diff --git a/Java/dynamic_programming/race_car.java b/Java/dynamic_programming/race_car.java
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+/*
+ * Title: Race Car (LeetCode #818)
+ * Problem: 
+ * The car starts at position 0 and speed +1. 
+ * In each move, you can either:
+ *   - "A": Accelerate -> position += speed; speed *= 2
+ *   - "R": Reverse -> if speed > 0 then speed = -1 else speed = 1
+ * Find the minimum number of instructions to reach the target position.
+ *
+ * Approach:
+ * Dynamic Programming (DP)
+ * - dp[i] stores the minimum number of steps required to reach position i.
+ * - Use bit manipulation to find the smallest k such that (2^k - 1) >= i.
+ * - Compute based on overshooting or undershooting target.
+ *
+ * Time Complexity: O(n log n)
+ * Space Complexity: O(n)
+ *
+ * Author: Mohammed Sufiyan Ahmed
+ * Date: October 2025
+ */
+
+class Solution {
+    public int racecar(int target) {
+        int[] dp = new int[target + 1];
+
+        for (int i = 1; i <= target; ++i) {
+            int k = 32 - Integer.numberOfLeadingZeros(i);
+
+            // Case 1: If target is exactly at position (2^k - 1)
+            if (i == (1 << k) - 1) {
+                dp[i] = k;
+                continue;
+            }
+
+            // Case 2: Overshoot and then reverse
+            dp[i] = dp[(1 << k) - 1 - i] + k + 1;
+
+            // Case 3: Undershoot and reverse earlier
+            for (int j = 0; j < k - 1; ++j) {
+                dp[i] = Math.min(dp[i],
+                        dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
+            }
+        }
+        return dp[target];
+    }
+
+    // Example test case (for contributors)
+    public static void main(String[] args) {
+        Solution sol = new Solution();
+        System.out.println("Minimum instructions for target=3: " + sol.racecar(3));  // Output: 2
+        System.out.println("Minimum instructions for target=6: " + sol.racecar(6));  // Output: 5
+    }
+}