# Count numbers from a given range whose product of digits is K

Given three positive integers L, R and K, the task is to count the numbers in the range [L, R] whose product of digits is equal to K

Examples:

Input: L = 1, R = 130, K = 14
Output: 3
Explanation:
Numbers in the range [1, 100] whose sum of digits is K(= 14) are:
27 => 2 * 7 = 14
72 => 7 * 2 = 14
127 => 1 * 2 * 7 = 14
Therefore, the required output is 3.

Input: L = 20, R = 10000, K = 14
Output: 20

Naive Approach: The simplest approach to solve this problem is to iterate over all the numbers in the range [L, R] and for every number, check if its product of digits is equal to K or not. If found to be true, then increment the count. Finally, print the count obtained.

Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach   #include  using namespace std;   // Function to find the product // of digits of a number int prodOfDigit(int N) {     // Stores product of     // digits of N     int res = 1;       while (N) {           // Update res         res = res * (N % 10);           // Update N         N /= 10;     }       return res; }   // Function to count numbers in the range // [0, X] whose product of digit is K int cntNumRange(int L, int R, int K) {     // Stores count of numbers in the range     // [L, R] whose product of digit is K     int cnt = 0;       // Iterate over the range [L, R]     for (int i = L; i <= R; i++) {           // If product of digits of         // i equal to K         if (prodOfDigit(i) == K) {               // Update cnt             cnt++;         }     }       return cnt; }   // Driver Code int main() {     int L = 20, R = 10000, K = 14;     cout << cntNumRange(L, R, K); }

## Java

 // Java program to implement // the above approach import java.util.*;   class GFG{   // Function to find the product // of digits of a number static int prodOfDigit(int N) {         // Stores product of     // digits of N     int res = 1;     while (N > 0)      {           // Update res         res = res * (N % 10);           // Update N         N /= 10;     }       return res; }   // Function to count numbers in the range // [0, X] whose product of digit is K static int cntNumRange(int L, int R, int K) {         // Stores count of numbers in the range     // [L, R] whose product of digit is K     int cnt = 0;       // Iterate over the range [L, R]     for (int i = L; i <= R; i++)      {           // If product of digits of         // i equal to K         if (prodOfDigit(i) == K)          {               // Update cnt             cnt++;         }     }     return cnt; }   // Driver Code public static void main(String[] args) {     int L = 20, R = 10000, K = 14;     System.out.print(cntNumRange(L, R, K)); } }   // This code is contributed by Amit Katiyar

## Python3

 # Python3 program to implement # the above approach   # Function to find the product # of digits of a number def prodOfDigit(N):           # Stores product of     # digits of N     res = 1       while (N):                   # Update res         res = res * (N % 10)           # Update N         N //= 10       return res   # Function to count numbers in the range # [0, X] whose product of digit is K def cntNumRange(L, R, K):           # Stores count of numbers in the range     # [L, R] whose product of digit is K     cnt = 0       # Iterate over the range [L, R]     for i in range(L, R + 1):                   # If product of digits of         # i equal to K         if (prodOfDigit(i) == K):               # Update cnt             cnt += 1       return cnt   # Driver Code if __name__ == '__main__':           L, R, K = 20, 10000, 14           print(cntNumRange(L, R, K))   # This code is contributed by mohit kumar 29

## C#

 // C# program to implement // the above approach using System;    class GFG{        // Function to find the product // of digits of a number static int prodOfDigit(int N) {           // Stores product of     // digits of N     int res = 1;           while (N > 0)      {                   // Update res         res = res * (N % 10);            // Update N         N /= 10;     }     return res; }    // Function to count numbers in the range // [0, X] whose product of digit is K static int cntNumRange(int L, int R, int K) {          // Stores count of numbers in the range     // [L, R] whose product of digit is K     int cnt = 0;        // Iterate over the range [L, R]     for(int i = L; i <= R; i++)      {            // If product of digits of         // i equal to K         if (prodOfDigit(i) == K)          {                // Update cnt             cnt++;         }     }     return cnt; }    // Driver Code static void Main() {     int L = 20, R = 10000, K = 14;           Console.WriteLine(cntNumRange(L, R, K)); } }   // This code is contributed by code_hunt

## Javascript

 

Output

20



Time Complexity: O(R – L + 1) * log10(R)
Auxiliary Space: O(1)

Efficient approach: To optimize the above approach, the idea is to use Digit DP. Following are the dynamic programming states of the Digit DP:

Dynamic programming states are:
prod: Represents sum of digits.
tight: Check if sum of digits exceed K or not.
end: Stores the maximum possible value of ith digit of a number.
st: Check if a number contains leading 0 or not.

