Given two integers ‘n’ and ‘sum’, find count of all n digit numbers with sum of digits as ‘sum’. Leading 0’s are not counted as digits.
Constrains:
1 <= n <= 100 and
1 <= sum <= 500
Example:
Input: n = 2, sum = 2
Output: 2
Explanation: Numbers are 11 and 20Input: n = 2, sum = 5
Output: 5
Explanation: Numbers are 14, 23, 32, 41 and 50Input: n = 3, sum = 6
Output: 21
Naive approach:
The idea is simple, we subtract all values from 0 to 9 from given sum and recur for sum minus that digit. Below is recursive formula.
countRec(n, sum) = ?countRec(n-1, sum-x)
where 0 =< x = 0
One important observation is, leading 0's must be
handled explicitly as they are not counted as digits.
So our final count can be written as below.
finalCount(n, sum) = ?countRec(n-1, sum-x)
where 1 =< x = 0
Below is a simple recursive solution based on above recursive formula.
// A C++ program using recursion to count numbers // with sum of digits as given 'sum' #include<bits/stdc++.h> using namespace std;
// Recursive function to count 'n' digit numbers // with sum of digits as 'sum'. This function // considers leading 0's also as digits, that is // why not directly called unsigned long long int countRec( int n, int sum)
{ // Base case
if (n == 0)
return sum == 0;
if (sum == 0)
return 1;
// Initialize answer
unsigned long long int ans = 0;
// Traverse through every digit and count
// numbers beginning with it using recursion
for ( int i=0; i<=9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining digits. unsigned long long int finalCount( int n, int sum)
{ // Initialize final answer
unsigned long long int ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
} // Driver program int main()
{ int n = 2, sum = 5;
cout << finalCount(n, sum);
return 0;
} |
// A Java program using recursion to count numbers // with sum of digits as given 'sum' import java.io.*;
public class sum_dig
{ // Recursive function to count 'n' digit numbers
// with sum of digits as 'sum'. This function
// considers leading 0's also as digits, that is
// why not directly called
static int countRec( int n, int sum)
{
// Base case
if (n == 0 )
return sum == 0 ? 1 : 0 ;
if (sum == 0 )
return 1 ;
// Initialize answer
int ans = 0 ;
// Traverse through every digit and count
// numbers beginning with it using recursion
for ( int i= 0 ; i<= 9 ; i++)
if (sum-i >= 0 )
ans += countRec(n- 1 , sum-i);
return ans;
}
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining digits.
static int finalCount( int n, int sum)
{
// Initialize final answer
int ans = 0 ;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( int i = 1 ; i <= 9 ; i++)
if (sum-i >= 0 )
ans += countRec(n- 1 , sum-i);
return ans;
}
/* Driver program to test above function */
public static void main (String args[])
{
int n = 2 , sum = 5 ;
System.out.println(finalCount(n, sum));
}
} /* This code is contributed by Rajat Mishra */
|
# A python 3 program using recursion to count numbers # with sum of digits as given 'sum' # Recursive function to count 'n' digit # numbers with sum of digits as 'sum' # This function considers leading 0's # also as digits, that is why not # directly called def countRec(n, sum ) :
# Base case
if (n = = 0 ) :
return ( sum = = 0 )
if ( sum = = 0 ) :
return 1
# Initialize answer
ans = 0
# Traverse through every digit and
# count numbers beginning with it
# using recursion
for i in range ( 0 , 10 ) :
if ( sum - i > = 0 ) :
ans = ans + countRec(n - 1 , sum - i)
return ans
# This is mainly a wrapper over countRec. It # explicitly handles leading digit and calls # countRec() for remaining digits. def finalCount(n, sum ) :
# Initialize final answer
ans = 0
# Traverse through every digit from 1 to
# 9 and count numbers beginning with it
for i in range ( 1 , 10 ) :
if ( sum - i > = 0 ) :
ans = ans + countRec(n - 1 , sum - i)
return ans
# Driver program n = 2
sum = 5
print (finalCount(n, sum ))
# This code is contributed by Nikita tiwari. |
// A C# program using recursion to count numbers // with sum of digits as given 'sum' using System;
class GFG {
// Recursive function to
// count 'n' digit numbers
// with sum of digits as
// 'sum'. This function
// considers leading 0's
// also as digits, that is
// why not directly called
static int countRec( int n, int sum)
{
// Base case
if (n == 0)
return sum == 0 ? 1 : 0;
if (sum == 0)
return 1;
// Initialize answer
int ans = 0;
// Traverse through every
// digit and count numbers
// beginning with it using
// recursion
for ( int i = 0; i <= 9; i++)
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
return ans;
}
// This is mainly a
// wrapper over countRec. It
// explicitly handles leading
// digit and calls countRec()
// for remaining digits.
