diff --git a/C++/13_RomanToInteger.cpp b/C++/13_RomanToInteger.cpp
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+// LeetCode Problem 13: Roman to Integer (Easy)
+// Language: C++
+// Contributor: Nihhaar0002
+//
+// Approach: Value Comparison (Single-Pass)
+// ----------------------------------------
+// Each Roman numeral has a fixed integer value:
+//   I=1, V=5, X=10, L=50, C=100, D=500, M=1000
+//
+// Rule:
+// - Normally we add values from left to right.
+// - But when a smaller numeral appears before a larger numeral,
+//   we must subtract the smaller value instead of adding it.
+//
+// Example:
+//   IV => 4  (because 1 < 5 → subtract 1)
+//   IX => 9  (because 1 < 10)
+//   XL => 40 (10 < 50)
+//
+// Algorithm:
+// - Traverse the string from left to right.
+// - Add each value to sum.
+// - If current value > previous value → subtract two times previous value
+//   because it was added once earlier.
+//
+// Time Complexity:  O(N)
+// Space Complexity: O(1)  (no extra data structures)
+
+#include <string>
+using namespace std;
+
+class Solution {
+public:
+    int romanToInt(string s) {
+        int current = 0, last = 0;
+        int sum = 0;
+
+        for(int i = 0; i < s.length(); i++){
+            // Convert Roman char to integer value
+            switch(s[i]){
+                case 'I': current = 1; break;
+                case 'V': current = 5; break;
+                case 'X': current = 10; break;
+                case 'L': current = 50; break;
+                case 'C': current = 100; break;
+                case 'D': current = 500; break;
+                case 'M': current = 1000; break;
+                default: current = 0;
+            }
+
+            sum += current;
+
+            // Check for subtractive combination
+            if(i != 0 && current > last)
+                sum -= 2 * last;
+
+            last = current;
+        }
+
+        return sum;
+    }
+};