Minimum numbers with one’s place as 9 to be added to get N

Given an integer N, the task is to find the minimum count of numbers which needs to be added to get N. (All such numbers must have 9 as one’s digit)

Examples:

Input: N = 27
Output: 3
27 = 9 + 9 + 9

Input: N = 109
Output: 1
109 itself has 9 as one’s digit.

Approach:



  • Check the one’s digit of N, based on one’s digit the minimum count of numbers which needs to be added can be easily found.
  • If one’s digit is 9: The answer will 1 as the number itself has 9 as its one’s place digit.
  • If one’s digit is:
    1. 1: 9 has to be added 9 times i.e. (9 * 9 = 81).
    2. 2: 9 has to be added 8 times i.e. (9 * 8 = 72).
    3. 3: 9 has to be added 7 times i.e. (9 * 7 = 63).
    4. 4: 9 has to be added 6 times i.e. (9 * 6 = 54).
    5. 5: 9 has to be added 5 times i.e. (9 * 5 = 45).
    6. 6: 9 has to be added 4 times i.e. (9 * 4 = 36).
    7. 7: 9 has to be added 3 times i.e. (9 * 3 = 27).
    8. 8: 9 has to be added 2 times i.e. (9 * 2 = 18).
    9. 0: 9 has to be added 10 times i.e. (9 * 10 = 90).
  • Observation here is that only the minimum multiple count for all the cases above mentioned have to be added. This is because say for one’s digit as 4, all 9 (one’s place) from 6 numbers can be used and the result can be subtracted from N say it is M. Now, M will have 0 in one’s place. So, just use 5 numbers as 9 and sixth number as (M + 9).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find minimum count
// of numbers(with one's digit 9)
// that sum up to N
int findMin(int N)
{
    // Fetch one's digit
    int digit = N % 10;
  
    // Apply Cases mentioned in approach
    switch (digit) {
    case 0:
        if (N >= 90)
            return 10;
        break;
    case 1:
        if (N >= 81)
            return 9;
        break;
    case 2:
        if (N >= 72)
            return 8;
        break;
    case 3:
        if (N >= 63)
            return 7;
        break;
    case 4:
        if (N >= 54)
            return 6;
        break;
    case 5:
        if (N >= 45)
            return 5;
        break;
    case 6:
        if (N >= 36)
            return 4;
        break;
    case 7:
        if (N >= 27)
            return 3;
        break;
    case 8:
        if (N >= 18)
            return 2;
        break;
    case 9:
        if (N >= 9)
            return 1;
        break;
    }
  
    // If no possible answer exists
    return -1;
}
  
// Driver code
int main()
{
    int N = 27;
  
    cout << findMin(N);
}

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Java

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// Java implementation of the approach 
class GFG
{
      
    // Function to find minimum count 
    // of numbers(with one's digit 9) 
    // that sum up to N 
    static int findMin(int N) 
    
        // Fetch one's digit 
        int digit = N % 10
      
        // Apply Cases mentioned in approach 
        switch (digit)
        
            case 0
                if (N >= 90
                    return 10
                break
            case 1
                if (N >= 81
                    return 9
                break
            case 2
                if (N >= 72
                    return 8
                break
            case 3
                if (N >= 63
                    return 7
                break
            case 4
                if (N >= 54
                    return 6
                break
            case 5
                if (N >= 45
                    return 5
                break
            case 6
                if (N >= 36
                    return 4
                break
            case 7
                if (N >= 27
                    return 3
                break
            case 8
                if (N >= 18
                    return 2
                break
            case 9
                if (N >= 9
                    return 1
                break
        
      
        // If no possible answer exists 
        return -1
    
      
    // Driver code 
    public static void main (String[] args)
    
        int N = 27
      
        System.out.println(findMin(N)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
  
# Function to find minimum count
# of numbers(with one's digit 9)
# that sum up to N
def findMin(N: int):
  
    # Fetch one's digit
    digit = N % 10
  
    # Apply Cases mentioned in approach
    if digit == 0 and N >= 90:
        return 10
    elif digit == 1 and N >= 81:
        return 9
    elif digit == 2 and N >= 72:
        return 8
    elif digit == 3 and N >= 63:
        return 7
    elif digit == 4 and N >= 54:
        return 6
    elif digit == 5 and N >= 45:
        return 5
    elif digit == 6 and N >= 36:
        return 4
    elif digit == 7 and N >= 27:
        return 3
    elif digit == 8 and N >= 18:
        return 2
    elif digit == 9 and N >= 9:
        return 1
  
    # If no possible answer exists
    return -1
  
# Driver Code
if __name__ == "__main__":
    N = 27
    print(findMin(N))
  
# This code is contributed by
# sanjeev2552

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C#

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// C# implementation of the approach 
using System;         
      
class GFG
{
      
    // Function to find minimum count 
    // of numbers(with one's digit 9) 
    // that sum up to N 
    static int findMin(int N) 
    
        // Fetch one's digit 
        int digit = N % 10; 
      
        // Apply Cases mentioned in approach 
        switch (digit)
        
            case 0: 
                if (N >= 90) 
                    return 10; 
                break
            case 1: 
                if (N >= 81) 
                    return 9; 
                break
            case 2: 
                if (N >= 72) 
                    return 8; 
                break
            case 3: 
                if (N >= 63) 
                    return 7; 
                break
            case 4: 
                if (N >= 54) 
                    return 6; 
                break
            case 5: 
                if (N >= 45) 
                    return 5; 
                break
            case 6: 
                if (N >= 36) 
                    return 4; 
                break
            case 7: 
                if (N >= 27) 
                    return 3; 
                break
            case 8: 
                if (N >= 18) 
                    return 2; 
                break
            case 9: 
                if (N >= 9) 
                    return 1; 
                break
        
      
        // If no possible answer exists 
        return -1; 
    
      
    // Driver code 
    public static void Main (String[] args)
    
        int N = 27; 
      
        Console.WriteLine(findMin(N)); 
    
}
  
// This code is contributed by 29AjayKumar

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Output:

3

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