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.

    Let,

  • 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.

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:

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// 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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