# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

Let,

• dp[i][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][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.

After computing the above dp, the desired answer is dp[L][K] + dp[L][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 ` `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][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 = 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][k] + dp[n][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; ` `} `

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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