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

• Last Updated : 24 Nov, 2021

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:

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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][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++

 `// 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;``}`

## Java

 `// Java implementation of the above approach``class` `Geeks{` `// Function to find number less than N``// having k non-zero digits``static` `int` `k_nonzero_numbers(String s, ``int` `n, ``int` `k)``{``    ` `    ``// Store the memorised values``    ``int` `dp[][][] = ``new` `int``[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 !=``                       ``0` `? ``9` `:s.charAt(i) - ``'0'``))``                ``{``                    ``if` `(j + (x > ``0` `? ``1` `: ``0``) < k + ``1``)``                    ``{``                        ``dp[i + ``1``][(sm != ``0` `|| x <``                                  ``(s.charAt(i) - ``'0'``)) ?``                                   ``1` `: ``0``][j + (x > ``0` `?``                                   ``1` `: ``0``)] += dp[i][sm][j];``                    ``}``                    ``++x;``                ``}``            ``}``            ``++sm;``        ``}``    ``}` `    ``// Return the required answer``    ``return` `dp[n][``0``][k] + dp[n][``1``][k];``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String s = ``"25"``;` `    ``int` `k = ``2``;` `    ``int` `n = s.length();` `    ``// Function call``    ``System.out.println(k_nonzero_numbers(s, n, k));``}``}` `// This code is contributed by Rajnis09`

## Python3

 `# Python3 implementation of the above approach` `# Function to find number less than N``# having k non-zero digits``def` `k_nonzero_numbers(s, n, k):``    ` `    ``# Store the memorised values``    ``dp ``=` `[[[ ``0` `for` `i ``in` `range``(k ``+` `2``)]``               ``for` `i ``in` `range``(``2``)]``               ``for` `i ``in` `range``(n ``+` `2``)]` `    ``# Initialise``    ``for` `i ``in` `range``(n ``+` `1``):``        ``for` `j ``in` `range``(``2``):``            ``for` `x ``in` `range``(k ``+` `1``):``                ``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` `i ``in` `range``(n):``        ``sm ``=` `0``        ` `        ``while` `(sm < ``2``):``            ``for` `j ``in` `range``(k ``+` `1``):``                ``x ``=` `0``                ``y ``=` `0``                ``if` `sm:``                    ``y ``=` `9``                ``else``:``                    ``y ``=` `ord``(s[i]) ``-` `ord``(``'0'``)` `                ``while` `(x <``=` `y):``                    ``dp[i ``+` `1``][(sm ``or` `x < (``                    ``ord``(s[i]) ``-` `ord``(``'0'``)))][j ``+``                     ``(x > ``0``)] ``+``=` `dp[i][sm][j]``                    ``x ``+``=` `1``                    ` `            ``sm ``+``=` `1` `    ``# Return the required answer``    ``return` `dp[n][``0``][k] ``+` `dp[n][``1``][k]` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``s ``=` `"25"` `    ``k ``=` `2` `    ``n ``=` `len``(s)` `    ``# Function call``    ``print``(k_nonzero_numbers(s, n, k))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the above approach``using` `System;``using` `System.Collections;` `class` `GFG{` `// Function to find number less than N``// having k non-zero digits``static` `int` `k_nonzero_numbers(``string` `s, ``int` `n,``                                       ``int` `k)``{``    ` `    ``// Store the memorised values``    ``int` `[,,]dp = ``new` `int``[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 !=``                       ``0 ? 9 : s[i]- ``'0'``))``                ``{``                    ``if` `(j + (x > 0 ? 1 : 0) < k + 1)``                    ``{``                        ``dp[i + 1, ((sm != 0 || x <``                        ``(s[i] - ``'0'``)) ? 1 : 0),``                           ``j + (x > 0 ? 1 : 0)] +=``                         ``dp[i, sm, j];``                    ``}``                    ``++x;``                ``}``            ``}``            ``++sm;``        ``}``    ``}` `    ``// Return the required answer``    ``return` `dp[n, 0, k] + dp[n, 1, k];``}` `// Driver code``public` `static` `void` `Main(``string``[] args)``{``    ``string` `s = ``"25"``;``    ``int` `k = 2;``    ``int` `n = s.Length;` `    ``// Function call``    ``Console.Write(k_nonzero_numbers(s, n, k));``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``
Output:
`14`

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

Auxiliary Space: O(N * K * 2)
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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