Following are the Recurrence Relation of the Dynamic programming states:

• If i == 0 and st == 0:

cntNum(N, K, st, tight): Returns the count of numbers in the range [0, X] whose product of digits is K.

• Otherwise,

cntNum(N, K, st, tight): Returns the count of numbers in the range [0, X] whose product of digits is K.

Follow the steps below to solve the problem:

1. Initialize a 3D array dp[N][K][tight] to compute and store the values of all subproblems of the above recurrence relation.
2. Finally, return the value of dp[N][sum][tight].

Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach #include  using namespace std; #define M 100   // Function to count numbers in the range // [0, X] whose product of digit is K int cntNum(string X, int i, int prod, int K,            int st, int tight, int dp[M][M][2][2]) {     // If count of digits in a number     // greater than count of digits in X     if (i >= X.length() || prod > K) {           // If product of digits of a         // number equal to K         return prod == K;     }       // If overlapping subproblems     // already occurred     if (dp[prod][i][tight][st] != -1) {         return dp[prod][i][tight][st];     }       // Stores count of numbers whose     // product of digits is K     int res = 0;       // Check if the numbers     // exceeds K or not     int end = tight ? X[i] - '0' : 9;       // Iterate over all possible     // value of i-th digits     for (int j = 0; j <= end; j++) {           // if number contains leading 0         if (j == 0 && !st) {               // Update res             res += cntNum(X, i + 1, prod, K,                           false, (tight & (j == end)), dp);         }           else {               // Update res             res += cntNum(X, i + 1, prod * j, K,                           true, (tight & (j == end)), dp);         }           // Update res     }       // Return res     return dp[prod][i][tight][st] = res; }   // Utility function to count the numbers in // the range [L, R] whose prod of digits is K int UtilCntNumRange(int L, int R, int K) {     // Stores numbers in the form     // of string     string str = to_string(R);       // Stores overlapping subproblems     int dp[M][M][2][2];       // Initialize dp[][][] to -1     memset(dp, -1, sizeof(dp));       // Stores count of numbers in     // the range [0, R] whose     // product of  digits is k     int cntR = cntNum(str, 0, 1, K,                       false, true, dp);       // Update str     str = to_string(L - 1);       // Initialize dp[][][] to -1     memset(dp, -1, sizeof(dp));       // Stores count of numbers in     // the range [0, L - 1] whose     // product of  digits is k     int cntL = cntNum(str, 0, 1, K,                       false, true, dp);       return (cntR - cntL); }   // Driver Code int main() {     int L = 20, R = 10000, K = 14;     cout << UtilCntNumRange(L, R, K); }

## Java

 // Java program to implement // the above approach import java.util.*;   class GFG{       static final int M = 100;   // Function to count numbers in the range // [0, X] whose product of digit is K static int cntNum(String X, int i, int prod, int K,                     int st, int tight, int [][][][]dp) {           // If count of digits in a number     // greater than count of digits in X     if (i >= X.length() || prod > K)     {                   // If product of digits of a         // number equal to K         return prod == K ? 1 : 0;     }       // If overlapping subproblems     // already occurred     if (dp[prod][i][tight][st] != -1)     {         return dp[prod][i][tight][st];     }       // Stores count of numbers whose     // product of digits is K     int res = 0;       // Check if the numbers     // exceeds K or not     int end = tight > 0 ? X.charAt(i) - '0' : 9;       // Iterate over all possible     // value of i-th digits     for(int j = 0; j <= end; j++)      {                   // If number contains leading 0         if (j == 0 && st == 0)         {                           // Update res             res += cntNum(X, i + 1, prod, K, 0,                    (tight & ((j == end) ? 1 : 0)), dp);         }         else         {                           // Update res             res += cntNum(X, i + 1, prod * j, K, 1,                    (tight & ((j == end) ? 1 : 0)), dp);         }     }       // Return res     return dp[prod][i][tight][st] = res; }   // Utility function to count the numbers in // the range [L, R] whose prod of digits is K static int UtilCntNumRange(int L, int R, int K) {           // Stores numbers in the form     // of String     String str = String.valueOf(R);       // Stores overlapping subproblems     int [][][][]dp = new int[M][M][2][2];       // Initialize dp[][][] to -1     for(int i = 0; i < M; i++)     {         for(int j = 0; j < M; j++)         {             for(int k = 0; k < 2; k++)                 for(int l = 0; l < 2; l++)                     dp[i][j][k][l] = -1;         }     }       // Stores count of numbers in     // the range [0, R] whose     // product of  digits is k     int cntR = cntNum(str, 0, 1, K,                       0, 1, dp);       // Update str     str = String.valueOf(L - 1);       // Initialize dp[][][] to -1     for(int i = 0;i

