Count of n digit numbers whose sum of digits equals to given sum
Last Updated :
23 Jul, 2025
Given two integers n and sum, the task is to find the count of all n digit numbers with sum of digits equal to sum.
Note: Leading 0's are not counted as digits. If there exist no n digit number with sum of digits equal to given sum, print -1.
Example:
Input: n = 2, sum= 2
Output: 2
Explanation: The numbers are 11 and 20
Input: n = 2, sum= 5
Output: 5
Explanation: The numbers are 14, 23, 32, 41 and 50
[Naive Approach - 1] - Using Recursion - O(n*(9^n)) Time and O(n) Space
The idea is to find all n-digit numbers where the sum of digits equals the sum. It uses recursion to try every digit from 0 to 9 for each position, reducing the given sum at each step. The first digit is taken from 1 to 9 to ensure valid n-digit numbers. The recursion stops when there are no digits left, checking if the sumbecomes zero.
C++
// A C++ program to count numbers with sum of
// digits as a given target using Recursion.
#include <bits/stdc++.h>
using namespace std;
// Recursive function to count n-digit numbers
// with sum of digits as the target.
int countRec(int n, int sum) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
int countWays(int n, int sum) {
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
if(ans == 0) return -1;
return ans;
}
int main() {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
cout<<ans;
return 0;
}
// A Java program to count numbers with sum of
// digits as a given target using Recursion.
import java.util.*;
class GfG {
// Recursive function to count n-digit numbers
// with sum of digits as the target.
static int countRec(int n, int sum) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
static int countWays(int n, int sum) {
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
if(ans == 0) return -1;
return ans;
}
public static void main(String[] args) {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
System.out.println(ans);
}
}
# A Python program to count numbers with sum of
# digits as a given target using Recursion.
def countRec(n, sum):
# Base case: If there are no digits left,
# check if the target is also zero.
if n == 0:
return 1 if sum == 0 else 0
# If the target becomes zero,
# there's exactly one valid number.
if sum == 0:
return 1
ans = 0
# Traverse through digits 0-9 to calculate
# the count of numbers recursively.
for i in range(10):
if sum - i >= 0:
ans += countRec(n - 1, sum - i)
return ans
# Function to count n-digit numbers
# with sum of digits as the target.
def countWays(n, sum):
ans = 0
# Traverse through digits 1-9 as the first
# digit cannot be zero for n-digit numbers.
for i in range(1, 10):
if sum - i >= 0:
ans += countRec(n - 1, sum - i)
if(ans == 0):
return -1;
return ans
if __name__ == "__main__":
n = 2
sum = 5
ans = countWays(n, sum)
print(ans)
// A C# program to count numbers with sum of
// digits as a given target using Recursion.
using System;
class GfG {
// Recursive function to count n-digit numbers
// with sum of digits as the target.
static int CountRec(int n, int sum) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += CountRec(n - 1, sum - i);
}
}
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
static int CountWays(int n, int sum) {
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += CountRec(n - 1, sum - i);
}
}
if(ans == 0) return -1;
return ans;
}
public static void Main(string[] args) {
int n = 2;
int sum = 5;
int ans = CountWays(n, sum);
Console.WriteLine(ans);
}
}
// A Javascript program to count numbers with sum of
// digits as a given target using Recursion.
function countRec(n, sum) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n === 0) {
return sum === 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum === 0) {
return 1;
}
let ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (let i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
function countWays(n, sum) {
let ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (let i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i);
}
}
if(ans == 0) return -1;
return ans;
}
const n = 2;
const sum = 5;
const ans = countWays(n, sum);
console.log(ans);
Time Complexity: O(n * (9 ^ n)) , for each of the n digits, we are exploring all other possible 9 digits.
Space Complexity: O(n), considering the recursive stack.
[Naive Approach - 2] - Using Iterative Approach - O(n*10^n) Time and O(1) Space
To count all n-digit numbers whose sum of digits equals a given target sum, we iterate through each number within the range of n-digit numbers. For each number, we calculate the sum of its digits. If the sum matches the given sum, we increment the count. While we considered optimizing the process by incrementing by 9 after finding a valid number, this step could skip some valid numbers where the sum of digits changes slightly. Therefore, it's safer to increment by 1 and check all possible numbers in the range to ensure no valid numbers are missed. Finally, we return the total count of numbers with the desired sum of digits.
Below is given the implementation:
C++
// C++ program to count n-digit numbers with
// the sum of digits as the target using iteration.