static int finalCount( int n, int sum)
{
// Initialize final answer
int ans = 0;
// Traverse through every
// digit from 1 to 9 and
// count numbers beginning
// with it
for ( int i = 1; i <= 9; i++)
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
return ans;
}
// Driver Code
public static void Main ()
{
int n = 2, sum = 5;
Console.Write(finalCount(n, sum));
}
} // This code is contributed by nitin mittal. |
<script> // A JavaScript program using // recursion to count numbers // with sum of digits as given 'sum' // Recursive function to
// count 'n' digit numbers
// with sum of digits as 'sum'.
//This function
// considers leading 0's also as digits,
//that is why not directly called
function countRec(n, sum) {
// Base case
if (n == 0)
return sum == 0;
if (sum == 0)
return 1;
// Initialize answer
let ans = 0;
// Traverse through every
// digit and count
// numbers beginning with
// it using recursion
for (let i = 0; i <= 9; i++) {
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
}
return ans;
}
// This is mainly a wrapper over countRec.
// It explicitly handles leading digit
// and calls countRec() for remaining digits.
function finalCount(n, sum) {
// Initialize final answer
let ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (let i = 1; i <= 9; i++) {
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
}
return ans;
}
// Driver program
let n = 2, sum = 5;
document.write(finalCount(n, sum));
//This code is contributed by Surbhi Tyagi </script> |
<?php // A PHP program using recursion to count numbers // with sum of digits as given 'sum' // Recursive function to count 'n' digit numbers // with sum of digits as 'sum'. This function // considers leading 0's also as digits, that is // why not directly called function countRec( $n , $sum )
{ // Base case
if ( $n == 0)
return $sum == 0;
if ( $sum == 0)
return 1;
// Initialize answer
$ans = 0;
// Traverse through every
// digit and count
// numbers beginning with
// it using recursion
for ( $i = 0; $i <= 9; $i ++)
if ( $sum - $i >= 0)
$ans += countRec( $n -1, $sum - $i );
return $ans ;
} // This is mainly a wrapper // over countRec. It // explicitly handles leading // digit and calls // countRec() for remaining digits. function finalCount( $n , $sum )
{ // Initialize final answer
$ans = 0;
// Traverse through every
// digit from 1 to
// 9 and count numbers
// beginning with it
for ( $i = 1; $i <= 9; $i ++)
if ( $sum - $i >= 0)
$ans += countRec( $n - 1, $sum - $i );
return $ans ;
} // Driver Code
$n = 2;
$sum = 5;
echo finalCount( $n , $sum );
// This code is contributed by ajit ?> |
5
Time Complexity: O(2n)
Auxiliary Space: O(n)
Approach using Memoization:
// A C++ memoization based recursive program to count // numbers with sum of n as given 'sum' #include<bits/stdc++.h> using namespace std;
// A lookup table used for memoization unsigned long long int lookup[101][501];
// Memoization based implementation of recursive // function unsigned long long int countRec( int n, int sum)
{ // Base case
if (n == 0)
return sum == 0;
// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n][sum] != -1)
return lookup[n][sum];
// Initialize answer
unsigned long long int ans = 0;
// Traverse through every digit and
// recursively count numbers beginning
// with it
for ( int i=0; i<10; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return lookup[n][sum] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining n. unsigned long long int finalCount( int n, int sum)
{ // Initialize all entries of lookup table
