# Find the numbers from 1 to N that contains exactly k non-zero digits

**Prerequisites:** Dynamic Programming, DigitDP

Given two integers **N** and **K**. The task is to find the number of integers between **1** and **N** (inclusive) that contains exactly **K** non-zero digits when written in base ten.

**Examples:**

Input:N = 100, K = 1

Output:19

Explanation:

The digits with exactly 1 non zero digits between 1 and 100 are:

1, 2, 3, 4, 5, 6, 7, 8, 9,

10, 20, 30, 40, 50, 60, 70, 80, 90, 100

Input:N = 25, K = 2

Output:14

Explanation:

The digits with exactly 2 non zero digits between 1 and 25 are:

11, 12, 13, 14, 15, 16, 17,

18, 19, 21, 22, 23, 24, 25

**Approach:** It is enough to consider the integers of **N** digits, by filling the higher digits with **0** if necessary. This problem can be solved by applying the method called digit DP.

**dp[i][0][j]**= The higher**i**digits have already been decided, and there are**j**non-zero digits, and it has already been determined that it is less than**N**.**dp[i][1][j]**= The higher**i**digits have already been decided, and there are**j**non-zero digits, and it has not yet been determined that it is less than**N**.

Let,

After computing the above dp, the desired answer is **dp[L][0][K] + dp[L][1][K]**, where **L** is the number of digits of **N**.

Below is the implementation of the above approach:

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find number less than N ` `// having k non-zero digits ` `int` `k_nonzero_numbers(string s, ` `int` `n, ` `int` `k) ` `{ ` ` ` `// Store the memorised values ` ` ` `int` `dp[n + 1][2][k + 1]; ` ` ` ` ` `// Initialise ` ` ` `for` `(` `int` `i = 0; i <= n; i++) ` ` ` `for` `(` `int` `j = 0; j < 2; j++) ` ` ` `for` `(` `int` `x = 0; x <= k; x++) ` ` ` `dp[i][j][x] = 0; ` ` ` ` ` `// Base ` ` ` `dp[0][0][0] = 1; ` ` ` ` ` `// Calculate all states ` ` ` `// For every state, from numbers 1 to N, ` ` ` `// the count of numbers which contain exactly j ` ` ` `// non zero digits is being computed and updated ` ` ` `// in the dp array. ` ` ` `for` `(` `int` `i = 0; i < n; ++i) { ` ` ` `int` `sm = 0; ` ` ` `while` `(sm < 2) { ` ` ` `for` `(` `int` `j = 0; j < k + 1; ++j) { ` ` ` `int` `x = 0; ` ` ` `while` `(x <= (sm ? 9 : s[i] - ` `'0'` `)) { ` ` ` `dp[i + 1][sm || x < (s[i] - ` `'0'` `)][j + (x > 0)] ` ` ` `+= dp[i][sm][j]; ` ` ` `++x; ` ` ` `} ` ` ` `} ` ` ` `++sm; ` ` ` `} ` ` ` `} ` ` ` ` ` `// Return the required answer ` ` ` `return` `dp[n][0][k] + dp[n][1][k]; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `string s = ` `"25"` `; ` ` ` ` ` `int` `k = 2; ` ` ` ` ` `int` `n = s.size(); ` ` ` ` ` `// Function call ` ` ` `cout << k_nonzero_numbers(s, n, k); ` ` ` ` ` `return` `0; ` `} ` |

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**Output:**

14

**Time Complexity:** O(LK) where L is the number of digits in N.

**Note:** The two for loops used to calculate the state which from [0, 1] and [0, 9] respectively are considered as a constant multiplication.

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