Count numbers (smaller than or equal to N) with given digit sum
Last Updated :
11 Jul, 2025
Given a number N and a sum S, find the count of numbers upto N that have digit sum equal to S.
Examples:
Input : N = 100, S = 4
Output : 5
Upto 100 only 5 numbers(4, 13, 22, 31, 40)
can produce 4 as their sum of digits.
Input : N = 1000, S = 1
Output : 4
Upto 1000 only 4 numbers(1, 10, 100 and 1000)
can produce 1 as their sum of digits.
We can do a digit DP having state (current index, whether the currently constructed number of i digits is equal or less than the number formed by first i digits of N, the sum of digits of currently constructed number). Let dp[i][tight][sum_so_far] denotes the count of numbers whose first i digits have been considered and tight denotes whether the currently constructed number is equal or less than number formed by first i digits of N. If tight is true, then it means that currently constructed number is equal to the number constructed by first i digits of N. If it is false then it means that currently constructed number is less than the number constructed by first i digits of N. sum_so_far denotes the sum of digits of currently constructed number. Base Case: If i = number of digits in N, sum_so_far = Sum, then answer = 1 else answer is 0. Transitions: For filling (i+1)th digit we can consider following - If tight is true, then it means that our constructed number is still equal to number formed by first i digits of N. We can try our current possible digit value from 0 to (i+1)th digit of N. If we try the digit more than (i+1)th digit, then the constructed number will become greater than number formed by first i digits of N, which will violate the property that our constructed number should be <= N. If tight is false, then it means that number constructed from first i - 1 digits has become less than number constructed from the first i - 1 digit of N, So it means that our number can never exceed N, so we can choose the any digit from 0 to 9. nsum_so_far can be obtained by adding sum_so_far and current digit(currdigit). Finally we will return answer which is count of numbers upto N that have digit sum equal to S.
C++
#include <bits/stdc++.h>
using namespace std;
// N can be max 10^18 and hence digitsum will be 162 maximum.
long long dp[18][2][162];
long long solve(int i, bool tight, int sum_so_far,
int Sum, string number, int len)
{
if (i == len) {
// If sum_so_far equals to given sum then
// return 1 else 0
if (sum_so_far == Sum)
return 1;
else
return 0;
}
long long& ans = dp[i][tight][sum_so_far];
if (ans != -1) {
return ans;
}
ans = 0;
bool ntight;
int nsum_so_far;
for (char currdigit = '0'; currdigit <= '9'; currdigit++) {
// Our constructed number should not become
// greater than N.
if (!tight && currdigit > number[i]) {
break;
}
// If tight is true then it will also be true for (i+1) digit.
ntight = tight || currdigit < number[i];
nsum_so_far = sum_so_far + (currdigit - '0');
ans += solve(i + 1, ntight, nsum_so_far, Sum, number, len);
}
return ans;
}
// Driver code
int main()
{
long long count = 0;
long long sum = 4;
string number = "100";
memset(dp, -1, sizeof dp);
cout << solve(0, 0, 0, sum, number, number.size());
return 0;
}
// Java program to count 2s from 0 to n
import java.io.*;
class GFG
{
// N can be max 10^18 and hence
// digitsum will be 162 maximum.
static long dp[][][] = new long[18][2][162];
static long solve(int i, boolean tight, int sum_so_far,
int Sum, String number, int len)
{
if (i == len)
{
// If sum_so_far equals to given sum then
// return 1 else 0
if (sum_so_far == Sum)
return 1;
else
return 0;
}
long ans = dp[i][1][sum_so_far];
if (ans != -1)
{
return ans;
}
ans = 0;
boolean ntight;
int nsum_so_far;
for (char currdigit = '0'; currdigit <= '9'; currdigit++)
{
// Our constructed number should not become
// greater than N.
if (!tight && currdigit > number.charAt(i))
{
break;
}
// If tight is true then it will
// also be true for (i+1) digit.
ntight = tight || currdigit < number.charAt(i);
nsum_so_far = sum_so_far + (currdigit - '0');
ans += solve(i + 1, ntight, nsum_so_far,
Sum, number, len);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
long count = 0;
int sum = 4;
String number = "100";
for(int i = 0; i < 18; i++)
{
for(int j = 0; j < 2; j++)
{
for(int k = 0; k < 162; k++)
dp[i][j][k] = -1;
}
}
System.out.println( solve(0, false, 0, sum,
number, number.length()));
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation of the given approach.
