Count of N digit numbers not having given prefixes

Last Updated : 24 Nov, 2021

Given an integer N and a vector of strings prefixes[], the task is to calculate the total possible strings of length N from characters '0' to '9'. such that the given prefixes cannot be used in any of the strings.

Examples:

Input: N = 3, prefixes = {"42"}
Output: 990
Explanation: All string except{"420", "421", "422", "423", "424", "425", "426", "427", "428", "429"} are valid.

Input: N = 5, prefixes[] = { "0", "1", "911" }
Output: 79900

Approach: The total possible strings with length are (10^N) as for each place in a string there are 10 choices of the character. Instead of calculating total good strings, subtract total bad strings from total strings. Before iterating over the prefixes merge prefixes with the same starting character as the bigger length prefix might lead to subtraction of some repetitions. Follow the steps below to solve the problem:

Initialize the variable total as 10^N.
Initialize a map<int, vector<string>> mp[].
Iterate over the range [0, M) using the variable i and perform the following tasks:
- Push the value of prefixes[i] in the vector of map mp[prefixes[i]-'0'].
Initialize the vector new_prefixes[].
Traverse over the map mp[] using the variable x and perform the following tasks:
- Initialize the variable mn as N.
- Traverse the vector x.second using the variable p and perform the following tasks:
  - Set the value of mn as the minimum of mn or p.length().
- Traverse the vector x.second using the variable p and perform the following tasks:
  - If p.length() is less than equal to mn, then push p into the vector new_prefixes[].
Iterate over the range [0, new_prefixes.size()) using the variable i and perform the following tasks:
- Subtract the value int(pow(10, N - new_prefixes[i].length())+ 0.5) from the variable total.
After performing the above steps, print the value of total as the answer.

Below is the implementation of the above approach:

C++

// C++ Program to implement the above approach
#include <bits/stdc++.h>
using namespace std;

// Change int to long long in case of overflow!!
// Function to calculate total strings of length
// N without the given prefixes
int totalGoodStrings(int N, vector<string> prefixes)
{

    // Calculate total strings present
    int total = int(pow(10, N) + 0.5);

    // Make a map and insert the prefixes with same
    // character in a vector
    map<int, vector<string> > mp;
    for (int i = 0; i < prefixes.size(); i++) {
        mp[prefixes[i][0] - '0']
            .push_back(prefixes[i]);
    }

    // Make a new vector of prefixes strings
    vector<string> new_prefixes;

    // Iterate for each starting character
    for (auto x : mp) {

        int mn = N;

        // Iterate through the vector to calculate
        // minimum size prefix
        for (auto p : x.second) {
            mn = min(mn, int(p.length()));
        }

        // Take all the minimum prefixes in the new
        // vector of prefixes
        for (string p : x.second) {
            if (p.length() > mn) {
                continue;
            }
            new_prefixes.push_back(p);
        }
    }

    // Iterate through the new prefixes
    for (int i = 0; i < new_prefixes.size(); i++) {

        // Subtract bad strings
        total -= int(pow(10,
                         N - new_prefixes[i].length())
                     + 0.5);
    }
    return total;
}

// Driver Code
int main()
{
    int N = 5;

    vector<string> prefixes
        = { "1", "0", "911" };

    cout << totalGoodStrings(N, prefixes);

    return 0;
}

Java

// Java Program to implement the above approach
import java.util.ArrayList;
import java.util.HashMap;

class GFG
{
  
  // Change int to long long in case of overflow!!
  // Function to calculate total strings of length
  // N without the given prefixes
  public static int totalGoodStrings(int N, String[] prefixes) {

    // Calculate total strings present
    int total = (int) (Math.pow(10, N) + 0.5);

    // Make a map and insert the prefixes with same
    // character in a vector
    HashMap<Integer, ArrayList<String>> mp = new HashMap<Integer, ArrayList<String>>();

    for (int i = 0; i < prefixes.length; i++) {
      int key = (int) prefixes[i].charAt(0) - (int) '0';

      if (mp.containsKey(key)) {
        ArrayList<String> temp = mp.get(key);
        temp.add(prefixes[i]);
        mp.put(key, temp);
      } else {
        ArrayList<String> temp = new ArrayList<String>();
        temp.add(prefixes[i]);
        mp.put(key, temp);
      }
    }

    // Make a new vector of prefixes strings
    ArrayList<String> new_prefixes = new ArrayList<String>();

    // Iterate for each starting character
    for (Integer x : mp.keySet()) {

      int mn = N;

      // Iterate through the vector to calculate
      // minimum size prefix
      for (String p : mp.get(x)) {
        mn = Math.min(mn, p.length());
      }

      // Take all the minimum prefixes in the new
      // vector of prefixes
      for (String p : mp.get(x)) {
        if (p.length() > mn) {
          continue;
        }
        new_prefixes.add(p);
      }
    }

    // Iterate through the new prefixes
    for (int i = 0; i < new_prefixes.size(); i++) {

      // Subtract bad strings
      total -= (int) (Math.pow(10, N - new_prefixes.get(i).length()) + 0.5);
    }
    return total;
  }

  // Driver Code
  public static void main(String args[]) 
  {

    int N = 5;
    String[] prefixes = { "1", "0", "911" };
    System.out.println(totalGoodStrings(N, prefixes));
  }
}

// This code is contributed by gfgking.

