^{N}

Given an integer **N**, the task is to generate an **N**-digit number which is comprising only of digits **1** or **2** and is divisible by **2 ^{N}**.

**Examples:**

Input:N = 4Output:2112Explanation:Since 2112 is divisible by 2^{4}( = 16).

Input:N = 15Output:211111212122112

**Approach: **Follow the steps below to solve the problem:

- Iterate over all values in the range
**[1, 2**.^{N}] - For each integer
**i**in that range, generate its binary representation using bitset and store it in a string, say**S**. - Reduce the length of the string
**S**to**N**. - Traverse the bits of
**S**and if**S**, set_{i}== ‘0’**S**_{i}= ‘2’. - Convert the obtained string to an integer, say
**res**using**stoll()**. - If
**res**is divisible by**2**, print the integer and break.^{N}

Below is the implementation of the above approach:

## C++14

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Utility function to find x^y in O(log(y))` `long` `long` `int` `power(` `long` `long` `int` `x,` ` ` `unsigned ` `long` `long` `int` `y)` `{` ` ` `// Stores the result` ` ` `long` `long` `int` `res = 1;` ` ` `// Update x if it is >= p` ` ` `while` `(y > 0) {` ` ` `// If y is odd` ` ` `if` `(y & 1)` ` ` `// Multiply x with res` ` ` `res = (res * x);` ` ` `// y must be even now` ` ` `// Set y = y/2` ` ` `y = y >> 1;` ` ` `x = (x * x);` ` ` `}` ` ` `return` `res;` `}` `// Function to generate the N digit number` `// satisfying the given conditions` `void` `printNth(` `int` `N)` `{` ` ` `// Find all possible integers upto 2^N` ` ` `for` `(` `long` `long` `int` `i = 1;` ` ` `i <= power(2, N); i++) {` ` ` `// Generate binary representation of i` ` ` `string s = bitset<64>(i).to_string();` ` ` `// Reduce the length of the string to N` ` ` `string s1 = s.substr(` ` ` `s.length() - N, s.length());` ` ` `for` `(` `long` `long` `int` `j = 0; s1[j]; j++) {` ` ` `// If current bit is '0'` ` ` `if` `(s1[j] == ` `'0'` `) {` ` ` `s1[j] = ` `'2'` `;` ` ` `}` ` ` `}` ` ` `// Convert string to equivalent integer` ` ` `long` `long` `int` `res = stoll(s1, nullptr, 10);` ` ` `// If condition satisfies` ` ` `if` `(res % power(2, N) == 0) {` ` ` `// Print and break` ` ` `cout << res << endl;` ` ` `break` `;` ` ` `}` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 15;` ` ` `printNth(N);` `}` |

## Python3

`# Python program for the above approach` `# Utility function to find x^y in O(log(y))` `def` `power(x, y):` ` ` ` ` `# Stores the result` ` ` `res ` `=` `1` ` ` ` ` `# Update x if it is >= p` ` ` `while` `(y > ` `0` `):` ` ` ` ` `# If y is odd` ` ` `if` `(y & ` `1` `):` ` ` ` ` `# Multiply x with res` ` ` `res ` `=` `(res ` `*` `x)` ` ` ` ` `# y must be even now` ` ` `# Set y = y/2` ` ` `y ` `=` `y >> ` `1` ` ` `x ` `=` `(x ` `*` `x)` ` ` `return` `res` `# Function to generate the N digit number` `# satisfying the given conditions` `def` `printNth(N):` ` ` ` ` `# Find all possible integers upto 2^N` ` ` `for` `i ` `in` `range` `(` `1` `,power(` `2` `, N) ` `+` `1` `):` ` ` ` ` `# Generate binary representation of i` ` ` `s ` `=` `"{:064b}"` `.` `format` `(i)` ` ` ` ` `# Reduce the length of the string to N` ` ` `s1 ` `=` `s[` `len` `(s)` `-` `N: ` `2` `*` `len` `(s)` `-` `N]` ` ` ` ` `j ` `=` `0` ` ` `while` `(j < ` `len` `(s1)):` ` ` ` ` `# If current bit is '0'` ` ` `if` `(s1[j] ` `=` `=` `'0'` `):` ` ` `s1 ` `=` `s1[:j] ` `+` `'2'` `+` `s1[j ` `+` `1` `:]` ` ` `j ` `+` `=` `1` ` ` ` ` `# Convert string to equivalent integer` ` ` `res ` `=` `int` `(s1)` ` ` ` ` `# If condition satisfies` ` ` `if` `(res ` `%` `power(` `2` `, N) ` `=` `=` `0` `):` ` ` ` ` `# Prand break` ` ` `print` `(res)` ` ` `break` `# Driver Code` `N ` `=` `15` `printNth(N)` `# This code is contributed by shubhamsingh10` |

**Output:**

211111212122112

**Time Complexity:** O(2^{N})**Auxiliary Space: **O(N)

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