienOrder(self, words: List[str]) -> str:
+        """Topological sort (Kahn's BFS) over letters.
+
+        Each adjacent pair of words gives at most one ordering fact: the first
+        position where they differ tells us c1 comes before c2. Everything after
+        that position is unconstrained, so we stop comparing there. The invalid
+        case is a prefix violation - ["abc", "ab"] - since a proper prefix must
+        sort first no matter what the alphabet is.
+
+        Time O(C) where C is total characters, space O(1) - at most 26 nodes and
+        26 * 26 edges.
+        """
+
+        adj = {c: set() for word in words for c in word}
+        indegree = {c: 0 for c in adj}
+
+        for first, second in zip(words, words[1:]):
+            for c1, c2 in zip(first, second):
+                if c1 != c2:
+                    if c2 not in adj[c1]:
+                        adj[c1].add(c2)
+                        indegree[c2] += 1
+                    break
+            else:
+                # No differing character: second must not be a strict prefix
+                if len(second) < len(first):
+                    return ""
+
+        q = deque([c for c in indegree if indegree[c] == 0])
+        order = []
+
+        while q:
+            c = q.popleft()
+            order.append(c)
+
+            for nxt in adj[c]:
+                indegree[nxt] -= 1
+                if indegree[nxt] == 0:
+                    q.append(nxt)
+
+        # A leftover letter means it sits on a cycle
+        return "".join(order) if len(order) == len(adj) else ""
+
+
+    def alienOrder_DFS(self, words: List[str]) -> str:
+        """Same graph, post-order DFS with cycle detection.
+
+        visiting[c] is True while c is on the current recursion stack (cycle) and
+        False once it is fully processed. Post-order emits a letter only after all
+        of its successors, so the reversed result is the topological order.
+        """
+
+        adj = defaultdict(set)
+        letters = {c for word in words for c in word}
+
+        for first, second in zip(words, words[1:]):
+            for c1, c2 in zip(first, second):
+                if c1 != c2:
+                    adj[c1].add(c2)
+                    break
+            else:
+                if len(second) < len(first):
+                    return ""
+
+        visiting = {}
+        order = []
+
+        def dfs(c: str) -> bool:
+            """Returns True if a cycle is reached from c."""
+            if c in visiting:
+                return visiting[c]
+
+            visiting[c] = True
+            for nxt in adj[c]:
+                if dfs(nxt):
+                    return True
+
+            visiting[c] = False
+            order.append(c)
+            return False
+
+        for c in letters:
+            if dfs(c):
+                return ""
+
+        return "".join(reversed(order))