Counting numbers with given digits and digit sum
Last Updated :
20 Mar, 2024
Given a number N, count the numbers X of length exactly N such that the number X and the sum of digits of the number X have digits A and B only in their decimal representation. The length of a number is defined as the number of digits in its decimal representation without leading zeroes.
Note: As this count may be large, return the remainder after dividing it by 1000000007 (109 + 7).
Example:
Input: A = 1, B = 2, N = 2
Output: 1
Explanation: The numbers of length 2 having only digits 1 and 2 are: 11, 12, 21, 22 out of which only 11 has the sum of digits = 2.
Input: A = 2, B = 4, N = 6
Output: 7
Explanation: The numbers of length 2 having only digits 2 and 4 are: 244444, 424444, 442444, 444244, 444424, 444442, 444444 and their sum of digits are: 22, 22, 22, 22, 22, 22, 24 respectively.
Approach: To solve the problem, follow the below idea:
The idea is to take only those numbers which have digits A and B and then determine if their sum also has digits A and B only. For a number of length N every i occurrence of A, will have (N - i) occurrences of B. If the sum (A * i + B * (N - i)) forms a valid number, then we can calculate all possible combinations of A and B using the formula N! / ((N − i)! * i!) , which represents the binomial coefficient. We precompute the factorial and inverse factorial of all numbers up to N in advance. The inverse factorial is used to calculate the denominator of the binomial coefficient in a modular arithmetic. In modular arithmetic, we can’t simply divide by a number. Instead, we multiply by its modular multiplicative inverse. The function power(fact[i], MOD - 2), using Fermat Little Theorem, calculates this inverse using Binary Exponentiation. The count of such numbers, given by the binomial coefficient, is then added to the final answer.
Step-by-step algorithm:
- Maintain a Function power(a, b) to calculate a^b under modulo MOD using binary exponentiation.
- Generate all numbers in the order of occurrences of digit A:
- 0 occurrence of digit A, N occurrences of digit B.
- 1 occurrence of digit A, (N - 1) occurrences of digit B.
- 2 occurrence of digit A, (N - 2) occurrence of digit B and so on.
- Check for the sum of digits of all the above numbers and if the sum has only digits A and B, count all the numbers that can be formed by rearranging all the digits.
- Sum of all the above numbers will be the final answer.
Below is the implementation of the algorithm:
C++
// C++ Code
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
// Function to calculate a^b under modulo MOD using
// binary exponentiation
long long power(long long a, long long b)
{
// If b = 0, whatever be the value of a,
// our result will be 1.
long long res = 1;
while (b > 0) {
// If b is an odd number, then
// (a^b) = (a * (a^(b–1)/2)^2)
if (b & 1) {
res = (res * a) % MOD;
}
// If b is an even number, then
// (a^b) = ((a^2)^(b/2))
a = (a * a) % MOD;
b >>= 1;
}
return res;
}
// Function to check if a number is good
bool check(long long N, long long A, long long B)
{
while (N) {
// If any digit is neither A nor B, the number is
// not valid
if (N % 10 != A && N % 10 != B)
return 0;
N /= 10;
}
// All digits are either A or B
return 1;
}
long long excellentno(long long A, long long B, long long N)
{
vector<long long> fact(N + 1), inv(N + 1);
fact[0] = inv[0] = 1;
// Precompute factorials and their inverses
for (long long i = 1; i <= N; i++) {
// Compute factorial
fact[i] = (fact[i - 1] * i) % MOD;
// Compute inverse factorial
inv[i] = power(fact[i], MOD - 2);
}
long long ans = 0;
for (long long i = 0; i <= N; i++) {
// If the sum of digits is a good number, add the
// count to the answer
if (check(A * i + B * (N - i), A, B)) {
// Total combinations = N! / (i!(N-i)!) = N! *
// inv[i] * inv[N-i]
ans = (ans
+ (fact[N] * inv[i] % MOD * inv[N - i]
% MOD))
% MOD;
}
}
return ans;
}
int main()
{
long long A, B, N;
// Example 1
A = 1, B = 2, N = 2;
cout << excellentno(A, B, N) << endl;
// Example 2
A = 2, B = 4, N = 6;
cout << excellentno(A, B, N) << endl;
return 0;
}
import java.util.*;
public class ExcellentNumber {
static final long MOD = 1000000007;
// Function to calculate a^b under modulo MOD using
// binary exponentiation
static long power(long a, long b)
{
// If b = 0, whatever be the value of a, our result
// will be 1.
long res = 1;
while (b > 0) {
// If b is an odd number, then (a^b) = (a *
// (a^(b–1)/2)^2)
if ((b & 1) == 1) {
res = (res * a) % MOD;
}
// If b is an even number, then (a^b) =
// ((a^2)^(b/2))
a = (a * a) % MOD;
b >>= 1;
}
return res;
}
// Function to check if a number is good
static boolean check(long N, long A, long B)
{
while (N > 0) {
// If any digit is neither A nor B, the number
// is not valid
if (N % 10 != A && N % 10 != B)
return false;
N /= 10;
}
// All digits are either A or B
return true;
}
static long excellentNo(long A, long B, long N)
{
long[] fact = new long[(int)(N + 1)];
long[] inv = new long[(int)(N + 1)];
fact[0] = inv[0] = 1;
// Precompute factorials and their inverses
for (long i = 1; i <= N; i++) {
// Compute factorial
fact[(int)i] = (fact[(int)(i - 1)] * i) % MOD;
// Compute inverse factorial
inv[(int)i] = power(fact[(int)i], MOD - 2);
}
long ans = 0;
for (long i = 0; i <= N; i++) {
// If the sum of digits is a good number, add
// the count to the answer
if (check(A * i + B * (N - i), A, B)) {
// Total combinations = N! / (i!(N-i)!) = N!
