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Count different numbers that can be generated such that there digits sum is equal to ‘n’
  • Difficulty Level : Medium
  • Last Updated : 20 Dec, 2018

Given an positive integer n. Count the different numbers that can be generated using digits 1, 2, 3 and 4 such that digits sum is the number ‘n’. Here digit ‘4’ will be treated as ‘1’. For instance,
32 = 3 + 2 = 5
1341 = 1 + 3 + 1 + 1 = 6
441 = 1 + 1 + 1 = 3
Note: Answer the value in mod = 109+7

Input: 2
Output: 5
Explanation
There are only '5' numbers that can 
be made:
11 = 1 + 1 = 2
14 = 1 + 1 = 2
41 = 1 + 1 = 2
44 = 1 + 1 = 2
2  = 2

Input: 3
Output: 13
Explanation
There are only '13' numbers that can 
be made i.e., 111, 114, 141, 144, 411, 
414, 441, 444, 12, 21, 42, 24, 3.

The approach is to use Dynamic programming. The problem is same as coin change and Ways to write n as sum of two or more positive integers problems. The only difference is that, instead of iterating up-to ‘n’, iterate only from 1 to 3 as according to question, only 1, 2, 3 and 4 digits are allowed. But since ‘4’ can be replaced with ‘1’ therefore iterate through 1, 2 and 3 and double the count of ‘1’ for compensation of digit ‘4’.

C++




// C++ program to count ways to write
// 'n' as sum of digits
#include<iostream>
using namespace std;
  
// Function to count 'num' as sum of
// digits(1, 2, 3, 4)
int countWays(int num)
{
    // Initialize dp[] array
    int dp[num+1];
  
    const int MOD = 1e9 + 7;
    // Base case
    dp[1] = 2;
  
    for(int i = 2; i <= num; ++i)
    {
        // Initialize the current dp[] 
        // array as '0'
        dp[i] = 0;
  
        for(int j = 1; j <= 3; ++j)
        {
            /* if i == j then there is only
               one way to write with element
               itself 'i' */
            if(i - j == 0)
               dp[i] += 1;
  
            /* If j == 1, then there exist
               two ways, one from '1' and
               other from '4' */
            else if (j == 1)
               dp[i] += dp[i-j] * 2;
  
            /* if i - j is positive then
               pick the element from 'i-j'
               element of dp[] array */
            else if(i - j > 0)
               dp[i] += dp[i-j];
  
        // Check for modulas
        if(dp[i] >= MOD)
            dp[i] %= MOD;
        }
  
    }
  
    // return the final answer
    return dp[num];
}
  
// Driver code
int main()
{
    int n = 3;
    cout << countWays(n);
      
    return 0;
}


Java




// Java program to count ways to 
// write 'n' as sum of digits
import java.io.*;
  
public class GFG
{
  
// Function to count 'num' as 
// sum of digits(1, 2, 3, 4)
static int countWays(int num)
{
      
    // Initialize dp[] array
    int []dp= new int[num + 1];
    int MOD = (int)1E9 + 7;
      
    // Base case
    dp[1] = 2;
  
    for(int i = 2; i <= num; ++i)
    {
        // Initialize the current
        // dp[] array as '0' 
        dp[i] = 0;
  
        for(int j = 1; j <= 3; ++j)
        {
            // if i == j then there is 
            // only one way to write with
            // element itself 'i'
            if(i - j == 0)
            dp[i] += 1;
  
            // If j == 1, then there exist
            // two ways, one from '1' and
            // other from '4' 
            else if (j == 1)
                dp[i] += dp[i - j] * 2;
  
            // if i - j is positive then
            // pick the element from 'i-j'
            // element of dp[] array 
            else if(i - j > 0)
                dp[i] += dp[i - j];
  
        // Check for modulas
        if(dp[i] >= MOD)
            dp[i] %= MOD;
        }
  
    }
  
    // return the final answer
    return dp[num];
}
  
    // Driver code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(countWays(n));
      