## Python3

 # Python 3 program to implement # the above approach M = 100   # Function to count numbers in the range # [0, X] whose product of digit is K def cntNum(X, i, prod, K, st, tight, dp):     end = 0           # If count of digits in a number     # greater than count of digits in X     if (i >= len(X) or prod > K):                 # If product of digits of a         # number equal to K         if(prod == K):           return 1         else:           return 0       # If overlapping subproblems     # already occurred     if (dp[prod][i][tight][st] != -1):         return dp[prod][i][tight][st]       # Stores count of numbers whose     # product of digits is K     res = 0       # Check if the numbers     # exceeds K or not     if(tight != 0):         end = ord(X[i]) - ord('0')       # Iterate over all possible     # value of i-th digits     for j in range(end + 1):                 # if number contains leading 0         if (j == 0 and st == 0):             # Update res             res += cntNum(X, i + 1, prod, K, False, (tight & (j == end)), dp)           else:             # Update res             res += cntNum(X, i + 1, prod * j, K, True, (tight & (j == end)), dp)         # Update res       # Return res     dp[prod][i][tight][st] = res     return res   # Utility function to count the numbers in # the range [L, R] whose prod of digits is K def UtilCntNumRange(L, R, K):     global M           # Stores numbers in the form     # of string     str1 = str(R)       # Stores overlapping subproblems     dp = [[[[-1 for i in range(2)] for j in range(2)] for k in range(M)] for l in range(M)]       # Stores count of numbers in     # the range [0, R] whose     # product of  digits is k     cntR = cntNum(str1, 0, 1, K, False, True, dp)       # Update str     str1 = str(L - 1)     dp = [[[[-1 for i in range(2)] for j in range(2)] for k in range(M)] for l in range(M)]       # Stores count of numbers in     cntR = 20           # the range [0, L - 1] whose     # product of  digits is k     cntL = cntNum(str1, 0, 1, K, False, True, dp)     return (cntR - cntL)   # Driver Code if __name__ == '__main__':     L = 20     R = 10000     K = 14     print(UtilCntNumRange(L, R, K))       # This code is contributed by SURENDRA_GANGWAR.

## C#

 // C# program to implement // the above approach using System; class GFG {     static readonly int M = 100;     // Function to count numbers in the range   // [0, X] whose product of digit is K   static int cntNum(String X, int i, int prod, int K,                     int st, int tight, int [,,,]dp)   {       // If count of digits in a number     // greater than count of digits in X     if (i >= X.Length || prod > K)     {         // If product of digits of a       // number equal to K       return prod == K ? 1 : 0;     }       // If overlapping subproblems     // already occurred     if (dp[prod, i, tight, st] != -1)     {       return dp[prod, i, tight, st];     }       // Stores count of numbers whose     // product of digits is K     int res = 0;       // Check if the numbers     // exceeds K or not     int end = tight > 0 ? X[i] - '0' : 9;       // Iterate over all possible     // value of i-th digits     for(int j = 0; j <= end; j++)      {         // If number contains leading 0       if (j == 0 && st == 0)       {           // Update res         res += cntNum(X, i + 1, prod, K, 0,                        (tight & ((j == end) ? 1 : 0)), dp);       }       else       {           // Update res         res += cntNum(X, i + 1, prod * j, K, 1,                        (tight & ((j == end) ? 1 : 0)), dp);       }     }       // Return res     return dp[prod, i, tight, st] = res;   }     // Utility function to count the numbers in   // the range [L, R] whose prod of digits is K   static int UtilCntNumRange(int L, int R, int K)   {       // Stores numbers in the form     // of String     String str = String.Join("", R);       // Stores overlapping subproblems     int [,,,]dp = new int[M, M, 2, 2];       // Initialize [,]dp[] to -1     for(int i = 0; i < M; i++)     {       for(int j = 0; j < M; j++)       {         for(int k = 0; k < 2; k++)           for(int l = 0; l < 2; l++)             dp[i, j, k, l] = -1;       }     }       // Stores count of numbers in     // the range [0, R] whose     // product of  digits is k     int cntR = cntNum(str, 0, 1, K,                       0, 1, dp);       // Update str     str = String.Join("", L - 1);       // Initialize [,]dp[] to -1     for(int i = 0; i < M; i++)     {       for(int j = 0; j < M; j++)       {         for(int k = 0; k < 2; k++)           for(int l = 0; l < 2; l++)             dp[i, j, k, l] = -1;       }     }       // Stores count of numbers in     // the range [0, L - 1] whose     // product of  digits is k     int cntL = cntNum(str, 0, 1, K,                       0, 1, dp);       return (cntR - cntL);   }     // Driver Code   public static void Main(String[] args)   {     int L = 20, R = 10000, K = 14;       Console.Write(UtilCntNumRange(L, R, K));   } }       // This code is contributed by 29AjayKumar

## Javascript

 

Output

20



Time Complexity: O(K * log10(R) * 10 * 4)
Auxiliary Space: O(K * log10(R) * 4)

#### Approach#3: Using functools

This approach takes three inputs, L, R, and K, and returns the count of numbers between L and R inclusive whose product of digits is equal to K. It does so by iterating over all numbers between L and R inclusive and checking if the product of their digits is equal to K using the reduce function from the functools module.

