Print all n-digit numbers whose sum of digits equals to given sum
Given number of digits n, print all n-digit numbers whose sum of digits adds upto given sum. Solution should not consider leading 0’s as digits.
Examples:
Input: N = 2, Sum = 3
Output: 12 21 30
Input: N = 3, Sum = 6
Output: 105 114 123 132 141 150 204
213 222 231 240 303 312 321
330 402 411 420 501 510 600
Input: N = 4, Sum = 3
Output: 1002 1011 1020 1101 1110 1200
2001 2010 2100 3000
A simple solution would be to generate all N-digit numbers and print numbers that have sum of their digits equal to given sum. The complexity of this solution would be exponential.
A better solution is to generate only those N-digit numbers that satisfy the given constraints. The idea is to use recursion. We basically fill all digits from 0 to 9 into current position and maintain sum of digits so far. We then recurse for remaining sum and number of digits left. We handle leading 0’s separately as they are not counted as digits.
Below is a simple recursive implementation of above idea –
C++
// A C++ recursive program to print all n-digit // numbers whose sum of digits equals to given sum #include <bits/stdc++.h> using namespace std; // Recursive function to print all n-digit numbers // whose sum of digits equals to given sum // n, sum --> value of inputs // out --> output array // index --> index of next digit to be filled in // output array void findNDigitNumsUtil( int n, int sum, char * out, int index) { // Base case if (index > n || sum < 0) return ; // If number becomes N-digit if (index == n) { // if sum of its digits is equal to given sum, // print it if (sum == 0) { out[index] = '\0' ; cout << out << " " ; } return ; } // Traverse through every digit. Note that // here we're considering leading 0's as digits for ( int i = 0; i <= 9; i++) { // append current digit to number out[index] = i + '0' ; // recurse for next digit with reduced sum findNDigitNumsUtil(n, sum - i, out, index + 1); } } // This is mainly a wrapper over findNDigitNumsUtil. // It explicitly handles leading digit void findNDigitNums( int n, int sum) { // output array to store N-digit numbers char out[n + 1]; // fill 1st position by every digit from 1 to 9 and // calls findNDigitNumsUtil() for remaining positions for ( int i = 1; i <= 9; i++) { out[0] = i + '0' ; findNDigitNumsUtil(n, sum - i, out, 1); } } // Driver program int main() { int n = 2, sum = 3; findNDigitNums(n, sum); return 0; } |
Java
// Java recursive program to print all n-digit // numbers whose sum of digits equals to given sum import java.io.*; class GFG { // Recursive function to print all n-digit numbers // whose sum of digits equals to given sum // n, sum --> value of inputs // out --> output array // index --> index of next digit to be // filled in output array static void findNDigitNumsUtil( int n, int sum, char out[], int index) { // Base case if (index > n || sum < 0 ) return ; // If number becomes N-digit if (index == n) { // if sum of its digits is equal to given sum, // print it if (sum == 0 ) { out[index] = '\0' ; System.out.print(out); System.out.print( " " ); } return ; } // Traverse through every digit. Note that // here we're considering leading 0's as digits for ( int i = 0 ; i <= 9 ; i++) { // append current digit to number out[index] = ( char )(i + '0' ); // recurse for next digit with reduced sum findNDigitNumsUtil(n, sum - i, out, index + 1 ); } } // This is mainly a wrapper over findNDigitNumsUtil. // It explicitly handles leading digit static void findNDigitNums( int n, int sum) { // output array to store N-digit numbers char [] out = new char [n + 1 ]; // fill 1st position by every digit from 1 to 9 and // calls findNDigitNumsUtil() for remaining positions for ( int i = 1 ; i <= 9 ; i++) { out[ 0 ] = ( char )(i + '0' ); findNDigitNumsUtil(n, sum - i, out, 1 ); } } // driver program to test above function public static void main (String[] args) { int n = 2 , sum = 3 ; findNDigitNums(n, sum); } } // This code is contributed by Pramod Kumar |
Python 3
# Python 3 recursive program to print # all n-digit numbers whose sum of # digits equals to given sum # Recursive function to print all # n-digit numbers whose sum of # digits equals to given sum # n, sum --> value of inputs # out --> output array # index --> index of next digit to be # filled in output array def findNDigitNumsUtil(n, sum , out,index): # Base case if (index > n or sum < 0 ): return f = "" # If number becomes N-digit if (index = = n): # if sum of its digits is equal # to given sum, print it if ( sum = = 0 ): out[index] = "\0" for i in out: f = f + i print (f, end = " " ) return # Traverse through every digit. Note # that here we're considering leading # 0's as digits for i in range ( 10 ): # append current digit to number out[index] = chr (i + ord ( '0' )) # recurse for next digit with reduced sum findNDigitNumsUtil(n, sum - i, out, index + 1 ) # This is mainly a wrapper over findNDigitNumsUtil. # It explicitly handles leading digit def findNDigitNums( n, sum ): # output array to store N-digit numbers out = [ False ] * (n + 1 ) # fill 1st position by every digit # from 1 to 9 and calls findNDigitNumsUtil() # for remaining positions for i in range ( 1 , 10 ): out[ 0 ] = chr (i + ord ( '0' )) findNDigitNumsUtil(n, sum - i, out, 1 ) # Driver Code if __name__ = = "__main__" : n = 2 sum = 3 findNDigitNums(n, sum ) # This code is contributed # by ChitraNayal |
C#
