Leetcode 238

Author: Kondareddy1209

Leetcode/Leetcode_238.py

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+"""
+Problem: Product of Array Except Self
+
+Goal:
+Given an integer array nums, return an array answer such that:
+answer[i] = product of all elements except nums[i]
+
+⚠️ Constraint:
+- Do NOT use division
+- Must run in O(n)
+
+---------------------------------------------------
+🧠 HOW IT WORKS (Prefix + Postfix Concept)
+---------------------------------------------------
+
+Instead of division, we use two passes:
+
+👉 Prefix pass:
+For each index i, store product of all elements BEFORE i
+
+👉 Postfix pass:
+Multiply with product of all elements AFTER i
+
+Example:
+nums = [1, 2, 3, 4]
+
+Prefix:
+[1, 1, 2, 6]
+
+Postfix:
+[24, 12, 4, 1]
+
+Final Answer:
+[24, 12, 8, 6]
+
+---------------------------------------------------
+"""
+
+from typing import List
+import time
+
+
+# -------------------------------------------------
+# Approach 1: Prefix + Postfix (Optimal)
+# -------------------------------------------------
+class SolutionOptimal:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        result = [1] * n
+
+        # Step 1: Prefix pass
+        prefix = 1
+        for i in range(n):
+            result[i] = prefix
+            prefix *= nums[i]
+
+        # Step 2: Postfix pass
+        postfix = 1
+        for i in range(n - 1, -1, -1):
+            result[i] *= postfix
+            postfix *= nums[i]
+
+        return result
+
+
+# -------------------------------------------------
+# Approach 2: Brute Force (for understanding only)
+# -------------------------------------------------
+class SolutionBrute:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        n = len(nums)
+        result = []
+
+        for i in range(n):
+            prod = 1
+            for j in range(n):
+                if i != j:
+                    prod *= nums[j]
+            result.append(prod)
+
+        return result
+
+
+# -------------------------------------------------
+# Example + Runtime Measurement
+# -------------------------------------------------
+if __name__ == "__main__":
+    nums = [1, 2, 3, 4]
+
+    # Optimal Approach
+    start = time.time()
+    result1 = SolutionOptimal().productExceptSelf(nums)
+    end = time.time()
+    print("Optimal Result :", result1)
+    print("Time Taken     :", (end - start) * 1000, "ms\n")
+
+    # Brute Force Approach
+    start = time.time()
+    result2 = SolutionBrute().productExceptSelf(nums)
+    end = time.time()
+    print("Brute Result   :", result2)
+    print("Time Taken     :", (end - start) * 1000, "ms")
+
+
+"""
+---------------------------------------------------
+⏱️ TIME COMPLEXITY
+---------------------------------------------------
+
+Optimal Approach:
+- O(n) → two passes
+
+Brute Force:
+- O(n^2) → nested loops
+
+---------------------------------------------------
+💾 SPACE COMPLEXITY
+---------------------------------------------------
+
+Optimal:
+- O(1) extra space (excluding output array)
+
+Brute:
+- O(1)
+
+---------------------------------------------------
+⚡ NOTE ON RUNTIME (ms)
+---------------------------------------------------
+- Measured using time.time()
+- Depends on system performance
+- Optimal is MUCH faster than brute
+
+---------------------------------------------------
+🔥 FINAL TAKEAWAY
+---------------------------------------------------
+
+Prefix + Postfix Pattern = No division needed
+
+Used in:
+- Product problems
+- Range queries
+- Accumulation problems
+
+---------------------------------------------------
+"""