# Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M

Last Updated : 08 Mar, 2023

Given a range [L, R] and two positive integers N and M. The task is to count the numbers in the range containing only non-zero digits whose sum of digits is equal to N and the number is divisible by M.
Examples:

Input: L = 1, R = 100, N = 8, M = 2
Output:
Only 8, 26, 44 and 62 are valid numbers

Input: L = 1, R = 200, N = 4, M = 11
Output:
Only 22 and 121 are valid numbers

Prerequisites : Digit DP

Approach: Firstly, if we are able to count the required numbers up to R i.e. in the range [0, R], we can easily reach our answer in the range [L, R] by solving for from zero to R and then subtracting the answer we get after solving for from zero to L â€“ 1. Now, we need to define the DP states.
DP States

• Since we can consider our number as a sequence of digits, one state is the position at which we are currently in. This position can have values from 0 to 18 if we are dealing with numbers up to 1018. In each recursive call, we try to build the sequence from left to right by placing a digit from 0 to 9.
• Second state is the sum of the digits we have placed so far.
• Third state is the remainder which defines the modulus of the number we have made so far modulo M.
• Another state is the boolean variable tight which tells the number we are trying to build has already become smaller than R so that in the upcoming recursive calls we can place any digit from 0 to 9. If the number has not become smaller, the maximum limit of digit we can place is digit at the current position in R.

