Added 28-10-2024 Solution and 29-10-2024 Question

Added 28-10-2024 Solution and 29-10-2024 Question

28-10-2024/Explanation.md

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+### Explanation
+
+**Why did the keyboard get kicked out of the typing club?**  
+Because it kept _shifting_ between rows! 😂
+
+In this code:
+
+1. **Input Collection:** We first take the number of words and store each word in `word_list`.
+2. **Function Definition:** `word_row_type` checks if each word can be typed on a single row by:
+   - Converting the word to uppercase.
+   - Using a list of keyboard rows to check if all letters of the word are contained within any one row.
+3. **Filtering Typable Words:** Using list comprehension, we select words that meet the typing criteria.
+4. **Output:** The resulting list of row-typable words is printed.
+
+This solution uses list comprehension and built-in functions for an efficient and beginner-friendly approach to the problem.

28-10-2024/Solution.py

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+# Taking Input
+N = int(input("Enter Number of Words: "))
+
+# List to Store Words
+word_list = []
+
+# Loop to collect words
+for i in range(N):
+    word = input(f"Enter Word {i+1}: ")
+    word_list.append(word)
+
+
+def word_row_type(w):
+    # Convert the word to uppercase to match keyboard row format
+    w = w.upper()
+
+    # Rows on American keyboard
+    rows = ["QWERTPOIUY", "ASDFLKJGH", "ZXCVMNB"]
+
+    # Check if all characters of the word can be typed using only one row
+    if any(all(char in row for char in w) for row in rows):
+        return 1  # Return 1 if the word can be typed on a single row
+
+
+# List of words that can be typed from a single row
+word_typable = [w for w in word_list if (word_row_type(w) == 1)]
+
+print("Words that can be typed from One Row Of American Keyboard are:")
+print(word_typable)

29-10-2024/Question.md

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+### **Question: Unique Element**
+
+**Difficulty Level:** 🟡 Intermediate  
+**Domain:** Algorithms and Data Structures (Binary Search)
+
+### **Objective:**
+
+Given a sorted array where every element appears exactly twice except for one unique element that appears only once, identify this unique element. The challenge requires achieving this in **O(log n)** time and **O(1)** space.
+
+### **Conceptual Background:**
+
+The array is sorted and every element except one appears in pairs, allowing the unique element to be located efficiently with a binary search approach. Since binary search works well with sorted data and can achieve logarithmic time complexity, it's a suitable technique here.
+
+### **Steps to Approach:**
+
+1. **Binary Search Setup:**  
+   Initialize two pointers, `left` and `right`, representing the bounds of the search range.
+
+2. **Binary Search to Locate Unique Element:**  
+   Perform binary search:
+
+   - Calculate the midpoint `mid`.
+   - Determine if the midpoint index is even or odd.
+   - If `nums[mid]` equals `nums[mid + 1]` or `nums[mid - 1]`, the unique element is located further to the right or left respectively.
+
+3. **Narrowing Down the Search:**  
+   Adjust `left` and `right` pointers based on comparisons to ensure you converge towards the unique element efficiently. This process will conclude with the unique element at the `left` pointer once the range is narrowed down.
+
+4. **Output the Unique Element:**  
+   Once found, the element at the `left` pointer is the unique element in the array.
+
+### **Example Execution:**
+
+- **Input:** `nums = [3,3,7,7,10,11,11]`
+- **Processing:**
+  - Binary search will adjust `left` and `right` pointers based on the pairing position.
+  - As the binary search narrows down, it isolates the unique element.
+- **Output:** `10`
+
+This approach, leveraging binary search, ensures the solution runs in optimal time and space complexity.