// C# recursive program to print all n-digit // numbers whose sum of digits equals to // given sum using System; class GFG { // Recursive function to print all n-digit // numbers whose sum of digits equals to // given sum // n, sum --> value of inputs // out --> output array // index --> index of next digit to be // filled in output array static void findNDigitNumsUtil( int n, int sum, char []ou, int index) { // Base case if (index > n || sum < 0) return ; // If number becomes N-digit if (index == n) { // if sum of its digits is equal to // given sum, print it if (sum == 0) { ou[index] = '\0' ; Console.Write(ou); Console.Write( " " ); } return ; } // Traverse through every digit. Note // that here we're considering leading // 0's as digits for ( int i = 0; i <= 9; i++) { // append current digit to number ou[index] = ( char )(i + '0' ); // recurse for next digit with // reduced sum findNDigitNumsUtil(n, sum - i, ou, index + 1); } } // This is mainly a wrapper over // findNDigitNumsUtil. It explicitly // handles leading digit static void findNDigitNums( int n, int sum) { // output array to store N-digit // numbers char []ou = new char [n + 1]; // fill 1st position by every digit // from 1 to 9 and calls // findNDigitNumsUtil() for remaining // positions for ( int i = 1; i <= 9; i++) { ou[0] = ( char )(i + '0' ); findNDigitNumsUtil(n, sum - i, ou, 1); } } // driver program to test above function public static void Main () { int n = 2, sum = 3; findNDigitNums(n, sum); } } // This code is contributed by nitin mittal. |
Javascript
<script> // Javascript recursive program to print all n-digit // numbers whose sum of digits equals to given sum // Recursive function to print all n-digit numbers // whose sum of digits equals to given sum // n, sum --> value of inputs // out --> output array // index --> index of next digit to be // filled in output array function findNDigitNumsUtil(n, sum, out, index) { // Base case if (index > n || sum < 0) return ; // If number becomes N-digit if (index == n) { // if sum of its digits is equal to given sum, // print it if (sum == 0) { out[index] = '\0' ; for (let i = 0; i < out.length; i++) document.write(out[i]); document.write( " " ); } return ; } // Traverse through every digit. Note that // here we're considering leading 0's as digits for (let i = 0; i <= 9; i++) { // append current digit to number out[index] = String.fromCharCode(i + '0' .charCodeAt(0)); // recurse for next digit with reduced sum findNDigitNumsUtil(n, sum - i, out, index + 1); } } // This is mainly a wrapper over findNDigitNumsUtil. // It explicitly handles leading digit function findNDigitNums(n,sum) { // output array to store N-digit numbers let out = new Array(n+1); for (let i=0;i<n+1;i++) { out[i]= false ; } // fill 1st position by every digit from 1 to 9 and // calls findNDigitNumsUtil() for remaining positions for (let i = 1; i <= 9; i++) { out[0] = String.fromCharCode(i + '0' .charCodeAt(0)); findNDigitNumsUtil(n, sum - i, out, 1); } } // driver program to test above function let n = 2, sum = 3; findNDigitNums(n, sum); // This code is contributed by avanitrachhadiya2155 </script> |
PHP
<?php // A PHP recursive program to print all // n-digit numbers whose sum of digits // equals to given sum // Recursive function to print all n-digit // numbers whose sum of digits equals to // given sum // n, sum --> value of inputs // out --> output array // index --> index of next digit to be // filled in output array function findNDigitNumsUtil( $n , $sum , $out , $index ) { // Base case if ( $index > $n || $sum < 0) return ; // If number becomes N-digit if ( $index == $n ) { // if sum of its digits is equal // to given sum, print it if ( $sum == 0) { $out [ $index ] = '' ; foreach ( $out as & $value ) print ( $value ); print ( " " ); } return ; } // Traverse through every digit. Note // that here we're considering leading // 0's as digits for ( $i = 0; $i <= 9; $i ++) { // append current digit to number $out [ $index ] = chr ( $i + ord( '0' )); // recurse for next digit with // reduced sum findNDigitNumsUtil( $n , $sum - $i , $out , $index + 1); } } // This is mainly a wrapper over findNDigitNumsUtil. // It explicitly handles leading digit function findNDigitNums( $n , $sum ) { // output array to store N-digit numbers $out = array_fill (0, $n + 1, false); // fill 1st position by every digit from // 1 to 9 and calls findNDigitNumsUtil() // for remaining positions for ( $i = 1; $i <= 9; $i ++) { $out [0] = chr ( $i + ord( '0' )); findNDigitNumsUtil( $n , $sum - $i , $out , 1); } } // Driver Code $n = 2; $sum = 3; findNDigitNums( $n , $sum ); // This code is contributed // by chandan_jnu ?> |
Output:
12 21 30
Time Complexity: O(n*n!)
Auxiliary Space: O(n)
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