For the number to have only non-zero digits, we maintain a variable nonz whose value if 1 tells the first digit in the number we have placed is a non-zero digit and thus, now we can’t place any zero digit in upcoming calls. Otherwise, we can place a zero digit as a leading zero so as to make number of digits in current number smaller than number of digits in upper limit.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `const` `int` `M = 20;`   `// states - position, sum, rem, tight` `// sum can have values upto 162, if we` `// are dealing with numbers upto 10^18` `// when all 18 digits are 9, then sum` `// is 18 * 9 = 162` `int` `dp[M][165][M][2];`   `// n is the sum of digits and number should` `// be divisible by m` `int` `n, m;`   `// Function to return the count of` `// required numbers from 0 to num` `int` `count(``int` `pos, ``int` `sum, ``int` `rem, ``int` `tight,` `          ``int` `nonz, vector<``int``> num)` `{` `    ``// Last position` `    ``if` `(pos == num.size()) {` `        ``if` `(rem == 0 && sum == n)` `            ``return` `1;` `        ``return` `0;` `    ``}`   `    ``// If this result is already computed` `    ``// simply return it` `    ``if` `(dp[pos][sum][rem][tight] != -1)` `        ``return` `dp[pos][sum][rem][tight];`   `    ``int` `ans = 0;`   `    ``// Maximum limit upto which we can place` `    ``// digit. If tight is 1, means number has` `    ``// already become smaller so we can place` `    ``// any digit, otherwise num[pos]` `    ``int` `limit = (tight ? 9 : num[pos]);`   `    ``for` `(``int` `d = 0; d <= limit; d++) {`   `        ``// If the current digit is zero` `        ``// and nonz is 1, we can't place it` `        ``if` `(d == 0 && nonz)` `            ``continue``;` `        ``int` `currSum = sum + d;` `        ``int` `currRem = (rem * 10 + d) % m;` `        ``int` `currF = tight || (d < num[pos]);` `        ``ans += count(pos + 1, currSum, currRem,` `                     ``currF, nonz || d, num);` `    ``}` `    ``return` `dp[pos][sum][rem][tight] = ans;` `}`   `// Function to convert x into its digit vector` `// and uses count() function to return the` `// required count` `int` `solve(``int` `x)` `{` `    ``vector<``int``> num;` `    ``while` `(x) {` `        ``num.push_back(x % 10);` `        ``x /= 10;` `    ``}` `    ``reverse(num.begin(), num.end());`   `    ``// Initialize dp` `    ``memset``(dp, -1, ``sizeof``(dp));` `    ``return` `count(0, 0, 0, 0, 0, num);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `L = 1, R = 100;` `    ``n = 8, m = 2;` `    ``cout << solve(R) - solve(L);` `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach ` `import` `java.util.*;`   `class` `GFG` `{` `    `  `static` `int` `M = ``20``; `   `// states - position, sum, rem, tight ` `// sum can have values upto 162, if we ` `// are dealing with numbers upto 10^18 ` `// when all 18 digits are 9, then sum ` `// is 18 * 9 = 162 ` `static` `int` `dp[][][][] = ``new` `int` `[M][``165``][M][``2``]; `   `// n is the sum of digits and number should ` `// be divisible by m ` `static` `int` `n, m; `   `// Function to return the count of ` `// required numbers from 0 to num ` `static` `int` `count(``int` `pos, ``int` `sum, ``int` `rem, ``int` `tight, ` `        ``int` `nonz, Vector num) ` `{ ` `    ``// Last position ` `    ``if` `(pos == num.size())` `    ``{ ` `        ``if` `(rem == ``0` `&& sum == n) ` `            ``return` `1``; ` `        ``return` `0``; ` `    ``} `   `    ``// If this result is already computed ` `    ``// simply return it ` `    ``if` `(dp[pos][sum][rem][tight] != -``1``) ` `        ``return` `dp[pos][sum][rem][tight]; `   `    ``int` `ans = ``0``; `   `    ``// Maximum limit upto which we can place ` `    ``// digit. If tight is 1, means number has ` `    ``// already become smaller so we can place ` `    ``// any digit, otherwise num[pos] ` `    ``int` `limit = (tight != ``0` `? ``9` `: num.get(pos)); `   `    ``for` `(``int` `d = ``0``; d <= limit; d++)` `    ``{ `   `        ``// If the current digit is zero ` `        ``// and nonz is 1, we can't place it ` `        ``if` `(d == ``0` `&& nonz != ``0``) ` `            ``continue``; ` `        ``int` `currSum = sum + d; ` `        ``int` `currRem = (rem * ``10` `+ d) % m; ` `        ``int` `currF = (tight != ``0` `|| (d < num.get(pos))) ? ``1` `: ``0``; ` `        ``ans += count(pos + ``1``, currSum, currRem, ` `                    ``currF, (nonz != ``0` `|| d != ``0``) ? ``1` `: ``0``, num); ` `    ``} ` `    ``return` `dp[pos][sum][rem][tight] = ans; ` `} `   `// Function to convert x into its digit vector ` `// and uses count() function to return the ` `// required count ` `static` `int` `solve(``int` `x) ` `{ ` `    ``Vector num = ``new` `Vector(); ` `    ``while` `(x != ``0``) ` `    ``{ ` `        ``num.add(x % ``10``); ` `        ``x /= ``10``; ` `    ``} ` `    ``Collections.reverse(num); `   `    ``// Initialize dp ` `    ``for``(``int` `i = ``0``; i < M; i++)` `        ``for``(``int` `j = ``0``; j < ``165``; j++)` `            ``for``(``int` `k = ``0``; k < M; k++)` `                ``for``(``int` `l = ``0``; l < ``2``; l++)` `                    ``dp[i][j][k][l]=-``1``;` `    `  `    ``return` `count(``0``, ``0``, ``0``, ``0``, ``0``, num); ` `} `   `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `L = ``1``, R = ``100``; ` `    ``n = ``8``; m = ``2``; ` `    ``System.out.print( solve(R) - solve(L)); ` `} ` `}`   `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation of the approach `   `# Function to return the count of ` `# required numbers from 0 to num ` `def` `count(pos, ``Sum``, rem, tight, nonz, num): `   `    ``# Last position ` `    ``if` `pos ``=``=` `len``(num): ` `        ``if` `rem ``=``=` `0` `and` `Sum` `=``=` `n: ` `            ``return` `1` `        ``return` `0`   `    ``# If this result is already computed ` `    ``# simply return it` `    ``if` `dp[pos][``Sum``][rem][tight] !