#include <bits/stdc++.h>
using namespace std;
int countWays(int n, int sum) {
// Calculate the range for n-digit numbers
int start = pow(10, n - 1);
int end = pow(10, n) - 1;
int count = 0;
int i = start;
// Iterate through all n-digit numbers
while (i <= end) {
int currentSum = 0;
int temp = i;
// Calculate the sum of digits of the number
while (temp != 0) {
currentSum += temp % 10;
temp /= 10;
}
// Check if the sum of digits matches the target
if (currentSum == sum) {
count++;
i += 9;
} else {
i++;
}
}
if(count == 0) return -1;
return count;
}
int main() {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
cout<<ans;
return 0;
}
// Java program to count n-digit numbers with
// the sum of digits as a given target
// using iteration.
class GfG {
// Function to count n-digit numbers with
// the sum of digits as the target.
static int countWays(int n, int sum) {
// Calculate the range for n-digit numbers.
int start = (int) Math.pow(10, n - 1);
int end = (int) Math.pow(10, n) - 1;
int count = 0;
int i = start;
// Iterate through all n-digit numbers.
while (i <= end) {
int currentSum = 0;
int temp = i;
// Calculate the sum of digits of the number.
while (temp != 0) {
currentSum += temp % 10;
temp /= 10;
}
// Check if the sum of digits matches the target.
if (currentSum == sum) {
count++;
i += 9;
} else {
i++;
}
}
if(count == 0) return -1;
return count;
}
public static void main(String[] args) {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
System.out.println(ans);
}
}
# Python program to count n-digit numbers with
# the sum of digits as the target using iteration.
# Function to count n-digit numbers with the
# sum of digits as the target using iteration.
def countWays(n, sum):
# Calculate the range for n-digit numbers.
start = 10 ** (n - 1)
end = 10 ** n - 1
count = 0
i = start
# Iterate through all n-digit numbers.
while i <= end:
current_sum = 0
temp = i
# Calculate the sum of digits of the number.
while temp != 0:
current_sum += temp % 10
temp //= 10
# Check if the sum of digits matches the target.
if current_sum == sum:
count += 1
i += 9
else:
i += 1
if(count == 0):
return -1;
return count
if __name__ == "__main__":
n = 2
sum = 5
ans = countWays(n, sum)
print(ans)
// C# program to count n-digit numbers with
// the sum of digits as the target using iteration.
using System;
class GfG {
// Function to count n-digit numbers
// with sum of digits as the target.
static int countWays(int n, int sum) {
// Calculate the range for n-digit numbers.
int start = (int)Math.Pow(10, n - 1);
int end = (int)Math.Pow(10, n) - 1;
int count = 0;
int i = start;
// Iterate through all n-digit numbers.
while (i <= end) {
int currentSum = 0;
int temp = i;
// Calculate the sum of digits of the number.
while (temp != 0) {
currentSum += temp % 10;
temp /= 10;
}
// Check if the sum of digits matches the target.
if (currentSum == sum) {
count++;
i += 9;
} else {
i++;
}
}
if(count == 0) return -1;
return count;
}
static void Main(string[] args) {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
Console.WriteLine(ans);
}
}
// Javascript program to count n-digit numbers with
// the sum of digits as the target using iteration.
// Function to count n-digit numbers
// with sum of digits as the target.
function countWays(n, sum) {
// Calculate the range for n-digit numbers.
const start = Math.pow(10, n - 1);
const end = Math.pow(10, n) - 1;
let count = 0;
let i = start;
// Iterate through all n-digit numbers.
while (i <= end) {
let currentSum = 0;
let temp = i;
// Calculate the sum of digits of the number.
while (temp !== 0) {
currentSum += temp % 10;
temp = Math.floor(temp / 10);
}
// Check if the sum of digits matches the target.
if (currentSum === sum) {
count++;
i += 9;
} else {
i++;
}
}
if(count == 0) return -1;
return count;
}
//driver code
const n = 2;
const sum = 5;
const ans = countWays(n, sum);
console.log(ans);
Time Complexity: O(n * 10^n), for each digit we are iterating through all possible 10 digits.
Space Complexity: O(1)
[Expected Approach] - Using Memoization - O(n * sum) Time and O(n * sum) Space
The idea is to count n-digit numbers with a sum of digits equal to the given sum using recursion with memoization to optimize repeated calculations. A 2d array memo[][] is used to store results for a given number of digits and given sum to avoid redundant computations. Starting from digits 1 to 9 for the first position, recursively explores all possible digits while reducing the given sum, ensuring efficiency by reusing previously computed results.
Below is given the implementation:
C++
// A C++ program to count numbers with sum of
// digits as a given target using Memoization.
