Count N digits numbers with sum divisible by K
Last Updated :
23 Jul, 2025
Given two integers N and K, the task is to count all N digits numbers whose sum of digits is divisible by K.
Examples:
Input: N = 2, K = 7
Output: 12
Explanation: 2 digits numbers with sum divisible by 7 are: {16, 25, 34, 43, 52, 59, 61, 68, 70, 77, 86, 95}.
Therefore, the required output is 12.
Input: N = 1, K = 2
Output: 4
Naive Approach: The simplest approach is to traverse all the numbers from the range [10(N - 1), 10N - 1] and check if the sum of all the digits of a number that lies within the range is divisible by K or not. For every number for which the condition is found to be true, increase count. Finally, print the count.
Time Complexity: O(10N - 10N - 1 - 1)
Auxiliary Space: O(1)
Efficient Approach: The idea is to use Digit DP technique to optimize the above approach. Below is the recurrence relation:
CountNum(N, sum, st) = \sum^{9}_{i=0} countNum(N - 1, (sum + i)\mod K, st)
sum: represents sum of digits
st: check if a number contains any leading 0.
Follow the steps below to solve the problem:
- Initialize a 3D array dp[N][K][st] to compute and store the values of all subproblems of the above recurrence relation.
- Finally, return the value of dp[N][sum%K][st].
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
#define M 1000
// Function to count the N digit numbers
// whose sum is divisible by K
int countNum(int N, int sum, int K,
bool st, int dp[M][M][2])
{
// Base case
if (N == 0 and sum == 0) {
return 1;
}
if (N < 0) {
return 0;
}
// If already computed
// subproblem occurred
if (dp[N][sum][st] != -1) {
return dp[N][sum][st];
}
// Store the count of N digit numbers
// whose sum is divisible by K
int res = 0;
// Check if the number does not contain
// any leading 0.
int start = st == 1 ? 0 : 1;
// Recurrence relation
for (int i = start; i <= 9; i++) {
res += countNum(N - 1, (sum + i) % K,
K, (st | i > 0), dp);
}
return dp[N][sum][st] = res;
}
// Driver Code
int main()
{
int N = 2, K = 7;
// Stores the values of
// overlapping subproblems
int dp[M][M][2];
memset(dp, -1, sizeof(dp));
cout << countNum(N, 0, K, 0, dp);
}
// Java program to implement
// the above approach
import java.util.*;
import java.lang.*;
class GFG {
static final int M = 1000;
// Function to count the N digit numbers
// whose sum is divisible by K
static int countNum(int N, int sum, int K,
int st, int dp[][][])
{
// Base case
if (N == 0 && sum == 0) {
return 1;
}
if (N < 0) {
return 0;
}
// If already computed
// subproblem occurred
if (dp[N][sum][st] != -1) {
return dp[N][sum][st];
}
// Store the count of N digit numbers
// whose sum is divisible by K
int res = 0;
// Check if the number does not contain
// any leading 0.
int start = st == 1 ? 0 : 1;
// Recurrence relation
for (int i = start; i <= 9; i++) {
res += countNum(N - 1, (sum + i) % K,
K, ((st | i) > 0) ? 1 : 0, dp);
}
return dp[N][sum][st] = res;
}
// Driver code
public static void main(String[] args)
{
int N = 2, K = 7;
// Stores the values of
// overlapping subproblems
int[][][] dp = new int[M][M][2];
for (int[][] i : dp)
for (int[] j : i)
Arrays.fill(j, -1);
System.out.print(countNum(N, 0, K, 0, dp));
}
}
// This code is contributed by offbeat
# Python3 program to implement
# the above approach
# Function to count the N digit
# numbers whose sum is divisible by K
def countNum(N, sum, K, st, dp):
# Base case
if (N == 0 and sum == 0):
return 1
if (N < 0):
return 0
# If already computed
# subproblem occurred
if (dp[N][sum][st] != -1):
return dp[N][sum][st]