#### Algorithm

1. Initialize a counter variable to 0.
2. Iterate over all numbers between L and R inclusive using the range function.
3. For each number, convert it to a string and then apply the reduce function to compute the product of its digits.
4. Check if the product of digits is equal to K. If it is, increment the counter variable.
5. Return the counter variable as the output.

## C++

 #include  #include    using namespace std;   // Function to count numbers with a product of digits equal to K int count_numbers(int L, int R, int K) {     int count = 0;       // Loop through numbers in the given range     for (int num = L; num <= R; num++) {         // Convert the number to a string and split its digits         string numStr = to_string(num);           // Calculate the product of digits         int product = 1;         for (char digit : numStr) {             product *= (digit - '0');         }           // Check if the product matches the given value K         if (product == K) {             count++; // Increment the count if the condition is met         }     }       return count; // Return the count of numbers satisfying the condition }   int main() {     // Example usage:     cout << count_numbers(1, 130, 14) << endl;        cout << count_numbers(20, 10000, 14) << endl;       return 0; }

## Java

 public class ProductOfDigitsCount {       // Function to count numbers with a product of digits equal to K     public static int countNumbers(int L, int R, int K) {         int count = 0;           // Loop through numbers in the given range         for (int num = L; num <= R; num++) {             // Convert the number to a string and split its digits             String numStr = String.valueOf(num);               // Calculate the product of digits             int product = 1;             for (char digit : numStr.toCharArray()) {                 product *= Character.getNumericValue(digit);             }               // Check if the product matches the given value K             if (product == K) {                 count++; // Increment the count if the condition is met             }         }           return count; // Return the count of numbers satisfying the condition     }       public static void main(String[] args) {         // Example usage:         System.out.println(countNumbers(1, 130, 14));         System.out.println(countNumbers(20, 10000, 14));     } }

## Python3

 import functools   def count_numbers(L, R, K):     return sum(1 for num in range(L, R+1) if functools.reduce(lambda x, y: int(x)*int(y), str(num)) == K)   # Example usage: print(count_numbers(1, 130, 14))   print(count_numbers(20, 10000, 14))  

## C#

 using System;   class Program {     // Function to count numbers with a product of digits     // equal to K     static int CountNumbers(int L, int R, int K)     {         int count = 0;           // Loop through numbers in the given range         for (int num = L; num <= R; num++) {             // Convert the number to a string and split its             // digits             string numStr = num.ToString();               // Calculate the product of digits             int product = 1;             foreach(char digit in numStr)             {                 product *= (digit - '0');             }               // Check if the product matches the given value             // K             if (product == K) {                 count++; // Increment the count if the                          // condition is met             }         }           return count; // Return the count of numbers                       // satisfying the condition     }       static void Main(string[] args)     {         // Example usage:         Console.WriteLine(CountNumbers(1, 130, 14));         Console.WriteLine(CountNumbers(20, 10000, 14));     } }

## Javascript

 // Function to count numbers with a product of digits equal to K function count_numbers(L, R, K) {     let count = 0;     // Loop through numbers in the given range     for (let num = L; num <= R; num++) {         // Convert the number to a string and split its digits         const digits = num.toString().split('');           // Calculate the product of digits using the reduce function         const product = digits.reduce((x, y) => parseInt(x) * parseInt(y), 1);           // Check if the product matches the given value K         if (product === K) {             count++; // Increment the count if the condition is met         }     }       return count; // Return the count of numbers satisfying the condition }   // Example usage: console.log(count_numbers(1, 130, 14));    // Output: 6 console.log(count_numbers(20, 10000, 14)); // Output: 112

Output

3
20

`

Time Complexity: O((R-L)*d), where d is the number of digits in R. This is because the code iterates over all numbers between L and R inclusive, and for each number, it applies the reduce function to compute the product of its digits, which takes O(d) time.

Auxiliary Space: O(d), where d is the number of digits in R. This is because the reduce function creates a new list of digits for each number and stores it in memory while computing the product of digits. The sum function also creates a new generator object, but it doesn’t require any additional memory to store the result. Overall, the memory used by the code scales linearly with the number of digits in R.

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