From 3e5c6c9ecd577bc20346c42421d83cc59c8774ea Mon Sep 17 00:00:00 2001
From: Nihhaar Saini <msnrk2006@hmail.com>
Date: Sun, 23 Nov 2025 11:05:42 +0530
Subject: [PATCH] Enhance Two Sum Python solution with explanations and
 formatting

---
 Python/1_TwoSum.py | 112 ++++++++++++++++++++++++++++++---------------
 1 file changed, 75 insertions(+), 37 deletions(-)

diff --git a/Python/1_TwoSum.py b/Python/1_TwoSum.py
index cb9359c9..1b16da26 100644
--- a/Python/1_TwoSum.py
+++ b/Python/1_TwoSum.py
@@ -1,56 +1,94 @@
-#Difficulty = Easy
-#Submission Speed = 64.69% 
+from typing import List
+
+# Difficulty = Easy
+# Submission Speed ≈ 64.69%
+
 '''
-Solution-1:
-Example: nums = [3,2,1,4]       target = 6
-Step1) Create a dictionary and populate it with elements of nums as the key and their corresponding index as values.
-        
+Solution-1: Two-pass Hash Table
+
+Example:
+    nums   = [3, 2, 1, 4]
+    target = 6
+
+Step 1)
+Create a dictionary and populate it with elements of nums as the key
+and their corresponding index as values.
+
     d = {
-            3:0,
-            2:1
-            1:2
-            4:3
-        }
-Step2) Traverse the array and check for every element, if complement (target-element) exists in dictionary. Also check,
-the index of (target-element) is not same as that of element (using dictionary as we have saved index as values.)
+        3: 0,
+        2: 1,
+        1: 2,
+        4: 3
+    }
+
+Step 2)
+Traverse the array and for every element, check if the complement
+(target - element) exists in the dictionary.
+
+Also check that the index of (target - element) is not the same as
+the current index (we use the dictionary values to get indices).
+
 Traversing the array:
-#   i=0         3,2,1,4
-                ^
-    checking if (6-3) exists in dictionary?
-    Yes it exists
-    but d[6-3]==i (We are adding the same element to get the target which is invalid according to the question)
+    i = 0         nums = [3, 2, 1, 4]
+                   ^
+    Check if (6 - 3) exists in dictionary:
+        Yes, it exists (3 is in d),
+        but d[6 - 3] == i
+    This means we are using the same element twice, which is invalid.
 
-Step3) If we found a valid pair, then we return [i,d[complement]]
-'''
-'''
-Time Complexity =  O(N)
-Space Complexity = O(N)
+Step 3)
+If we find a valid pair, we return [i, d[complement]].
+
+Time Complexity:  O(N)
+Space Complexity: O(N)
 '''
 
-#Two-pass
+# Two-pass
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
-        d = {v:i for i,v in enumerate(nums)}
+        # First pass: build value -> index mapping
+        d = {v: i for i, v in enumerate(nums)}
+
+        # Second pass: check for each element if its complement exists
         for i in range(len(nums)):
             complement = target - nums[i]
-            if complement in d and d[complement]!=i:
-                return [i,d[complement]]
+            # Ensure complement exists and is not the same index
+            if complement in d and d[complement] != i:
+                return [i, d[complement]]
+
 
 '''
-Solution-2: Single Pass:
-Instead of doing double passes (one pass while populating the dictionary and other while checking for complement),
-We do this in Single Pass. that is, checking for complement while populating the dictionary.
-Before inserting the element in dictionary, we check whether the complement already exists, if exists, we return both indexes
-(index of current element and index of its complement) and if not exists, we insert the element in dictionary.
-This way we wont have to check if the index of element and its complement is same or not.
+Solution-2: Single Pass Hash Table (Optimized)
+
+Instead of doing two passes (one to populate the dictionary and
+another to check for complements), we do this in a single pass.
+
+Idea:
+    While traversing the array:
+        1) For the current element nums[i], compute complement = target - nums[i]
+        2) Check if complement is already present in the dictionary:
+            - If yes → we have found our answer: [index of complement, i]
+        3) If not present → store nums[i] with its index in the dictionary.
+
+This way:
+    - We don't need to worry about using the same index twice.
+    - We only traverse the list once.
+
+Time Complexity:  O(N)
+Space Complexity: O(N)
 '''
 
-#Single Pass
+# Single-pass (this class overrides the previous one in LeetCode-style usage)
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
-        d = {}
+        d = {}  # value -> index
+
         for i in range(len(nums)):
             complement = target - nums[i]
+
+            # If we have already seen the complement, return the pair of indices
             if complement in d:
-                return [d[complement],i]
-            d[nums[i]]=i
+                return [d[complement], i]
+
+            # Otherwise, store the current value and its index
+            d[nums[i]] = i