# Store the count of N digit
# numbers whose sum is divisible by K
res = 0
start = 1
# Check if the number does not contain
# any leading 0.
if (st == 1):
start = 0
else:
start = 1
# Recurrence relation
for i in range(start, 10):
min = 0
if ((st | i) > 0):
min = 1
else:
min = 0
res += countNum(N - 1, (sum + i) % K,
K, min, dp)
dp[N][sum][st] = res
return dp[N][sum][st]
# Driver code
if __name__ == '__main__':
N = 2
K = 7
M = 100
# Stores the values of
# overlapping subproblems
dp = [[[-1 for i in range(2)]
for j in range(M)]
for j in range(M)]
print(countNum(N, 0, K, 0, dp))
# This code is contributed by shikhasingrajput
// C# program to implement
// the above approach
using System;
class GFG{
static int M = 1000;
// Function to count the N digit numbers
// whose sum is divisible by K
static int countNum(int N, int sum, int K,
int st, int[,, ] dp)
{
// Base case
if (N == 0 && sum == 0)
{
return 1;
}
if (N < 0)
{
return 0;
}
// If already computed
// subproblem occurred
if (dp[N, sum, st] != -1)
{
return dp[N, sum, st];
}
// Store the count of N digit numbers
// whose sum is divisible by K
int res = 0;
// Check if the number does not contain
// any leading 0.
int start = (st == 1 ? 0 : 1);
// Recurrence relation
for(int i = start; i <= 9; i++)
{
res += countNum(N - 1, (sum + i) % K,
K, ((st | i) > 0) ? 1 : 0, dp);
}
return dp[N, sum, st] = res;
}
// Driver code
static public void Main()
{
int N = 2, K = 7;
// Stores the values of
// overlapping subproblems
int[,, ] dp = new int[M, M, 2];
for(int i = 0; i < M; i++)
for(int j = 0; j < M; j++)
for(int k = 0; k < 2; k++)
dp[i, j, k] = -1;
Console.WriteLine(countNum(N, 0, K, 0, dp));
}
}
// This code is contributed by offbeat
<script>
// JavaScript Program to implement
// the above approach
var M = 1000;
// Function to count the N digit numbers
// whose sum is divisible by K
function countNum(N, sum, K, st, dp)
{
// Base case
if (N == 0 && sum == 0) {
return 1;
}
if (N < 0) {
return 0;
}
// If already computed
// subproblem occurred
if (dp[N][sum][st] != -1) {
return dp[N][sum][st];
}
// Store the count of N digit numbers
// whose sum is divisible by K
var res = 0;
// Check if the number does not contain
// any leading 0.
var start = st == 1 ? 0 : 1;
// Recurrence relation
for (var i = start; i <= 9; i++) {
res += countNum(N - 1, (sum + i) % K,
K, (st | i > 0), dp);
}
return dp[N][sum][st] = res;
}
// Driver Code
var N = 2, K = 7;
// Stores the values of
// overlapping subproblems
var dp = Array.from(Array(M), ()=>Array(M));
for(var i =0; i<M; i++)
for(var j =0; j<M; j++)
dp[i][j] = new Array(2).fill(-1);
document.write( countNum(N, 0, K, 0, dp));
</script>
Time Complexity:O(10*N*K)
Auxiliary Space: O(N*K)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a table 3D DP table to store the solution of the subproblems which is of length N * K * 2 (true , false) .
- Initialize the table with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[N][sum][1] + dp[N][sum][0] .
Implementation :
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the N digit numbers
// whose sum is divisible by K
int countNum(int N, int sum, int K) {
// initialize DP to Store computations of subproblems
int dp[N + 1][K][2];
memset(dp, 0, sizeof(dp)); // fill DP with 0
// Base case
dp[0][0][0] = 1;
// Iterate over subproblems and get the solution
// of current problem with the help of recursion
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1); k <= 9; k++) { // skip leading zeroes
int rem = (j - k % K + K) % K;
// get current value from previous computation
dp[i][j][1] += dp[i - 1][rem][1];
if (k > 0)
dp[i][j][1] += dp[i - 1][rem][0];
else
dp[i][j][0] += dp[i - 1][rem][0];
}
}
}
// return answer
return dp[N][sum][1] + dp[N][sum][0];
}
// Drivre Code
int main() {
int N = 2, K = 7;
// function call
cout << countNum(N, 0, K) << endl;
return 0;
}
// this code is contributed by bhardwajji
import java.util.*;
public class Main {
// Function to count the N digit numbers
// whose sum is divisible by K
public static int countNum(int N, int sum, int K) {