Python3

# python Program to implement the above approach
import math

# Change int to long long in case of overflow!!
# Function to calculate total strings of length
# N without the given prefixes
def totalGoodStrings(N,  prefixes):

        # Calculate total strings present
    total = int(math.pow(10, N) + 0.5)

    # Make a map and insert the prefixes with same
    # character in a vector
    mp = {}
    for i in range(0, len(prefixes)):
        if (ord(prefixes[i][0]) - ord('0')) in mp:
            mp[ord(prefixes[i][0]) - ord('0')].append(prefixes[i])

        else:
            mp[ord(prefixes[i][0]) - ord('0')] = [prefixes[i]]

        # Make a new vector of prefixes strings
    new_prefixes = []

    # Iterate for each starting character
    for x in mp:

        mn = N

        # Iterate through the vector to calculate
        # minimum size prefix
        for p in mp[x]:
            mn = min(mn, len(p))

        # Take all the minimum prefixes in the new
        # vector of prefixes
        for p in mp[x]:
            if (len(p) > mn):
                continue

            new_prefixes.append(p)

        # Iterate through the new prefixes
    for i in range(0, len(new_prefixes)):

                # Subtract bad strings
        total -= int(pow(10, N - len(new_prefixes[i])) + 0.5)

    return total

# Driver Code
if __name__ == "__main__":

    N = 5
    prefixes = ["1", "0", "911"]
    print(totalGoodStrings(N, prefixes))

    # This code is contributed by rakeshsahni

// C# Program to implement the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Change int to long long in case of overflow!!
    // Function to calculate total strings of length
    // N without the given prefixes
    public static int totalGoodStrings(int N,
                                       string[] prefixes)
    {

        // Calculate total strings present
        int total = (int)(Math.Pow(10, N) + 0.5);

        // Make a map and insert the prefixes with same
        // character in a vector
        Dictionary<int, List<string> > mp
            = new Dictionary<int, List<string> >();

        for (int i = 0; i < prefixes.Length; i++) {
            int key = (int)prefixes[i][0] - (int)'0';

            if (mp.ContainsKey(key)) {
                List<string> temp = mp[key];
                temp.Add(prefixes[i]);
                mp[key] = temp;
            }
            else {
                List<string> temp = new List<string>();
                temp.Add(prefixes[i]);
                mp[key] = temp;
            }
        }

        // Make a new vector of prefixes strings
        List<string> new_prefixes = new List<string>();

        // Iterate for each starting character
        foreach(int x in mp.Keys)
        {

            int mn = N;

            // Iterate through the vector to calculate
            // minimum size prefix
            foreach(string p in mp[x])
            {
                mn = Math.Min(mn, p.Length);
            }

            // Take all the minimum prefixes in the new
            // vector of prefixes
            foreach(string p in mp[x])
            {
                if (p.Length > mn) {
                    continue;
                }
                new_prefixes.Add(p);
            }
        }

        // Iterate through the new prefixes
        for (int i = 0; i < new_prefixes.Count; i++) {

            // Subtract bad strings
            total
                -= (int)(Math.Pow(
                             10, N - new_prefixes[i].Length)
                         + 0.5);
        }
        return total;
    }

    // Driver Code
    public static void Main(string[] args)
    {

        int N = 5;
        string[] prefixes = { "1", "0", "911" };
        Console.WriteLine(totalGoodStrings(N, prefixes));
    }
}

// This code is contributed by ukasp.

JavaScript

// Javascript Program to implement the above approach

// Change int to long long in case of overflow!!
// Function to calculate total strings of length
// N without the given prefixes
function totalGoodStrings(N, prefixes) {

  // Calculate total strings present
  let total = Math.floor(Math.pow(10, N) + 0.5);

  // Make a map and insert the prefixes with same
  // character in a vector
  let mp = new Map();
  for (let i = 0; i < prefixes.length; i++) {
    let key = prefixes[i][0] - '0'.charCodeAt(0);

    if (mp.has(key)) {
      let temp = mp.get(key);
      temp.push(prefixes[i])
      mp.set(key, temp)
    } else {
      mp.set(key, [prefixes[i]])
    }
  }

  // Make a new vector of prefixes strings
  let new_prefixes = [];

  // Iterate for each starting character
  for (x of mp) {

    let mn = N;

    // Iterate through the vector to calculate
    // minimum size prefix
    for (p of x[1]) {
      mn = Math.min(mn, p.length);
    }

    // Take all the minimum prefixes in the new
    // vector of prefixes
    for (p of x[1]) {
      if (p.length > mn) {
        continue;
      }
      new_prefixes.push(p);
    }
  }

  // Iterate through the new prefixes
  for (let i = 0; i < new_prefixes.length; i++) {

    // Subtract bad strings
    total -= Math.floor(Math.pow(10, N - new_prefixes[i].length) + 0.5);
  }
  return total;
}

// Driver Code

let N = 5;

let prefixes = ["1", "0", "911"];

document.write(totalGoodStrings(N, prefixes))

// This code is contributed by saurabh_jaiswal.

Output

Time Complexity: O(M), where M is the size of the vector prefixes[]
Auxiliary Space: O(M*K), where K is the maximum length of a string

Count of repeating digits in a given Number

kartikmodi

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