memset (lookup, -1, sizeof lookup);
// Initialize final answer
unsigned long long int ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
} // Driver program int main()
{ int n = 3, sum = 5;
cout << finalCount(n, sum);
return 0;
} |
// A Java memoization based recursive program to count // numbers with sum of n as given 'sum' import java.io.*;
public class sum_dig
{ // A lookup table used for memoization
static int lookup[][] = new int [ 101 ][ 501 ];
// Memoization based implementation of recursive
// function
static int countRec( int n, int sum)
{
// Base case
if (n == 0 )
return sum == 0 ? 1 : 0 ;
// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n][sum] != - 1 )
return lookup[n][sum];
// Initialize answer
int ans = 0 ;
// Traverse through every digit and
// recursively count numbers beginning
// with it
for ( int i= 0 ; i< 10 ; i++)
if (sum-i >= 0 )
ans += countRec(n- 1 , sum-i);
return lookup[n][sum] = ans;
}
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
static int finalCount( int n, int sum)
{
// Initialize all entries of lookup table
for ( int i = 0 ; i <= 100 ; ++i){
for ( int j = 0 ; j <= 500 ; ++j){
lookup[i][j] = - 1 ;
}
}
// Initialize final answer
int ans = 0 ;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( int i = 1 ; i <= 9 ; i++)
if (sum-i >= 0 )
ans += countRec(n- 1 , sum-i);
return ans;
}
/* Driver program to test above function */
public static void main (String args[])
{
int n = 3 , sum = 5 ;
System.out.println(finalCount(n, sum));
}
} /* This code is contributed by Rajat Mishra */
|
# A Python3 memoization based recursive # program to count numbers with Sum of n # as given 'Sum' # A lookup table used for memoization lookup = [[ - 1 for i in range ( 501 )]
for i in range ( 101 )]
# Memoization based implementation # of recursive function def countRec(n, Sum ):
# Base case
if (n = = 0 ):
return Sum = = 0
# If this subproblem is already evaluated,
# return the evaluated value
if (lookup[n][ Sum ] ! = - 1 ):
return lookup[n][ Sum ]
# Initialize answer
ans = 0
# Traverse through every digit and
# recursively count numbers beginning
# with it
for i in range ( 10 ):
if ( Sum - i > = 0 ):
ans + = countRec(n - 1 , Sum - i)
lookup[n][ Sum ] = ans
return lookup[n][ Sum ]
# This is mainly a wrapper over countRec. It # explicitly handles leading digit and calls # countRec() for remaining n. def finalCount(n, Sum ):
# Initialize final answer
ans = 0
# Traverse through every digit from 1 to
# 9 and count numbers beginning with it
for i in range ( 1 , 10 ):
if ( Sum - i > = 0 ):
ans + = countRec(n - 1 , Sum - i)
return ans
# Driver Code n, Sum = 3 , 5
print (finalCount(n, Sum ))
# This code is contributed by mohit kumar 29 |
// A C# memoization based recursive program to count // numbers with sum of n as given 'sum' using System;
class sum_dig
{ // A lookup table used for memoization
static int [,]lookup = new int [101,501];
// Memoization based implementation of recursive
// function
static int countRec( int n, int sum)
{
// Base case
if (n == 0)
return sum == 0 ? 1 : 0;
// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n,sum] != -1)
return lookup[n,sum];
// Initialize answer
int ans = 0;
// Traverse through every digit and
// recursively count numbers beginning
// with it
for ( int i=0; i<10; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return lookup[n,sum] = ans;
}
// This is mainly a wrapper over countRec. It
// explicitly handles leading digit and calls
// countRec() for remaining n.
