adding algo

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

@@ -0,0 +1,41 @@
+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(s)
+
+
+        # 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