#include <bits/stdc++.h>
using namespace std;
// Recursive function to count n-digit numbers
// with sum of digits as the target.
int countRec(int n, int sum,
vector<vector<int>> &memo) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
// Check if already computed
if (memo[n][sum] != -1) {
return memo[n][sum];
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1,
sum - i, memo);
}
}
// Store and return the result
return memo[n][sum] = ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
int countWays(int n, int sum) {
// Create a memoization table
vector<vector<int>> memo(n + 1,
vector<int>(sum + 1, -1));
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i, memo);
}
}
if(ans == 0) return -1;
return ans;
}
int main() {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
cout<<ans;
return 0;
}
// A Java program to count numbers with sum of
// digits as a given target using Memoization.
import java.util.*;
class GfG {
// Recursive function to count n-digit numbers
// with sum of digits as the target.
static int countRec(
int n, int sum, int[][] memo) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
// If already computed, return the result.
if (memo[n][sum] != -1) {
return memo[n][sum];
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(
n - 1, sum - i, memo);
}
}
// Store and return the result.
return memo[n][sum] = ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
static int countWays(int n, int sum) {
// Create a memoization table.
int[][] memo = new int[n + 1]
[sum + 1];
// Initialize memo with -1 to indicate
// uncomputed states.
for (int[] row : memo) {
Arrays.fill(row, -1);
}
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(
n - 1, sum - i, memo);
}
}
if(ans == 0) return -1;
return ans;
}
public static void main(String[] args) {
int n = 2;
int sum = 5;
int ans = countWays(n, sum);
System.out.println(ans);
}
}
# A Python program to count numbers with sum of
# digits as a given target using Memoization.
def countRec(n, sum, memo):
# Base case: If there are no digits left,
# check if the target is also zero.
if n == 0:
return 1 if sum == 0 else 0
# If the target becomes zero,
# there's exactly one valid number.
if sum == 0:
return 1
# If already computed, return the result.
if memo[n][sum] != -1:
return memo[n][sum]
ans = 0
# Traverse through digits 0-9 to calculate
# the count of numbers recursively.
for i in range(10):
if sum - i >= 0:
ans += countRec(n - 1, sum - i, memo)
# Store and return the result.
memo[n][sum] = ans
return ans
# Function to count n-digit numbers
# with sum of digits as the target.
def countWays(n, sum):
# Create a memoization table initialized to -1.
memo = [[-1] * (sum + 1) for _ in range(n + 1)]
ans = 0
# Traverse through digits 1-9 as the first
# digit cannot be zero for n-digit numbers.
for i in range(1, 10):
if sum - i >= 0:
ans += countRec(n - 1, sum - i, memo)
if(ans == 0):
return -1;
return ans
if __name__ == "__main__":
n = 2
sum = 5
ans = countWays(n, sum)
print(ans)
// A C# program to count numbers with sum of
// digits as a given target using Memoization.
using System;
class GfG {
static int CountRec(int n, int sum, int[,] memo) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n == 0) {
return sum == 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum == 0) {
return 1;
}
// If already computed, return the result.
if (memo[n, sum] != -1) {
return memo[n, sum];
}
int ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (int i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += CountRec(n - 1, sum - i, memo);
}
}
// Store and return the result.
memo[n, sum] = ans;
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
static int CountWays(int n, int sum) {
// Initialize a memoization table with -1.
int[,] memo = new int[n + 1, sum + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= sum; j++) {
memo[i, j] = -1;
}
}
int ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (int i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += CountRec(n - 1, sum - i, memo);
}
}
if(ans == 0) return -1;
return ans;
}
public static void Main(string[] args) {
int n = 2;
int sum = 5;
int ans = CountWays(n, sum);
Console.WriteLine(ans);
}
}
// A JavaScript program to count numbers with sum of
// digits as a given target using Memoization.
function countRec(n, sum, memo) {
// Base case: If there are no digits left,
// check if the target is also zero.
if (n === 0) {
return sum === 0 ? 1 : 0;
}
// If the target becomes zero,
// there's exactly one valid number.
if (sum === 0) {
return 1;
}
// If already computed, return the result.
if (memo[n][sum] !== -1) {
return memo[n][sum];
}
let ans = 0;
// Traverse through digits 0-9 to calculate
// the count of numbers recursively.
for (let i = 0; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i, memo);
}
}
// Store and return the result.
memo[n][sum] = ans;
return ans;
}
// Function to count n-digit numbers
// with sum of digits as the target.
function countWays(n, sum) {
// Initialize a memoization table with -1.
const memo = Array.from({ length: n + 1 }, () =>
Array(sum + 1).fill(-1)
);
let ans = 0;
// Traverse through digits 1-9 as the first
// digit cannot be zero for n-digit numbers.
for (let i = 1; i <= 9; i++) {
if (sum - i >= 0) {
ans += countRec(n - 1, sum - i, memo);
}
}
if(ans == 0) return -1;
return ans;
}
//driver code
const n = 2;
const sum = 5;
const ans = countWays(n, sum);
console.log(ans);
Time Complexity: O(n * sum), for each of the n digits, we are considering the sum up to given sum.
Space Complexity: O(n * sum), to store the elements in memo.
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