// initialize DP to Store computations of subproblems
int[][][] dp = new int[N + 1][K][2];
for (int[][] row : dp)
for (int[] innerRow : row)
Arrays.fill(innerRow, 0); // fill DP with 0
// Base case
dp[0][0][0] = 1;
// Iterate over subproblems and get the solution
// of current problem with the help of recursion
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1 ? 1 : 0); k <= 9; k++) { // skip leading zeroes
int rem = (j - k % K + K) % K;
// get current value from previous computation
dp[i][j][1] += dp[i - 1][rem][1];
if (k > 0)
dp[i][j][1] += dp[i - 1][rem][0];
else
dp[i][j][0] += dp[i - 1][rem][0];
}
}
}
// return answer
return dp[N][sum][1] + dp[N][sum][0];
}
// Driver Code
public static void main(String[] args) {
int N = 2, K = 7;
// function call
System.out.println(countNum(N, 0, K));
}
}
def count_num(n, sum_, k):
# initialize DP to Store computations of subproblems
dp = [[[0] * 2 for _ in range(k)] for _ in range(n + 1)]
# Base case
dp[0][0][0] = 1
# Iterate over subproblems and get the solution
# of current problem with the help of recursion
for i in range(1, n + 1):
for j in range(k):
for l in range(10):
if i == 1 and l == 0:
continue
rem = (j - l % k + k) % k
# get current value from previous computation
dp[i][j][1] += dp[i - 1][rem][1]
if l > 0:
dp[i][j][1] += dp[i - 1][rem][0]
else:
dp[i][j][0] += dp[i - 1][rem][0]
# return answer
return dp[n][sum_][1] + dp[n][sum_][0]
# Drivre Code
if __name__ == '__main__':
n = 2
k = 7
# function call
print(count_num(n, 0, k))
using System;
class GFG {
// Function to count the N digit numbers
// whose sum is divisible by K
static int countNum(int N, int sum, int K)
{
// initialize DP to Store computations of
// subproblems
int[, , ] dp = new int[N + 1, K, 2];
Array.Clear(dp, 0, dp.Length); // fill DP with 0
// Base case
dp[0, 0, 0] = 1;
// Iterate over subproblems and get the solution
// of current problem with the help of recursion
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1 ? 1 : 0); k <= 9;
k++) // skip leading zeroes
{
int rem = (j - k % K + K) % K;
// get current value from previous
// computation
dp[i, j, 1] += dp[i - 1, rem, 1];
if (k > 0)
dp[i, j, 1] += dp[i - 1, rem, 0];
else
dp[i, j, 0] += dp[i - 1, rem, 0];
}
}
}
// return answer
return dp[N, sum, 1] + dp[N, sum, 0];
}
// Drivre Code
static void Main()
{
int N = 2;
int K = 7;
Console.WriteLine(countNum(N, 0, K));
}
}
function countNum(N, sum, K) {
// initialize DP to Store computations of subproblems
let dp = new Array(N + 1);
for (let i = 0; i <= N; i++) {
dp[i] = new Array(K);
for (let j = 0; j < K; j++) {
dp[i][j] = new Array(2).fill(0);
}
}
// Base case
dp[0][0][0] = 1;
// Iterate over subproblems and get the solution
// of current problem with the help of recursion
for (let i = 1; i <= N; i++) {
for (let j = 0; j < K; j++) {
for (let k = (i == 1 ? 1 : 0); k <= 9; k++) { // skip leading zeroes
let rem = (j - k % K + K) % K;
// get current value from previous computation
dp[i][j][1] += dp[i - 1][rem][1];
if (k > 0)
dp[i][j][1] += dp[i - 1][rem][0];
else
dp[i][j][0] += dp[i - 1][rem][0];
}
}
}
// return answer
return dp[N][sum][1] + dp[N][sum][0];
}
// Driver Code
let N = 2, K = 7;
console.log(countNum(N, 0, K));
Time Complexity: O(10*N*K)
Auxiliary Space: O(N*K)
Optmized approach: Using DP Tabulation method (Iterative approach) with Space Optimization
The approach to solve this problem remains the same, but we will utilize DP tabulation (bottom-up) method with a space-efficient optimization using "prev" and "curr" arrays.
Steps to solve this problem using space optimization:
- Create two 2D arrays: "prev" and "curr", each of length [K][2].
- Initialize the "prev" array with base cases.
- Iterate over the subproblems to compute the current problem's value using the previous computations stored in the "prev" array.
- Update the "curr" array with the calculated values for the current iteration.
- Copy the values of the "curr" array to the "prev" array, and reset the "curr" array to all zeros for the next iteration.
- Finally, return the final solution stored in prev[sum][1] + prev[sum][0].
