diff --git a/src/main/java/com/thealgorithms/searches/BinarySearch.java b/src/main/java/com/thealgorithms/searches/BinarySearch.java
index 0cac484d56b4..7a5361b280ea 100644
--- a/src/main/java/com/thealgorithms/searches/BinarySearch.java
+++ b/src/main/java/com/thealgorithms/searches/BinarySearch.java
@@ -3,14 +3,32 @@
 import com.thealgorithms.devutils.searches.SearchAlgorithm;
 
 /**
- * Binary search is one of the most popular algorithms The algorithm finds the
- * position of a target value within a sorted array
- * IMPORTANT
- * This algorithm works correctly only if the input array is sorted
- * in ascending order.
- * <p>
- * Worst-case performance O(log n) Best-case performance O(1) Average
- * performance O(log n) Worst-case space complexity O(1)
+ * Binary Search Algorithm Implementation
+ *
+ * <p>Binary search is one of the most efficient searching algorithms for finding a target element
+ * in a SORTED array. It works by repeatedly dividing the search space in half, eliminating half of
+ * the remaining elements in each step.
+ *
+ * <p>IMPORTANT: This algorithm ONLY works correctly if the input array is sorted in ascending
+ * order.
+ *
+ * <p>Algorithm Overview: 1. Start with the entire array (left = 0, right = array.length - 1) 2.
+ * Calculate the middle index 3. Compare the middle element with the target: - If middle element
+ * equals target: Found! Return the index - If middle element is less than target: Search the right
+ * half - If middle element is greater than target: Search the left half 4. Repeat until element is
+ * found or search space is exhausted
+ *
+ * <p>Performance Analysis: - Best-case time complexity: O(1) - Element found at middle on first
+ * try - Average-case time complexity: O(log n) - Most common scenario - Worst-case time
+ * complexity: O(log n) - Element not found or at extreme end - Space complexity: O(1) - Only uses
+ * a constant amount of extra space
+ *
+ * <p>Example Walkthrough: Array: [1, 3, 5, 7, 9, 11, 13, 15, 17, 19] Target: 7
+ *
+ * <p>Step 1: left=0, right=9, mid=4, array[4]=9 (9 &gt; 7, search left half) Step 2: left=0,
+ * right=3, mid=1, array[1]=3 (3 &lt; 7, search right half) Step 3: left=2, right=3, mid=2,
+ * array[2]=5 (5 &lt; 7, search right half) Step 4: left=3, right=3, mid=3, array[3]=7 (Found!
+ * Return index 3)
  *
  * @author Varun Upadhyay (https://github.com/varunu28)
  * @author Podshivalov Nikita (https://github.com/nikitap492)
@@ -20,41 +38,89 @@
 class BinarySearch implements SearchAlgorithm {
 
     /**
-     * @param array is an array where the element should be found
-     * @param key is an element which should be found
-     * @param <T> is any comparable type
-     * @return index of the element
+     * Generic method to perform binary search on any comparable type. This is the main entry point
+     * for binary search operations.
+     *
+     * <p>Example Usage:
+     * <pre>
+     * Integer[] numbers = {1, 3, 5, 7, 9, 11};
+     * int result = new BinarySearch().find(numbers, 7);
+     * // result will be 3 (index of element 7)
+     *
+     * int notFound = new BinarySearch().find(numbers, 4);
+     * // notFound will be -1 (element 4 does not exist)
+     * </pre>
+     *
+     * @param <T> The type of elements in the array (must be Comparable)
+     * @param array The sorted array to search in (MUST be sorted in ascending order)
+     * @param key The element to search for
+     * @return The index of the key if found, -1 if not found or if array is null/empty
      */
     @Override
     public <T extends Comparable<T>> int find(T[] array, T key) {
+        // Handle edge case: empty array
         if (array == null || array.length == 0) {
             return -1;
         }
+
+        // Delegate to the core search implementation
         return search(array, key, 0, array.length - 1);
     }
 
     /**
-     * This method implements the Generic Binary Search
+     * Core recursive implementation of binary search algorithm. This method divides the problem
+     * into smaller subproblems recursively.
+     *
+     * <p>How it works:
+     * <ol>
+     * <li>Calculate the middle index to avoid integer overflow</li>
+     * <li>Check if middle element matches the target</li>
+     * <li>If not, recursively search either left or right half</li>
+     * <li>Base case: left &gt; right means element not found</li>
+     * </ol>
+     *
+     * <p>Time Complexity: O(log n) because we halve the search space each time.
+     * Space Complexity: O(log n) due to recursive call stack.
      *
-     * @param array The array to make the binary search
-     * @param key The number you are looking for
-     * @param left The lower bound
-     * @param right The upper bound
-     * @return the location of the key
+     * @param <T> The type of elements (must be Comparable)
+     * @param array The sorted array to search in
+     * @param key The element we're looking for
+     * @param left The leftmost index of current search range (inclusive)
+     * @param right The rightmost index of current search range (inclusive)
+     * @return The index where key is located, or -1 if not found
      */
     private <T extends Comparable<T>> int search(T[] array, T key, int left, int right) {
+        // Base case: Search space is exhausted
+        // This happens when left pointer crosses right pointer
         if (right < left) {
-            return -1; // this means that the key not found
+            return -1; // Key not found in the array
         }
-        // find median
-        int median = (left + right) >>> 1;
+
+        // Calculate middle index
+        // Using (left + right) / 2 could cause integer overflow for large arrays
+        // So we use: left + (right - left) / 2 which is mathematically equivalent
+        // but prevents overflow
+        int median = (left + right) >>> 1; // Unsigned right shift is faster division by 2
+
+        // Get the value at middle position for comparison
         int comp = key.compareTo(array[median]);
 
+        // Case 1: Found the target element at middle position
         if (comp == 0) {
-            return median;
-        } else if (comp < 0) {
+            return median; // Return the index where element was found
+        }
+        // Case 2: Target is smaller than middle element
+        // This means if target exists, it must be in the LEFT half
+        else if (comp < 0) {
+            // Recursively search the left half
+            // New search range: [left, median - 1]
             return search(array, key, left, median - 1);
-        } else {
+        }
+        // Case 3: Target is greater than middle element
+        // This means if target exists, it must be in the RIGHT half
+        else {
+            // Recursively search the right half
+            // New search range: [median + 1, right]
             return search(array, key, median + 1, right);
         }
     }