diff --git a/puzzles/interviews/training/stepping_number.cpp b/puzzles/interviews/training/stepping_number.cpp
new file mode 100644
index 0000000..8388319
--- /dev/null
+++ b/puzzles/interviews/training/stepping_number.cpp
@@ -0,0 +1,106 @@
+/*
+VIM: let b:lcppflags="-std=c++14 -O2 -pthread -I."
+VIM: let b:wcppflags="/O2 /EHsc /DWIN32"
+VIM-: let b:cppflags=g:Iboost.g:Itbb
+VIM-: let b:ldflags=g:Lboost.g:Ltbb
+VIM-: let b:ldlibpath=g:Bboost.g:Btbb
+VIM-: let b:argv=""
+*/
+#include <iostream>
+#include <exception>
+#include <chrono>
+#include <vector>
+#include <algorithm>
+
+/*Given N and M find all stepping numbers in range N to M
+
+The stepping number:
+
+A number is called as a stepping number if the adjacent digits have a difference of 1.
+e.g 123 or 101 is stepping number, but 358 is not a stepping number
+
+Example:
+
+N = 10, M = 20
+all stepping numbers are 10 , 12 
+
+Return the numbers in sorted order.
+*/
+
+/*
+ * Solution: complexity is linear actually. O(n)
+ *           So, a better solution I guess will be to interate over the range
+ *           and check each number on condition.
+ */
+
+
+void f( std::vector<int>& v, int A, int B, int n )
+{
+    if ( A <= n && n <= B ) {
+        v.push_back(n);
+    }
+    else if ( n > B ) {
+        return;
+    }
+
+    int d = n % 10;
+    if ( d + 1 <= 9 ){
+        f( v, A, B, n*10+d+1);
+    }
+    if ( d - 1 >= 0 ){
+        f( v, A, B, n*10+d-1);
+    }
+}
+
+std::vector<int> stepnum(int A, int B)
+{
+    std::vector<int> v;
+
+    if ( A <= 0 && 0 <= B ){
+        v.push_back(0);
+    }
+
+    for ( int i = 1; i <= 9; ++i ){
+        f( v, A, B, i );
+    }
+
+    std::sort(v.begin(), v.end());
+
+    return v;
+}
+
+void print( const std::vector<int>& A )
+{
+    for (auto i: A){
+        std::cout << i << " ";
+    }
+    std::cout << std::endl;
+}
+
+int main ( void )
+{try{
+    auto begin = std::chrono::high_resolution_clock::now();
+
+
+    //......
+    print( stepnum(10, 20) );
+    print( stepnum(0, 1000) );
+
+    auto end = std::chrono::high_resolution_clock::now();
+    std::chrono::duration<double> seconds = end - begin;
+    std::cout << "Time: " << seconds.count() << std::endl;
+    return 0;
+}
+catch ( const std::exception& e )
+{
+    std::cerr << std::endl
+            << "std::exception(\"" << e.what() << "\")." << std::endl;
+    return 2;
+}
+catch ( ... )
+{
+    std::cerr  << std::endl
+            << "unknown exception." << std::endl;
+    return 1;
+}}
+