static int finalCount( int n, int sum)
{
// Initialize all entries of lookup table
for ( int i = 0; i <= 100; ++i){
for ( int j = 0; j <= 500; ++j){
lookup[i,j] = -1;
}
}
// Initialize final answer
int ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( int i = 1; i <= 9; i++)
if (sum-i >= 0)
ans += countRec(n-1, sum-i);
return ans;
}
/* Driver program to test above function */
public static void Main ()
{
int n = 3, sum = 5;
Console.Write(finalCount(n, sum));
}
} |
<script> // A Javascript memoization based // recursive program to count numbers // with sum of n as given 'sum' // A lookup table used for memoization let lookup = new Array(101);
// Memoization based implementation // of recursive function function countRec(n, sum)
{ // Base case
if (n == 0)
return sum == 0 ? 1 : 0;
// If this subproblem is already evaluated,
// return the evaluated value
if (lookup[n][sum] != -1)
return lookup[n][sum];
// Initialize answer
let ans = 0;
// Traverse through every digit and
// recursively count numbers beginning
// with it
for (let i = 0; i < 10; i++)
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
return lookup[n][sum] = ans;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining n. function finalCount(n, sum)
{ // Initialize all entries of lookup table
for (let i = 0; i < 101; i++)
{
lookup[i] = new Array(501);
for (let j = 0; j < 501; j++)
{
lookup[i][j] = -1;
}
}
// Initialize final answer
let ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for (let i = 1; i <= 9; i++)
if (sum - i >= 0)
ans += countRec(n - 1, sum - i);
return ans;
} // Driver code let n = 3, sum = 5; document.write(finalCount(n, sum)); // This code is contributed by avanitrachhadiya2155 </script> |
<?php // A PHP memoization based recursive program // to count numbers with sum of n as given 'sum' // A lookup table used for memoization $lookup = array_fill (0, 101,
array_fill (0, 501, -1));
// Memoization based implementation // of recursive function function countRec( $n , $sum )
{ global $lookup ;
// Base case
if ( $n == 0)
return $sum == 0;
// If this subproblem is already evaluated,
// return the evaluated value
if ( $lookup [ $n ][ $sum ] != -1)
return $lookup [ $n ][ $sum ];
// Initialize answer
$ans = 0;
// Traverse through every digit and
// recursively count numbers beginning
// with it
for ( $i = 0; $i < 10; $i ++)
if ( $sum - $i >= 0)
$ans += countRec( $n - 1, $sum - $i );
return $lookup [ $n ][ $sum ] = $ans ;
} // This is mainly a wrapper over countRec. It // explicitly handles leading digit and calls // countRec() for remaining n. function finalCount( $n , $sum )
{ // Initialize all entries of lookup table
// Initialize final answer
$ans = 0;
// Traverse through every digit from 1 to
// 9 and count numbers beginning with it
for ( $i = 1; $i <= 9; $i ++)
if ( $sum - $i >= 0)
$ans += countRec( $n - 1, $sum - $i );
return $ans ;
} // Driver Code $n = 3;
$sum = 5;
echo finalCount( $n , $sum );
// This code is contributed by mits ?> |
15
Time Complexity: O(n*sum)
Auxiliary Space: O(101*501)
Another Method: We can easily count n digit numbers whose sum of digit equals to given sum by iterating all n digits and checking if current n digit number’s sum is equal to given sum, if it is then we will start increment number by 9 until it reaches to number whose sum of digit’s is greater than given sum, then again we will increment by 1 until we found another number with given sum.