Below is the implementation of the above approach:
C++
// C++ implementation of space optmized approach
#include <bits/stdc++.h>
using namespace std;
int countNum(int N, int sum, int K)
{
int prev[K][2] = { 0 };
int curr[K][2] = { 0 };
// Base case
prev[0][0] = 1;
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1); k <= 9;
k++) { // skip leading zeroes
int rem = (j - k % K + K) % K;
curr[j][1] += prev[rem][1];
if (k > 0)
curr[j][1] += prev[rem][0];
else
curr[j][0] += prev[rem][0];
}
}
// Copy curr to prev for the next iteration
for (int j = 0; j < K; j++) {
prev[j][0] = curr[j][0];
prev[j][1] = curr[j][1];
curr[j][0] = 0;
curr[j][1] = 0;
}
}
return prev[sum][1] + prev[sum][0];
}
int main()
{
int N = 2, K = 7;
cout << countNum(N, 0, K) << endl;
return 0;
}
// This code is contributed by Tapesh(tapeshdua420)
// Java implementation of space optimized approach
import java.util.*;
class Main {
public static int countNum(int N, int sum, int K) {
int[][] prev = new int[K][2];
int[][] curr = new int[K][2];
// Base case
prev[0][0] = 1;
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1) ? 1 : 0; k <= 9; k++) { // skip leading zeroes
int rem = (j - k % K + K) % K;
curr[j][1] += prev[rem][1];
if (k > 0)
curr[j][1] += prev[rem][0];
else
curr[j][0] += prev[rem][0];
}
}
// Copy curr to prev for the next iteration
for (int j = 0; j < K; j++) {
prev[j][0] = curr[j][0];
prev[j][1] = curr[j][1];
curr[j][0] = 0;
curr[j][1] = 0;
}
}
return prev[sum][1] + prev[sum][0];
}
public static void main(String[] args) {
int N = 2, K = 7;
System.out.println(countNum(N, 0, K));
}
}
// This code is contributed by Tapesh(tapeshdua420)
# Python implementation:
def countNum(N, sum, K):
prev = [[0, 0] for _ in range(K)]
curr = [[0, 0] for _ in range(K)]
# Base case
prev[0][0] = 1
for i in range(1, N+1):
for j in range(K):
for k in range(int(i == 1), 10): # skip leading zeroes
rem = (j - k % K + K) % K
curr[j][1] += prev[rem][1]
if k > 0:
curr[j][1] += prev[rem][0]
else:
curr[j][0] += prev[rem][0]
# Copy curr to prev for the next iteration
for j in range(K):
prev[j][0] = curr[j][0]
prev[j][1] = curr[j][1]
curr[j][0] = 0
curr[j][1] = 0
return prev[sum][1] + prev[sum][0]
N = 2
K = 7
print(countNum(N, 0, K))
# This code is contributed by Tapesh(tapeshdu420)
// C# implementation
using System;
class MainClass {
public static int CountNum(int N, int sum, int K) {
int[,] prev = new int[K, 2];
int[,] curr = new int[K, 2];
// Base case
prev[0, 0] = 1;
for (int i = 1; i <= N; i++) {
for (int j = 0; j < K; j++) {
for (int k = (i == 1) ? 1 : 0; k <= 9; k++) { // skip leading zeroes
int rem = (j - k % K + K) % K;
curr[j, 1] += prev[rem, 1];
if (k > 0)
curr[j, 1] += prev[rem, 0];
else
curr[j, 0] += prev[rem, 0];
}
}
// Copy curr to prev for the next iteration
for (int j = 0; j < K; j++) {
prev[j, 0] = curr[j, 0];
prev[j, 1] = curr[j, 1];
curr[j, 0] = 0;
curr[j, 1] = 0;
}
}
return prev[sum, 1] + prev[sum, 0];
}
public static void Main(string[] args) {
int N = 2, K = 7;
Console.WriteLine(CountNum(N, 0, K));
}
}
// This code is contributed by Sakshi
function countNum(N, sum, K) {
const prev = new Array(K).fill(0).map(() => new Array(2).fill(0));
const curr = new Array(K).fill(0).map(() => new Array(2).fill(0));
// Base case
prev[0][0] = 1;
for (let i = 1; i <= N; i++) {
for (let j = 0; j < K; j++) {
for (let k = (i === 1) ? 1 : 0; k <= 9; k++) { // skip leading zeroes
const rem = (j - (k % K) + K) % K;
curr[j][1] += prev[rem][1];
if (k > 0) {
curr[j][1] += prev[rem][0];
} else {
curr[j][0] += prev[rem][0];
}
}
}
// Copy curr to prev for the next iteration
for (let j = 0; j < K; j++) {
prev[j][0] = curr[j][0];
prev[j][1] = curr[j][1];
curr[j][0] = 0;
curr[j][1] = 0;
}
}
return prev[sum][1] + prev[sum][0];
}
const N = 2;
const K = 7;
console.log(countNum(N, 0, K));
// This code is contributed by shivamgupta0987654321
Time Complexity: O(10*N*K)
Auxiliary Space: O(K)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem