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Minimum count of numbers required with unit digit X that sums up to N

  • Last Updated : 11 May, 2021

Given two integers N and X, the task is to find the minimum count of integers with sum N and having unit digit X. If no such representation exists then print -1.
Examples: 
 

Input: N = 38, X = 9 
Output:
Explanation: 
Minimum two integers are required with unit digit as X to represent as a sum equal to 38. 
38 = 19 + 19 or 38 = 29 + 9
Input: N = 6, X = 4 
Output: -1 
Explanation: 
No such representation of 
 

 

Approach: 
Follow the steps below to solve the problem: 
 

  • Obtain the unit digit of N and check if it achieved by the sum of numbers whose unit digits are X.
  • If it is possible, check if N ? X * ( Minimum number of times a number with unit digit X needs to be added to get sum N).
  • If the above condition is satisfied then print the minimum number of times a number with unit digit X needs to be added to get sum N. Otherwise, print -1.

Below is the implementation of the above approach:
 



C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
int check(int unit_digit, int X)
{
    int times, digit;
 
    // Calculate the number of
    // additions required to get unit
    // digit of N
    for (int times = 1; times <= 10;
         times++) {
        digit = (X * times) % 10;
        if (digit == unit_digit)
            return times;
    }
 
    // If unit digit of N
    // cannot be obtained
    return -1;
}
 
// Function to return the minimum
// number required to represent N
int getNum(int N, int X)
{
    int unit_digit;
 
    // Stores unit digit of N
    unit_digit = N % 10;
 
    // Stores minimum addition
    // of X required to
    // obtain unit digit of N
    int times = check(unit_digit, X);
 
    // If unit digit of N
    // cannot be obtained
    if (times == -1)
        return times;
 
    // Otherwise
    else {
 
        // If N is greater than
        // or equal to (X*times)
        if (N >= (times * X))
 
            // Minimum count of numbers
            // that needed to represent N
            return times;
 
        // Representation not
        // possible
        else
            return -1;
    }
}
 
// Driver Code
int main()
{
    int N = 58, X = 7;
    cout << getNum(N, X) << endl;
    return 0;
}

Java




// Java Program to implement
// the above approach
class GFG{
     
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
    int times, digit;
 
    // Calculate the number of
    // additions required to get unit
    // digit of N
    for (times = 1; times <= 10;
                    times++)
    {
        digit = (X * times) % 10;
        if (digit == unit_digit)
            return times;
    }
 
    // If unit digit of N
    // cannot be obtained
    return -1;
}
 
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
    int unit_digit;
 
    // Stores unit digit of N
    unit_digit = N % 10;
 
    // Stores minimum addition
    // of X required to
    // obtain unit digit of N
    int times = check(unit_digit, X);
 
    // If unit digit of N
    // cannot be obtained
    if (times == -1)
        return times;
 
    // Otherwise
    else
    {
 
        // If N is greater than
        // or equal to (X*times)
        if (N >= (times * X))
 
            // Minimum count of numbers
            // that needed to represent N
            return times;
 
        // Representation not
        // possible
        else
            return -1;
    }
}
 
// Driver Code
public static void main(String []args)
{
    int N = 58, X = 7;
    System.out.println( getNum(N, X));
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 program to implement
# the above approach
 
# Function to calculate and return
# the minimum number of times a number
# with unit digit X needs to be added
# to get a sum N
def check(unit_digit, X):
     
    # Calculate the number of additions
    # required to get unit digit of N
    for times in range(1, 11):
        digit = (X * times) % 10
        if (digit == unit_digit):
            return times
             
    # If unit digit of N
    # cannot be obtained
    return -1
 
# Function to return the minimum
# number required to represent N
def getNum(N, X):
     
    # Stores unit digit of N
    unit_digit = N % 10
 
    # Stores minimum addition
    # of X required to
    # obtain unit digit of N
    times = check(unit_digit, X)
 
    # If unit digit of N
    # cannot be obtained
    if (times == -1):
        return times
 
    # Otherwise
    else:
 
        # If N is greater than
        # or equal to (X*times)
        if (N >= (times * X)):
 
            # Minimum count of numbers
            # that needed to represent N
            return times
 
        # Representation not
        # possible
        else:
            return -1
 
# Driver Code
N = 58
X = 7
 
print(getNum(N, X))
 
# This code is contributed by Sanjit_Prasad

C#




// C# Program to implement
// the above approach
using System;
class GFG{
     
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
static int check(int unit_digit, int X)
{
    int times, digit;
 
    // Calculate the number of
    // additions required to get unit
    // digit of N
    for (times = 1; times <= 10;
                    times++)
    {
        digit = (X * times) % 10;
        if (digit == unit_digit)
            return times;
    }
 
    // If unit digit of N
    // cannot be obtained
    return -1;
}
 
// Function to return the minimum
// number required to represent N
static int getNum(int N, int X)
{
    int unit_digit;
 
    // Stores unit digit of N
    unit_digit = N % 10;
 
    // Stores minimum addition
    // of X required to
    // obtain unit digit of N
    int times = check(unit_digit, X);
 
    // If unit digit of N
    // cannot be obtained
    if (times == -1)
        return times;
 
    // Otherwise
    else
    {
 
        // If N is greater than
        // or equal to (X*times)
        if (N >= (times * X))
 
            // Minimum count of numbers
            // that needed to represent N
            return times;
 
        // Representation not
        // possible
        else
            return -1;
    }
}
 
// Driver Code
public static void Main()
{
    int N = 58, X = 7;
    Console.Write(getNum(N, X));
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
 
// Javascript program implementation
// of the approach
 
// Function to calculate and return
// the minimum number of times a number
// with unit digit X needs to be added
// to get a sum N
function check(unit_digit, X)
{
    let times, digit;
   
    // Calculate the number of
    // additions required to get unit
    // digit of N
    for (times = 1; times <= 10;
                    times++)
    {
        digit = (X * times) % 10;
        if (digit == unit_digit)
            return times;
    }
   
    // If unit digit of N
    // cannot be obtained
    return -1;
}
   
// Function to return the minimum
// number required to represent N
function getNum(N, X)
{
    let unit_digit;
   
    // Stores unit digit of N
    unit_digit = N % 10;
   
    // Stores minimum addition
    // of X required to
    // obtain unit digit of N
    let times = check(unit_digit, X);
   
    // If unit digit of N
    // cannot be obtained
    if (times == -1)
        return times;
   
    // Otherwise
    else
    {
   
        // If N is greater than
        // or equal to (X*times)
        if (N >= (times * X))
   
            // Minimum count of numbers
            // that needed to represent N
            return times;
   
        // Representation not
        // possible
        else
            return -1;
    }
}
  
// Driver Code
     
    let N = 58, X = 7;
    document.write( getNum(N, X));
    
   // This code is contributed by splevel62.
</script>
Output: 
4

 

Time Complexity: O(1) 
Auxiliary Space: O(1)
 

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