def solve(i, tight, sum_so_far, Sum, number, length):
if i == length:
# If sum_so_far equals to given
# sum then return 1 else 0
if sum_so_far == Sum:
return 1
else:
return 0
ans = dp[i][tight][sum_so_far]
if ans != -1:
return ans
ans = 0
for currdigit in range(0, 10):
currdigitstr = str(currdigit)
# Our constructed number should
# not become greater than N.
if not tight and currdigitstr > number[i]:
break
# If tight is true then it will also be true for (i+1) digit.
ntight = tight or currdigitstr < number[i]
nsum_so_far = sum_so_far + currdigit
ans += solve(i + 1, ntight, nsum_so_far, Sum, number, length)
return ans
# Driver code
if __name__ == "__main__":
count, Sum = 0, 4
number = "100"
dp = [[[-1 for i in range(162)] for j in range(2)] for k in range(18)]
print(solve(0, 0, 0, Sum, number, len(number)))
# This code is contributed by Rituraj Jain
// C# program to count 2s from 0 to n
using System;
class GFG
{
// N can be max 10^18 and hence
// digitsum will be 162 maximum.
static long [ , , ]dp = new long[18,2,162];
static long solve(int i, bool tight, int sum_so_far,
int Sum, String number, int len)
{
if (i == len)
{
// If sum_so_far equals to given sum then
// return 1 else 0
if (sum_so_far == Sum)
return 1;
else
return 0;
}
long ans = dp[i,1,sum_so_far];
if (ans != -1)
{
return ans;
}
ans = 0;
bool ntight;
int nsum_so_far;
for (char currdigit = '0'; currdigit <= '9'; currdigit++)
{
// Our constructed number should not become
// greater than N.
if (!tight && currdigit > number[i])
{
break;
}
// If tight is true then it will
// also be true for (i+1) digit.
ntight = tight || currdigit < number[i];
nsum_so_far = sum_so_far + (currdigit - '0');
ans += solve(i + 1, ntight, nsum_so_far,
Sum, number, len);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int sum = 4;
String number = "100";
for(int i = 0; i < 18; i++)
{
for(int j = 0; j < 2; j++)
{
for(int k = 0; k < 162; k++)
dp[i,j,k] = -1;
}
}
Console.WriteLine( solve(0, false, 0, sum,
number, number.Length));
}
}
// This code has been contributed by Rajput-Ji
// Javascript code for the above approach
// N can be max 10^18 and hence
// digitsum will be 162 maximum.
let dp = new Array(18);
for (let i = 0; i < dp.length; i++) {
dp[i] = new Array(2);
for (let j = 0; j < 2; j++) {
dp[i][j] = new Array(162).fill(-1);
}
}
function solve(i, tight, sum_so_far, Sum, number, len) {
if (i === len)
{
// If sum_so_far equals to given sum then
// return 1 else 0
if (sum_so_far === Sum)
return 1;
else
return 0;
}
if (dp[i][1][sum_so_far] !== -1) {
return dp[i][1][sum_so_far];
}
let ans = 0;
let ntight;
let nsum_so_far;
for (let currdigit = '0'; currdigit <= '9'; currdigit++) {
// Our constructed number should not become
// greater than N.
if (!tight && currdigit > number.charAt(i)) {
break;
}
// If tight is true then it will
// also be true for (i+1) digit.
ntight = tight || currdigit < number.charAt(i);
nsum_so_far = sum_so_far + (currdigit - '0');
ans += solve(i + 1, ntight, nsum_so_far,
Sum, number, len);
}
return ans;
}
// Driver code
let count = 0;
let sum = 4;
let number = "100";
console.log(solve(0, false, 0, sum,
number, number.length))
// This code is contributed by lokeshpotta20.
Time Complexity: 2(tight) * Sum * log 10(N) * 10(transitions) = 20*log 10(N)*Sum.
Auxiliary Space: O(log N * 2 * 162).
Here log N is for the number of digits in the given number N, 2 is for the boolean tight value and 162 is for the maximum sum of digits as N can be of 10^18.
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