# Count numbers less than N containing digits from the given set : Digit DP

• Difficulty Level : Medium
• Last Updated : 29 Jun, 2022

Given an integer N and set of digits D[], which consists of digits from [1, 9]. The task is to count the numbers possible less than N, whose digits are from the given set of digits.

Examples:

Input: D = [“1”, “4”, “9”], N = 10
Output:
Explanation:
There are only 3 numbers possible less than 3 with given set of digits –
1, 4, 9

Input: D[] = {“1”, “3”, “5”, “7”}, N = 100
Output: 20
Explanation:
There are only 20 numbers possible less than 100 with given set of digits –
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Naive Approach:Check the digits of all the numbers of range [1, N], If all the digits of a number belong to the given digit set then increment the count by 1.

Efficient Approach: The idea is to use the concept of Digit DP and traverse the given set of digits and generate all the numbers which are strictly less than the given number N. Recursively choose the digit for all the possible position of the number and pass a boolean variable tight to check that by including that digit, the number falls into the given range or not.

Let’s think of the possible state for the DP –

1. pos: It tells about the position of the digit to be chosen, such that the number falls into the given range.
2. tight: This will help us know about the current digits are restricted or not. If the digits are restricted, then any digit can be chosen from the given set of digits. Otherwise, the digits can be chosen in range [1, N[pos]].
3. size: It will tells the number of the digits to be chosen.

Below is the illustration of the recursive function:

• Base Case: The base case for this problem will be when the position of the digit to be chosen is equal to the length of digits to be chosen, then there is only one possible number containing the digits which are chosen till yet.
if (position == countDigits(N))
return 1
• Recursive Case: For generating the number in the given range, use the tight variable to choose the possible digits in range as follows:
• If the value of tight is 0, denotes that by including that digit will give the number less than the given range.
• Otherwise, If the value of tight is 1, denotes that by including that digit, it will give the number greater than the given range. So we can remove all permutations after getting tight value 1 to avoid more number of recursive calls.
• Store the count of the numbers possible after choosing each digit for every position.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the// count of numbers possible less// than N, such that every digit// is from the given set of digits#include  using namespace std; int dp[15][2]; // Function to convert integer// into the stringstring convertToString(int num){    stringstream ss;    ss << num;    string s = ss.str();    return s;} // Recursive function to find the// count of numbers possible less// than N, such that every digit// is from the given set of digitsint calculate(int pos, int tight,    int D[], int sz, string& num){    // Base case    if (pos == num.length())        return 1;         // Condition when the subproblem    // is computed previously    if (dp[pos][tight] != -1)        return dp[pos][tight];     int val = 0;         // Condition when the number    // chosen till now is definitely    // smaller than the given number N    if (tight == 0) {                 // Loop to traverse all the        // digits of the given set        for (int i = 0; i < sz; i++) {                         if (D[i] < (num[pos] - '0')) {                val += calculate(pos + 1,                           1, D, sz, num);            }            else if (D[i] == num[pos] - '0')                val += calculate(pos + 1,                       tight, D, sz, num);        }    }    else {        // Loop to traverse all the        // digits from the given set        for (int i = 0; i < sz; i++) {            val += calculate(pos + 1,                    tight, D, sz, num);        }    }         // Store the solution for    // current subproblem    return dp[pos][tight] = val;} // Function to count the numbers// less than N from given set of digitsint countNumbers(int D[], int N, int sz){    // Converting the number to string    string num = convertToString(N);    int len = num.length();         // Initially no subproblem    // is solved till now    memset(dp, -1, sizeof(dp));         // Find the solution of all the    // number equal to the length of    // the given number N    int ans = calculate(0, 0, D, sz, num);         // Loop to find the number less in    // in the length of the given number    for (int i = 1; i < len; i++)        ans += calculate(i, 1, D, sz, num);     return ans;} // Driver Codeint main(){    int sz = 3;     int D[sz] = { 1, 4, 9 };    int N = 10;         // Function Call    cout << countNumbers(D, N, sz);    return 0;}

