# N digit numbers having difference between the first and last digits as K

Given two integer N and K, the task is to print all positive numbers made up of N digits whose difference between the first and last digits equal to K.

Examples:

Input: N = 2, K = 0
Output: 11, 22, 33, 44, 55, 66, 77, 88, 99

Input: N = 2, K = 9
Output: 90

Approach: The idea is to generate all possible 1-digit numbers to N-digit numbers using recursion and check if the difference between the first and the last digit of that number is equal to K or not. Below are the steps:

1. Generate all possible numbers with length 1.
2. At each step, keep on adding digits to the number until the length of the number becomes N.
3. When the length of the number becomes equal to N, calculate the difference between the first and last digit of the number, and check if the difference is equal to N or not. If found to be true, print that number and proceed to generate the next number.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to store and check the` `// difference of digits` `void` `findNumbers(string st, vector<``int``>& result,` `                 ``int` `prev, ``int` `n, ``int` `K)` `{` `    ``// Base Case` `    ``if` `(st.length() == n) {` `        ``result.push_back(stoi(st));` `        ``return``;` `    ``}`   `    ``// Last digit of the number to` `    ``// check the difference from the` `    ``// first digit` `    ``if` `(st.size() == n - 1) {`   `        ``// Condition to avoid` `        ``// repeated values` `        ``if` `(prev - K >= 0) {`   `            ``string pt = ``""``;`   `            ``// Update the string pt` `            ``pt += prev - K + 48;`   `            ``// Recursive Call` `            ``findNumbers(st + pt, result,` `                        ``prev - K, n, K);` `        ``}`   `        ``if` `(K != 0 && prev + K < 10) {`   `            ``string pt = ``""``;`   `            ``// Update the string pt` `            ``pt += prev + K + 48;`   `            ``// Recursive Call` `            ``findNumbers(st + pt, result,` `                        ``prev + K, n, K);` `        ``}` `    ``}`   `    ``// Any number can come in between` `    ``// first and last except the zero` `    ``else` `{` `        ``for` `(``int` `j = 1; j <= 9; j++) {`   `            ``string pt = ``""``;` `            ``pt += j + 48;`   `            ``// Recursive Call` `            ``findNumbers(st + pt, result,` `                        ``prev, n, K);` `        ``}` `    ``}` `}`   `// Function to place digit of the number` `vector<``int``> numDifference(``int` `N, ``int` `K)` `{` `    ``vector<``int``> res;` `    ``string st = ``""``;`   `    ``// When N is 1 and K > 0, then the` `    ``// single number will be the first` `    ``// and last digit it cannot have` `    ``// difference greater than 0` `    ``if` `(N == 1 && K == 0) {` `        ``res.push_back(0);` `    ``}`   `    ``else` `if` `(N == 1 && K > 0) {` `        ``return` `res;` `    ``}`   `    ``// This loop place the digit at the` `    ``// starting` `    ``for` `(``int` `i = 1; i < 10; i++) {`   `        ``string temp = ``""``;` `        ``temp += 48 + i;`   `        ``// Recursive Call` `        ``findNumbers(st + temp, res, i, N, K);` `        ``st = ``""``;` `    ``}`   `    ``return` `res;` `}`   `void` `numDifferenceUtil(``int` `N, ``int` `K)` `{`   `    ``// Vector to store results` `    ``vector<``int``> res;`   `    ``// Generate all the resultant number` `    ``res = numDifference(N, K);`   `    ``// Print the result` `    ``for` `(``int` `i = 0; i < res.size(); i++) {` `        ``cout << res[i] << ``" "``;` `    ``}` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 2, K = 9;`   `    ``// Function Call` `    ``numDifferenceUtil(N, K);`   `    ``return` `0;` `}`

## Python3

 `# Python 3 program for ` `# the above approach`   `# Function to store and ` `# check the difference` `# of digits` `result ``=` `[]` `def` `findNumbers(st, prev, ` `                ``n, K): ` `  `  `    ``global` `result` `    `  `    ``# Base Case` `    ``if` `(``len``(st) ``=``=` `n):` `        ``result.append(``int``(st))` `        ``return`   `    ``# Last digit of the number to` `    ``# check the difference from the` `    ``# first digit` `    ``if``(``len``(st) ``=``=` `n ``-` `1``):` `      `  `        ``# Condition to avoid` `        ``# repeated values` `        ``if` `(prev ``-` `K >``=` `0``):` `            ``pt ``=` `""`   `            ``# Update the string pt` `            ``pt ``+``=` `prev ``-` `K ``+` `48`   `            ``# Recursive Call` `            ``findNumbers(st ``+` `pt, ` `                        ``prev ``-` `K, ` `                        ``n, K)`   `        ``if` `(K !``=` `0` `and` `            ``prev ``+` `K < ``10``):` `            ``pt ``=` `""`   `            ``# Update the string pt` `            ``pt ``+``=` `prev ``+` `K ``+` `48`   `            ``# Recursive Call` `            ``findNumbers(st ``+` `pt, ` `                        ``prev ``+` `K, ` `                        ``n, K)`   `    ``# Any number can come in between` `    ``# first and last except the zero` `    ``else``:` `        ``for` `j ``in` `range``(``1``, ``10``, ``1``):` `            ``pt ``=` `""` `            ``pt ``+``=` `j ``+` `48`   `            ``# Recursive Call` `            ``findNumbers(st ``+` `pt, ` `                        ``prev, n, K)`   `# Function to place digit ` `# of the number` `def` `numDifference(N,K):` `    ``global` `result` `    ``st ``=` `""` `    `  `    ``# When N is 1 and K > 0, ` `    ``# then the single number ` `    ``# will be the first and ` `    ``# last digit it cannot have` `    ``# difference greater than 0` `    ``if` `(N ``=``=` `1` `and` `K ``=``=` `0``):` `        ``result.append(``0``)`   `    ``elif``(N ``=``=` `1` `and` `K > ``0``):` `        ``return` `result`   `    ``# This loop place the ` `    ``# digit at the starting` `    ``for` `i ``in` `range``(``1``, ``10``, ``1``):` `        ``temp ``=` `""` `        ``temp ``+``=` `str``(``48` `+` `i)`   `        ``# Recursive Call` `        ``findNumbers(st ``+` `temp, ` `                    ``i, N, K)` `        ``st ``=` `""` `    ``return` `result`   `def` `numDifferenceUtil(N, K):` `  `  `    ``# Vector to store results` `    ``res ``=` `[]`   `    ``# Generate all the` `    ``# resultant number` `    ``res ``=` `numDifference(N, K)`   `    ``# Print the result` `    ``for` `i ``in` `range``(``1``, ``len``(res)):` `        ``print``(res[i] ``+` `40``, ` `              ``end ``=` `" "``)` `        ``break`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``N ``=` `2` `    ``K ``=` `9` `    `  `    ``# Function Call` `    ``numDifferenceUtil(N, K)` `    `  `# This code is contributed by bgangwar59`

Output:

```90

```

Time Complexity: O(2N
Auxiliary Space: O(N)

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Improved By : bgangwar59