adding algo

Author: ivanpenaloza

src/my_project/interviews/amazon_high_frequency_23/round_2/minimum_number_of_keypresses.py

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+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+from collections import Counter
+
+class Solution:
+    def minimumKeypresses(self, s: str) -> int:
+        """
+        Greedy approach: Assign most frequent characters to 1-press slots
+        
+        Example: s = "apple"
+        Frequency: {'p': 2, 'a': 1, 'l': 1, 'e': 1}
+        
+        Assignment:
+        - 'p' (freq=2) → position 0 → 1 press → total: 2 * 1 = 2
+        - 'a' (freq=1) → position 1 → 1 press → total: 1 * 1 = 1
+        - 'l' (freq=1) → position 2 → 1 press → total: 1 * 1 = 1
+        - 'e' (freq=1) → position 3 → 1 press → total: 1 * 1 = 1
+        
+        Total: 2 + 1 + 1 + 1 = 5
+        """
+        
+        # Count character frequencies
+        freq = Counter(s)
+        print(freq)
+        
+        # Sort by frequency (descending) - greedy choice
+        counts = sorted(freq.values(), reverse=True)
+        print(counts)
+        
+        total_presses = 0
+        
+        # Calculate presses based on position
+        for position, frequency in enumerate(counts):
+            # Positions 0-8: 1 press (9 slots)
+            # Positions 9-17: 2 presses (9 slots)
+            # Positions 18-26: 3 presses (9 slots)
+            presses_per_char = (position // 9) + 1
+            total_presses += presses_per_char * frequency
+        
+        return total_presses        
+            

src/my_project/interviews/amazon_high_frequency_23/round_2/number_of_ways_to_select_building.py

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+from typing import List, Union, Collection, Mapping, Optional
+from abc import ABC, abstractmethod
+
+class Solution:
+    def numberOfWays(self, s: str) -> int:
+        """
+        Count valid 3-building selections forming "010" or "101" patterns.
+        
+        Key insight: Track how many 2-character patterns we can form,
+        then extend them with the appropriate third character.
+        
+        Time: O(n), Space: O(1)
+        """
+        count_0 = 0  # Count of '0' seen so far
+        count_1 = 0  # Count of '1' seen so far
+        count_01 = 0  # Count of "01" patterns (0 followed by 1)
+        count_10 = 0  # Count of "10" patterns (1 followed by 0)
+        
+        result = 0
+        
+        for char in s:
+            if char == '0':
+                # Current '0' can complete "010": we need "01" before this '0'
+                result += count_01
+                
+                # Current '0' can start new "01" patterns: pair with each previous '0'
+                # Wait no, "01" means 0 then 1, so we can't create "01" with another 0
+                
+                # Current '0' extends all previous '1' to form "10" pattern
+                count_10 += count_1
+                
+                count_0 += 1
+            else:  # char == '1'
+                # Current '1' can complete "101": we need "10" before this '1'
+                result += count_10
+                
+                # Current '1' extends all previous '0' to form "01" pattern
+                count_01 += count_0
+                
+                count_1 += 1
+        
+        return result