adding algo

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/common_algos/two_sum_round_12.py

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+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_12.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='[^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_4/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_4/jump_game_vii.py

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+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/top_150_questions_round_21/reverse_bits.py

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+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def reverseBits(self, n: int) -> int:
+        return int((('{0:032b}'.format(n))[::-1]),2)

tests/test_150_questions_round_21/test_reverse_bits_round_21.py

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+import unittest
+from src.my_project.interviews.top_150_questions_round_21\
+.reverse_bits import Solution
+
+class ReverseBitsTestCase(unittest.TestCase):
+
+    def test_reverse_bits(self):
+        solution = Solution()
+        output = solution.reverseBits(2)
+        target = 1073741824
+        self.assertEqual(output, target)