// * inv[i] * inv[N-i]
ans = (ans
+ (fact[(int)N] * inv[(int)i] % MOD
* inv[(int)(N - i)] % MOD))
% MOD;
}
}
return ans;
}
public static void main(String[] args)
{
long A, B, N;
// Example 1
A = 1;
B = 2;
N = 2;
System.out.println(excellentNo(A, B, N));
// Example 2
A = 2;
B = 4;
N = 6;
System.out.println(excellentNo(A, B, N));
}
}
// This code is contributed by akshitaguprzj3
using System;
using System.Numerics;
public class Program
{
const long MOD = 1000000007;
// Function to calculate a^b under modulo MOD using
// binary exponentiation
static long Power(long a, long b)
{
// If b = 0, whatever be the value of a,
// our result will be 1.
long res = 1;
while (b > 0)
{
// If b is an odd number, then
// (a^b) = (a * (a^(b–1)/2)^2)
if ((b & 1) == 1)
{
res = (res * a) % MOD;
}
// If b is an even number, then
// (a^b) = ((a^2)^(b/2))
a = (a * a) % MOD;
b >>= 1;
}
return res;
}
// Function to check if a number is good
static bool Check(long N, long A, long B)
{
while (N > 0)
{
// If any digit is neither A nor B, the number is
// not valid
if (N % 10 != A && N % 10 != B)
return false;
N /= 10;
}
// All digits are either A or B
return true;
}
static long ExcellentNo(long A, long B, long N)
{
long[] fact = new long[N + 1];
long[] inv = new long[N + 1];
fact[0] = inv[0] = 1;
// Precompute factorials and their inverses
for (long i = 1; i <= N; i++)
{
// Compute factorial
fact[i] = (fact[i - 1] * i) % MOD;
// Compute inverse factorial
inv[i] = Power(fact[i], MOD - 2);
}
long ans = 0;
for (long i = 0; i <= N; i++)
{
// If the sum of digits is a good number, add the
// count to the answer
if (Check(A * i + B * (N - i), A, B))
{
// Total combinations = N! / (i!(N-i)!) = N! *
// inv[i] * inv[N-i]
ans = (ans + (fact[N] * inv[i] % MOD * inv[N - i] % MOD)) % MOD;
}
}
return ans;
}
public static void Main(string[] args)
{
long A, B, N;
// Example 1
A = 1; B = 2; N = 2;
Console.WriteLine(ExcellentNo(A, B, N));
// Example 2
A = 2; B = 4; N = 6;
Console.WriteLine(ExcellentNo(A, B, N));
}
}
// Define the modulo constant
const MOD = BigInt(1000000007);
// Function to calculate a^b under modulo MOD using binary exponentiation
function power(a, b) {
let res = BigInt(1);
a = BigInt(a);
b = BigInt(b);
while (b > 0) {
if (b & 1n) {
res = (res * a) % MOD;
}
a = (a * a) % MOD;
b >>= 1n;
}
return res;
}
// Function to check if a number is good
function check(N, A, B) {
while (N) {
if (N % 10n != BigInt(A) && N % 10n != BigInt(B))
return false;
N /= 10n;
}
return true;
}
function excellentno(A, B, N) {
let fact = new Array(N + 1).fill(BigInt(1));
let inv = new Array(N + 1).fill(BigInt(1));
for (let i = 1; i <= N; i++) {
fact[i] = (fact[i - 1] * BigInt(i)) % MOD;
inv[i] = power(fact[i], MOD - 2n);
}
let ans = BigInt(0);
for (let i = 0; i <= N; i++) {
if (check(BigInt(A) * BigInt(i) + BigInt(B) * BigInt(N - i), A, B)) {
ans = (ans + (fact[N] * inv[i] % MOD * inv[N - i] % MOD)) % MOD;
}
}
return ans.toString();
}
let A = 1, B = 2, N = 2;
console.log(excellentno(A, B, N));
A = 2, B = 4, N = 6;
console.log(excellentno(A, B, N));
MOD = 1000000007
# Function to calculate a^b under modulo MOD using binary exponentiation
def power(a, b):
# If b = 0, whatever be the value of a, our result will be 1.
res = 1
while b > 0:
# If b is an odd number, then (a^b) = (a * (a^(b–1)/2)^2)
if b & 1:
res = (res * a) % MOD
# If b is an even number, then (a^b) = ((a^2)^(b/2))
a = (a * a) % MOD
b >>= 1
return res
# Function to check if a number is good
def check(N, A, B):
while N:
# If any digit is neither A nor B, the number is not valid
if N % 10 != A and N % 10 != B:
return False
N //= 10
# All digits are either A or B
return True
def excellentno(A, B, N):
fact = [0] * (N + 1)
inv = [0] * (N + 1)
fact[0] = inv[0] = 1
# Precompute factorials and their inverses
for i in range(1, N + 1):
# Compute factorial
fact[i] = (fact[i - 1] * i) % MOD
# Compute inverse factorial
inv[i] = power(fact[i], MOD - 2)
ans = 0
for i in range(N + 1):
# If the sum of digits is a good number, add the count to the answer
if check(A * i + B * (N - i), A, B):
# Total combinations = N! / (i!(N-i)!) = N! * inv[i] * inv[N-i]
ans = (ans + (fact[N] * inv[i] % MOD * inv[N - i] % MOD)) % MOD
return ans
# Example 1
A, B, N = 1, 2, 2
print(excellentno(A, B, N))
# Example 2
A, B, N = 2, 4, 6
print(excellentno(A, B, N))
Time Complexity: O(N * logN), where N is the length of numbers.
Auxiliary Space: O(N)
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