Merge pull request #1533 from ivanpenaloza/january31

adding algo

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/common_algos/two_sum_round_2.py

@@ -0,0 +1,16 @@
+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+
+        answer = dict()
+
+        for k, v in enumerate(nums):
+
+            if v in answer:
+                return [answer[v], k]
+            else: 
+                answer[target - v] = k
+
+        return []

src/my_project/interviews/amazon_high_frequency_23/common_algos/valid_palindrome_round_2.py

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+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+import re
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+
+        # To lowercase
+        s = s.lower()
+
+        # Remove non-alphanumeric characters
+        s = re.sub(pattern=r'[^a-zA-Z0-9]', repl='', string=s)
+
+        # Determine if s is palindrome or not
+
+        len_s = len(s)
+
+        for i in range(len_s//2):
+            
+            if s[i] != s[len_s - 1 - i]:
+                return False 
+            
+        return True 

src/my_project/interviews/amazon_high_frequency_23/round_5/jump_game_ii.py

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+from typing import List 
+
+class Solution:
+    def jump(self, nums: List[int]) -> int:
+        """
+        Greedy approach: At each position, jump to the farthest reachable index
+        
+        Example: [2,3,1,1,4]
+        - From index 0 (value=2): can reach indices 1,2
+        - Greedy choice: Jump to index 1 (value=3) because it reaches farthest
+        - From index 1: can reach indices 2,3,4 (end)
+        - Answer: 2 jumps
+        """
+
+        if len(nums) <= 1:
+            return 0
+        
+        jumps = 0
+        current_end = 0 # End of current jump range
+        farthest = 0    # The farthest position we can reach
+
+        for i in range(len(nums) - 1):
+            # Update farthest position reachable
+            farthest = max(farthest, i + nums[i])
+
+            # If we've reached the end of current jump range
+            if i == current_end:
+                jumps += 1
+                current_end = farthest  # Make the greedy choice
+
+                if current_end > len(nums) - 1:
+                    break
+        return jumps

src/my_project/interviews/amazon_high_frequency_23/round_5/jump_game_vii.py

@@ -0,0 +1,61 @@
+from typing import List, Union, Collection, Mapping, Optional
+
+
+class Solution:
+    def minCost(self, nums: List[int], costs: List[int]) -> int:
+        """
+        Example: nums = [3, 1, 2, 4], costs = [1, 2, 3, 4]
+        
+        From index 0 (value=3):
+          - Next smaller: index 1 (value=1)
+          - Next larger: index 3 (value=4)
+        
+        DP builds cost backwards:
+        - dp[3] = 0 (at end, no cost)
+        - dp[2] = dp[3] + costs[3] = 0 + 4 = 4 (can jump to 4)
+        - dp[1] = dp[2] + costs[2] = 4 + 3 = 7 (can jump to 2)
+        - dp[0] = min(dp[1]+costs[1], dp[3]+costs[3]) = min(7+2, 0+4) = 4
+        """
+        
+        smallStack = []  # Monotonic increasing (next smaller element)
+        largeStack = []  # Monotonic decreasing (next larger element)
+        dp = [0] * len(nums)  # dp[i] = min cost from i to end
+        
+        # Process backwards (right to left)
+        for i in range(len(nums) - 1, -1, -1):
+            
+            # Maintain monotonic increasing stack (for next smaller)
+            # Remove elements >= current (they can't be "next smaller")
+            while smallStack and nums[smallStack[-1]] >= nums[i]:
+                smallStack.pop()
+            
+            # Maintain monotonic decreasing stack (for next larger)
+            # Remove elements < current (they can't be "next larger")
+            while largeStack and nums[largeStack[-1]] < nums[i]:
+                largeStack.pop()
+            
+            # Calculate minimum cost for this position
+            nxtCost = []
+            
+            if largeStack:
+                lid = largeStack[-1]  # Next larger index
+                nxtCost.append(dp[lid] + costs[lid])
+            
+            if smallStack:
+                sid = smallStack[-1]  # Next smaller index
+                nxtCost.append(dp[sid] + costs[sid])
+            
+            # Min cost from current position to end
+            dp[i] = min(nxtCost) if nxtCost else 0
+            
+            # Add current index to both stacks
+            largeStack.append(i)
+            smallStack.append(i)
+
+        print(smallStack)
+        print(largeStack)
+        
+
+        print(dp)
+        
+        return dp[0]  # Minimum cost from start

src/my_project/interviews/amazon_high_frequency_23/round_5/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
+
+'''