    }
}
  
// This code is contributed by vt_m


Python3




# Python3 program to count ways to write 
# 'n' as sum of digits 
  
# Function to count 'num' as sum of 
# digits(1, 2, 3, 4) 
def countWays(num): 
  
    # Initialize dp[] array 
    dp = [0] * (num + 1); 
  
    MOD = 100000000 + 7
      
    # Base case 
    dp[1] = 2
  
    for i in range(2, num + 1):
          
        # Initialize the current dp[] 
        # array as '0' 
        dp[i] = 0
  
        for j in range(1, 4): 
              
            # if i == j then there is only 
            # one way to write with element 
            # itself 'i' 
            if(i - j == 0):
                dp[i] += 1
  
            # If j == 1, then there exist 
            # two ways, one from '1' and 
            # other from '4'
            elif (j == 1):
                dp[i] += dp[i - j] * 2
  
            # if i - j is positive then 
            # pick the element from 'i-j' 
            # element of dp[] array 
            elif(i - j > 0):
                dp[i] += dp[i - j]; 
  
        # Check for modulas 
        if(dp[i] >= MOD): 
            dp[i] %= MOD; 
  
    # return the final answer 
    return dp[num]; 
  
# Driver code 
n = 3
print(countWays(n)); 
  
# This code is contributed by mits


C#




// C# program to count ways to 
// write 'n' as sum of digits
using System;
  
public class GFG
{
  
// Function to count 'num' as 
// sum of digits(1, 2, 3, 4)
static int countWays(int num)
{
      
    // Initialize dp[] array
    int []dp= new int[num + 1];
    int MOD = (int)1E9 + 7;
      
    // Base case
    dp[1] = 2;
  
    for(int i = 2; i <= num; ++i)
    {
        // Initialize the current
        // dp[] array as '0' 
        dp[i] = 0;
  
        for(int j = 1; j <= 3; ++j)
        {
            // if i == j then there is 
            // only one way to write with
            // element itself 'i'
            if(i - j == 0)
            dp[i] += 1;
  
            // If j == 1, then there exist
            // two ways, one from '1' and
            // other from '4' 
            else if (j == 1)
                dp[i] += dp[i - j] * 2;
  
            // if i - j is positive then
            // pick the element from 'i-j'
            // element of dp[] array 
            else if(i - j > 0)
                dp[i] += dp[i - j];
  
        // Check for modulas
        if(dp[i] >= MOD)
            dp[i] %= MOD;
        }
  
    }
  
    // return the final answer
    return dp[num];
}
  
    // Driver code
    static public void Main (String []args)
    {
        int n = 3;
        Console.WriteLine(countWays(n));
      
    }
}
  
// This code is contributed by vt_m


PHP




<?php
// PHP program to count ways to write 
// 'n' as sum of digits 
  
// Function to count 'num' as sum of 
// digits(1, 2, 3, 4) 
function countWays($num
    // Initialize dp[] array 
    $dp[$num + 1] = array(); 
  
    $MOD = 100000000 + 7; 
      
    // Base case 
    $dp[1] = 2; 
  
    for($i = 2; $i <= $num; ++$i
    
        // Initialize the current dp[] 
        // array as '0' 
        $dp[$i] = 0; 
  
        for($j = 1; $j <= 3; ++$j
        
            /* if i == j then there is only 
            one way to write with element 
            itself 'i' */
            if($i - $j == 0) 
            $dp[$i] += 1; 
  
            /* If j == 1, then there exist 
            two ways, one from '1' and 
            other from '4' */
            else if ($j == 1) 
            $dp[$i] += $dp[$i - $j] * 2; 
  
            /* if i - j is positive then 
            pick the element from 'i-j' 
            element of dp[] array */
            else if($i - $j > 0) 
            $dp[$i] += $dp[$i - $j]; 
  
        // Check for modulas 
        if($dp[$i] >= $MOD
            $dp[$i] %= $MOD
        
    
  
    // return the final answer 
    return $dp[$num]; 
  
// Driver code 
$n = 3; 
echo countWays($n); 
  
// This code is contributed by jit_t
?>


Output

13

Time complexity: O(n)
Auxiliary space: O(n)

Note: Asked in Directi coding round(2014 and 2017)

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