``=` `-``1``: ` `        ``return` `dp[pos][``Sum``][rem][tight] ` `    `  `    ``ans ``=` `0`   `    ``# Maximum limit upto which we can place ` `    ``# digit. If tight is 1, means number has ` `    ``# already become smaller so we can place ` `    ``# any digit, otherwise num[pos]` `    ``if` `tight:` `        ``limit ``=` `9` `    ``else``: ` `        ``limit ``=` `num[pos]` `    `  `    ``for` `d ``in` `range``(``0``, limit ``+` `1``): `   `        ``# If the current digit is zero ` `        ``# and nonz is 1, we can't place it ` `        ``if` `d ``=``=` `0` `and` `nonz:` `            ``continue` `            `  `        ``currSum ``=` `Sum` `+` `d ` `        ``currRem ``=` `(rem ``*` `10` `+` `d) ``%` `m ` `        ``currF ``=` `int``(tight ``or` `(d < num[pos])) ` `        ``ans ``+``=` `count(pos ``+` `1``, currSum, currRem, ` `                     ``currF, nonz ``or` `d, num) ` `    `  `    ``dp[pos][``Sum``][rem][tight] ``=` `ans` `    ``return` `ans`   `# Function to convert x into its digit ` `# vector and uses count() function to ` `# return the required count ` `def` `solve(x): `   `    ``num ``=` `[]` `    ``global` `dp` `    `  `    ``while` `x > ``0``: ` `        ``num.append(x ``%` `10``) ` `        ``x ``/``/``=` `10` `    `  `    ``num.reverse() `   `    ``# Initialize dp ` `    ``dp ``=` `[[[[``-``1``, ``-``1``] ``for` `i ``in` `range``(M)] ` `                     ``for` `j ``in` `range``(``165``)] ` `                     ``for` `k ``in` `range``(M)]` `    ``return` `count(``0``, ``0``, ``0``, ``0``, ``0``, num) `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:`   `    ``L, R ``=` `1``, ``100` `    `  `    ``# n is the sum of digits and number ` `    ``# should be divisible by m` `    ``n, m, M ``=` `8``, ``2``, ``20` `    `  `    ``# States - position, sum, rem, tight ` `    ``# sum can have values upto 162, if we ` `    ``# are dealing with numbers upto 10^18 ` `    ``# when all 18 digits are 9, then sum ` `    ``# is 18 * 9 = 162 ` `    ``dp ``=` `[]`   `    ``print``(solve(R) ``-` `solve(L)) ` `    `  `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{` `    `  `static` `int` `M = 20; `   `// states - position, sum, rem, tight ` `// sum can have values upto 162, if we ` `// are dealing with numbers upto 10^18 ` `// when all 18 digits are 9, then sum ` `// is 18 * 9 = 162 ` `static` `int` `[,,,]dp = ``new` `int` `[M, 165, M, 2]; `   `// n is the sum of digits and number should ` `// be divisible by m ` `static` `int` `n, m; `   `// Function to return the count of ` `// required numbers from 0 to num ` `static` `int` `count(``int` `pos, ``int` `sum, ``int` `rem, ``int` `tight, ` `                            ``int` `nonz, List<``int``> num) ` `{ ` `    ``// Last position ` `    ``if` `(pos == num.Count)` `    ``{ ` `        ``if` `(rem == 0 && sum == n) ` `            ``return` `1; ` `        ``return` `0; ` `    ``} `   `    ``// If this result is already computed ` `    ``// simply return it ` `    ``if` `(dp[pos,sum,rem,tight] != -1) ` `        ``return` `dp[pos,sum,rem,tight]; `   `    ``int` `ans = 0; `   `    ``// Maximum limit upto which we can place ` `    ``// digit. If tight is 1, means number has ` `    ``// already become smaller so we can place ` `    ``// any digit, otherwise num[pos] ` `    ``int` `limit = (tight != 0 ? 9 : num[pos]); `   `    ``for` `(``int` `d = 0; d <= limit; d++)` `    ``{ `   `        ``// If the current digit is zero ` `        ``// and nonz is 1, we can't place it ` `        ``if` `(d == 0 && nonz != 0) ` `            ``continue``; ` `        ``int` `currSum = sum + d; ` `        ``int` `currRem = (rem * 10 + d) % m; ` `        ``int` `currF = (tight != 0 || (d < num[pos])) ? 1 : 0; ` `        ``ans += count(pos + 1, currSum, currRem, ` `                    ``currF, (nonz != 0 || d != 0) ? 1 : 0, num); ` `    ``} ` `    ``return` `dp[pos, sum, rem, tight] = ans; ` `} `   `// Function to convert x into its digit vector ` `// and uses count() function to return the ` `// required count ` `static` `int` `solve(``int` `x) ` `{ ` `    ``List<``int``> num = ``new` `List<``int``>(); ` `    ``while` `(x != 0) ` `    ``{ ` `        ``num.Add(x % 10); ` `        ``x /= 10; ` `    ``} ` `    ``num.Reverse(); `   `    ``// Initialize dp ` `    ``for``(``int` `i = 0; i < M; i++)` `        ``for``(``int` `j = 0; j < 165; j++)` `            ``for``(``int` `k = 0; k < M; k++)` `                ``for``(``int` `l = 0; l < 2; l++)` `                    ``dp[i, j, k, l] = -1;` `    `  `    ``return` `count(0, 0, 0, 0, 0, num); ` `} `   `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `L = 1, R = 100; ` `    ``n = 8; m = 2; ` `    ``Console.Write( solve(R) - solve(L)); ` `} ` `}`   `// This code has been contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(M*M)
Auxiliary Space: O(M*M)