// C++ program to Count of n digit numbers // whose sum of digits equals to given sum #include <bits/stdc++.h> #include <iostream> using namespace std;
void findCount( int n, int sum) {
//in case n = 2 start is 10 and end is (100-1) = 99
int start = pow (10, n-1);
int end = pow (10, n)-1;
int count = 0;
int i = start;
while (i <= end) {
int cur = 0;
int temp = i;
while ( temp != 0) {
cur += temp % 10;
temp = temp / 10;
}
if (cur == sum) {
count++;
i += 9;
} else
i++;
}
cout << count;
/* This code is contributed by Anshuman */
}
int main() {
int n = 3;
int sum = 5;
findCount(n,sum);
return 0;
} |
// Java program to Count of n digit numbers // whose sum of digits equals to given sum import java.io.*;
public class GFG {
public static void main(String[] args) {
int n = 3 ;
int sum = 5 ;
findCount(n,sum);
}
private static void findCount( int n, int sum) {
//in case n = 2 start is 10 and end is (100-1) = 99
int start = ( int ) Math.pow( 10 , n- 1 );
int end = ( int ) Math.pow( 10 , n)- 1 ;
int count = 0 ;
int i = start;
while (i < end) {
int cur = 0 ;
int temp = i;
while ( temp != 0 ) {
cur += temp % 10 ;
temp = temp / 10 ;
}
if (cur == sum) {
count++;
i += 9 ;
} else
i++;
}
System.out.println(count);
/* This code is contributed by Anshuman */
}
} |
# Python3 program to Count of n digit numbers # whose sum of digits equals to given sum import math
def findCount(n, sum ):
# in case n = 2 start is 10 and
# end is (100-1) = 99
start = math. pow ( 10 , n - 1 );
end = math. pow ( 10 , n) - 1 ;
count = 0 ;
i = start;
while (i < = end):
cur = 0 ;
temp = i;
while (temp ! = 0 ):
cur + = temp % 10 ;
temp = temp / / 10 ;
if (cur = = sum ):
count = count + 1 ;
i + = 9 ;
else :
i = i + 1 ;
print (count);
# Driver Code n = 3 ;
sum = 5 ;
findCount(n, sum );
# This code is contributed # by Akanksha Rai |
// C# program to Count of n digit numbers // whose sum of digits equals to given sum using System;
class GFG
{ private static void findCount( int n,
int sum)
{ // in case n = 2 start is 10 and
// end is (100-1) = 99
int start = ( int ) Math.Pow(10, n - 1);
int end = ( int ) Math.Pow(10, n) - 1;
int count = 0;
int i = start;
while (i < end)
{
int cur = 0;
int temp = i;
while ( temp != 0)
{
cur += temp % 10;
temp = temp / 10;
}
if (cur == sum)
{
count++;
i += 9;
}
else
i++;
}
Console.WriteLine(count);
} // Driver Code public static void Main()
{ int n = 3;
int sum = 5;
findCount(n,sum);
} } // This code is contributed // by Akanksha Rai |
<script> // Javascript program to Count of n digit numbers // whose sum of digits equals to given sum function findCount(n, sum) {
// in case n = 2 start is 10 and end is (100-1) = 99
let start = Math.pow(10, n-1);
let end = Math.pow(10, n)-1;
let count = 0;
let i = start;
while (i <= end)
{
let cur = 0;
let temp = i;
while ( temp != 0)
{
cur += temp % 10;
temp = parseInt(temp / 10);
}
if (cur == sum)
{
count++;
i += 9;
} else
i++;
}
document.write(count);
}
let n = 3;
let sum = 5;
findCount(n,sum);
// This code is contributed by souravmahato348. </script> |
<?php // PHP program to Count of n digit numbers // whose sum of digits equals to given sum function findCount( $n , $sum )
{ // In case n = 2 start is 10 and // end is (100-1) = 99 $start = (int)pow(10, $n - 1);
$end = (int)pow(10, $n ) - 1;
$count = 0;
$i = $start ;
while ( $i < $end )
{ $cur = 0;
$temp = $i ;
while ( $temp != 0)
{
$cur += $temp % 10;
$temp = (int) $temp / 10;
}
if ( $cur == $sum )
{
$count ++;
$i += 9;
}
else
$i ++;
} echo ( $count );
} // Driver Code $n = 3;
$sum = 5;
findCount( $n , $sum );
// This code is contributed // by jit_t ?> |
15
Time Complexity: O(log n)
Auxiliary Space: O(1)