## Java

 // Java implementation to find the// count of numbers possible less// than N, such that every digit// is from the given set of digitsimport java.util.*; class GFG{  static int [][]dp = new int[15][2];  // Function to convert integer// into the Stringstatic String convertToString(int num){    return String.valueOf(num);}  // Recursive function to find the// count of numbers possible less// than N, such that every digit// is from the given set of digitsstatic int calculate(int pos, int tight,    int D[], int sz, String num){    // Base case    if (pos == num.length())        return 1;          // Condition when the subproblem    // is computed previously    if (dp[pos][tight] != -1)        return dp[pos][tight];      int val = 0;          // Condition when the number    // chosen till now is definitely    // smaller than the given number N    if (tight == 0) {                  // Loop to traverse all the        // digits of the given set        for (int i = 0; i < sz; i++) {                          if (D[i] < (num.charAt(pos) - '0')) {                val += calculate(pos + 1,                           1, D, sz, num);            }            else if (D[i] == num.charAt(pos) - '0')                val += calculate(pos + 1,                       tight, D, sz, num);        }    }    else {        // Loop to traverse all the        // digits from the given set        for (int i = 0; i < sz; i++) {            val += calculate(pos + 1,                    tight, D, sz, num);        }    }          // Store the solution for    // current subproblem    return dp[pos][tight] = val;}  // Function to count the numbers// less than N from given set of digitsstatic int countNumbers(int D[], int N, int sz){    // Converting the number to String    String num = convertToString(N);    int len = num.length();          // Initially no subproblem    // is solved till now    for (int i = 0; i < 15; i++)        for (int j = 0; j < 2; j++)            dp[i][j] = -1;          // Find the solution of all the    // number equal to the length of    // the given number N    int ans = calculate(0, 0, D, sz, num);          // Loop to find the number less in    // in the length of the given number    for (int i = 1; i < len; i++)        ans += calculate(i, 1, D, sz, num);      return ans;}  // Driver Codepublic static void main(String[] args){    int sz = 3;      int D[] = { 1, 4, 9 };    int N = 10;          // Function Call    System.out.print(countNumbers(D, N, sz));}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 implementation to find the# count of numbers possible less# than N, such that every digit# is from the given set of digitsimport numpy as np;dp = np.ones((15,2))*-1; # Function to convert integer# into the stringdef convertToString(num) :    return str(num); # Recursive function to find the# count of numbers possible less# than N, such that every digit# is from the given set of digitsdef calculate(pos,tight,  D, sz, num) :     # Base case    if (pos == len(num)):        return 1;         # Condition when the subproblem    # is computed previously    if (dp[pos][tight] != -1) :        return dp[pos][tight];     val = 0;         # Condition when the number    # chosen till now is definitely    # smaller than the given number N    if (tight == 0) :                 # Loop to traverse all the        # digits of the given set        for i in range(sz) :                         if (D[i] < (ord(num[pos]) - ord('0'))) :                val += calculate(pos + 1, 1, D, sz, num);                         elif (D[i] == ord(num[pos]) - ord('0')) :                val += calculate(pos + 1, tight, D, sz, num);         else :        # Loop to traverse all the        # digits from the given set        for i in range(sz) :            val += calculate(pos + 1, tight, D, sz, num);                 # Store the solution for    # current subproblem    dp[pos][tight] = val;         return dp[pos][tight]; # Function to count the numbers# less than N from given set of digitsdef countNumbers(D, N, sz) :     # Converting the number to string    num = convertToString(N);    length = len(num);         # Initially no subproblem    # is solved till now    # dp = np.ones((15,2))*-1;         # Find the solution of all the    # number equal to the length of    # the given number N    ans = calculate(0, 0, D, sz, num);         # Loop to find the number less in    # in the length of the given number    for i in range(1,length) :        ans += calculate(i, 1, D, sz, num);     return ans;  # Driver Codeif __name__ == "__main__" :     sz = 3;     D = [ 1, 4, 9 ];    N = 10;         # Function Call    print(countNumbers(D, N, sz));     # This code is contributed by AnkitRai01

## C#

 // C# implementation to find the// count of numbers possible less// than N, such that every digit// is from the given set of digitsusing System; class GFG{   static int [,]dp = new int[15, 2];   // Function to convert integer// into the Stringstatic String convertToString(int num){    return String.Join("",num);}   // Recursive function to find the// count of numbers possible less// than N, such that every digit// is from the given set of digitsstatic int calculate(int pos, int tight,    int []D, int sz, String num){    // Base case    if (pos == num.Length)        return 1;           // Condition when the subproblem    // is computed previously    if (dp[pos,tight] != -1)        return dp[pos,tight];       int val = 0;           // Condition when the number    // chosen till now is definitely    // smaller than the given number N    if (tight == 0) {                   // Loop to traverse all the        // digits of the given set        for (int i = 0; i < sz; i++) {                           if (D[i] < (num[pos] - '0')) {                val += calculate(pos + 1,                           1, D, sz, num);            }            else if (D[i] == num[pos] - '0')                val += calculate(pos + 1,                       tight, D, sz, num);        }    }    else {        // Loop to traverse all the        // digits from the given set        for (int i = 0; i < sz; i++) {            val += calculate(pos + 1,                    tight, D, sz, num);        }    }           // Store the solution for    // current subproblem    return dp[pos,tight] = val;}   // Function to count the numbers// less than N from given set of digitsstatic int countNumbers(int []D, int N, int sz){    // Converting the number to String    String num = convertToString(N);    int len = num.Length;           // Initially no subproblem    // is solved till now    for (int i = 0; i < 15; i++)        for (int j = 0; j < 2; j++)            dp[i,j] = -1;           // Find the solution of all the    // number equal to the length of    // the given number N    int ans = calculate(0, 0, D, sz, num);           // Loop to find the number less in    // in the length of the given number    for (int i = 1; i < len; i++)        ans += calculate(i, 1, D, sz, num);       return ans;}   // Driver Codepublic static void Main(String[] args){    int sz = 3;       int []D = { 1, 4, 9 };    int N = 10;           // Function Call    Console.Write(countNumbers(D, N, sz));}} // This code is contributed by Princi Singh

## Javascript



Output:

3

Time complexity: O(Len(D))
Space complexity: O(12*2)

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