Short Implementation :

## C++

 `#include ` `#include ` `#include ` `#include ` `#include `   `int` `main()` `{` `    ``int` `l = 1, r = 100, n = 8, m = 2;` `    ``std::vector<``int``> output;`   `    ``for` `(``int` `x = l; x <= r; x++) {` `        ``int` `sum = 0;` `        ``std::string str = std::to_string(x);` `        ``for` `(``char` `c : str) {` `            ``sum += c - ``'0'``;` `        ``}` `        ``if` `(sum == n && x % m == 0` `            ``&& str.find(``"0"``) == std::string::npos) {` `            ``output.push_back(x);` `        ``}` `    ``}`   `    ``std::cout << output.size() << std::endl;` `    ``return` `0;` `} ``// this code is contributed by dany`

## Python3

 `# User Input` `l, r, n, m ``=` `1``, ``100``, ``8``, ``2`   `# Initialize Result` `output ``=` `[]`   `# Traverse through all numbers` `for` `x ``in` `range``(l, r``+``1``):`   `    ``# Check for all conditions in every number ` `    ``if` `sum``([``int``(k) ``for` `k ``in` `str``(x)]) ``=``=` `n ``and` `x ``%` `m ``=``=` `0` `and` `'0'` `not` `in` `str``(x): ``# Check conditions` `        ``output.append(x)` ` `  `print``(``len``(output))  ` ` `  `# This code is contributed by mailprakashindia`

## Java

 `import` `java.util.ArrayList;` `import` `java.util.List;`   `public` `class` `Main {` `    ``public` `static` `void` `main(String[] args) {` `        ``int` `l = ``1``, r = ``100``, n = ``8``, m = ``2``;`   `        ``List output = ``new` `ArrayList<>();`   `        ``// Traverse through all numbers` `        ``for` `(``int` `x = l; x <= r; x++) {`   `            ``// Check for all conditions in every number ` `            ``if` `(String.valueOf(x).chars().map(Character::getNumericValue).sum() == n && x % m == ``0` `&& !String.valueOf(x).contains(``"0"``)) {` `                ``output.add(x);` `            ``}` `        ``}`   `        ``System.out.println(output.size());` `    ``}` `}` `// this code is contributed by devendrasalunke`

## C#

 `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `MainClass {` `    ``public` `static` `void` `Main (``string``[] args) {` `        ``int` `l = 1, r = 100, n = 8, m = 2;`   `        ``List<``int``> output = ``new` `List<``int``>();`   `        ``// Traverse through all numbers` `        ``for` `(``int` `x = l; x <= r; x++) {`   `            ``// Check for all conditions in every number` `            ``if` `(x.ToString().ToCharArray().Select(c => (``int``)Char.GetNumericValue(c)).Sum() == n && x % m == 0 && !x.ToString().Contains(``"0"``)) {` `                ``output.Add(x);` `            ``}` `        ``}`   `        ``Console.WriteLine(output.Count);` `    ``}` `}`

## Javascript

 `// User Input` `let l = 1, r = 100, n = 8, m = 2;`   `// Initialize Result` `let output = [];`   `// Traverse through all numbers` `for` `(let x = l; x <= r; x++) {`   `    ``// Check for all conditions in every number ` `    ``if` `(x.toString().split(``''``).map(Number).reduce((a,b) => a + b) === n && x % m === 0 && !x.toString().includes(``'0'``)) { ``// Check conditions` `        ``output.push(x);` `    ``}` `}`   `console.log(output.length);`

Output

`4`

Time complexity of this solution would be O(n * k), where n is the number of elements in the given range (r-l) and k = no. of digit required to represent a number in decimal format. We traverse through all the elements in the given range so O(n) and for checking all the conditions we need O(k) for every possible value, as here we are using stl find() function which need O(k) time.
Auxiliary Space complexity of this solution would be O(1). We do not use any extra space for this solution.