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Count of minimum numbers having K as the last digit required to obtain sum N

  • Difficulty Level : Easy
  • Last Updated : 28 Jul, 2021
Geek Week

Given a positive integer N and a digit K, the task is to find the minimum count of numbers ending with digit K such that the sum of those numbers is N. If no such number exists whose sum is K, then print “-1”.

Examples:

Input: N = 42, K = 3
Output: 4
Explanation:
The given number N(= 43) can be expressed as 3 + 3 + 13 + 23, such all the numbers ends with digit K(= 3). Therefore, the count split numbers 4.

Input: N = 17, K = 3
Output: -1

Approach: The given problem can be solved by an observation that if a number can be expressed as the sum of numbers ending with digit K, then the result will be the maximum of 10. Follow the steps below to solve the problem:



  • If the digit K is even and the integer N is odd, then print “-1″ as it is not possible to obtain an odd sum with an even number.
  • For the number K, find the smallest number ending with digit i over the range [0, 9]. If it is not possible, then set the value as INT_MAX.
  • Also, for each number K find the minimum steps required to create a number that ends with digit i over the range [0, 9].
  • Now, if the smallest number ending with digit i is greater than N with unit digit i then print “-1” as it is no possible to make the sum of numbers as N.
  • Otherwise, the answer will be minimum steps to create a number that ends with digit i which is the same as the unit digit of N because the remaining digit can be obtained by inserting those digits in any of the numbers that contribute to the answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
int minCount(int N, int K)
{
    // Stores the smallest number that
    // ends with digit i (0, 9)
    int SmallestNumber[10];
 
    // Stores the minimum number of
    // steps to create a number ending
    // with digit i
    int MinimumSteps[10];
 
    // Initialize elements as infinity
    for (int i = 0; i <= 9; i++) {
        SmallestNumber[i] = INT_MAX;
        MinimumSteps[i] = INT_MAX;
    }
 
    for (int i = 1; i <= 10; i++) {
 
        int num = K * i;
 
        // Minimum number
        // ending with digit i
        SmallestNumber[num % 10]
            = min(
                SmallestNumber[num % 10],
                num);
 
        // Minimum steps to create a
        // number ending with digit i
        MinimumSteps[num % 10]
            = min(
                MinimumSteps[num % 10], i);
    }
 
    // If N < SmallestNumber then,
    // return -1
    if (N < SmallestNumber[N % 10]) {
        return -1;
    }
 
    // Otherwise, return answer
    else {
        return MinimumSteps[N % 10];
    }
}
 
// Driver Code
int main()
{
    int N = 42, K = 7;
    cout << minCount(N, K);
 
    return 0;
}

Java




// Java program for above approach
import java.util.*;
 
class GFG{
 
static int minCount(int N, int K)
{
     
    // Stores the smallest number that
    // ends with digit i (0, 9)
    int SmallestNumber[] = new int[10];
 
    // Stores the minimum number of
    // steps to create a number ending
    // with digit i
    int MinimumSteps[] = new int[10];
 
    // Initialize elements as infinity
    for(int i = 0; i <= 9; i++)
    {
        SmallestNumber[i] = Integer.MAX_VALUE;
        MinimumSteps[i] = Integer.MAX_VALUE;
    }
 
    for(int i = 1; i <= 10; i++)
    {
        int num = K * i;
         
        // Minimum number
        // ending with digit i
        SmallestNumber[num % 10] = Math.min(
            SmallestNumber[num % 10], num);
 
        // Minimum steps to create a
        // number ending with digit i
        MinimumSteps[num % 10] = Math.min(
            MinimumSteps[num % 10], i);
    }
 
    // If N < SmallestNumber then,
    // return -1
    if (N < SmallestNumber[N % 10])
    {
        return -1;
    }
 
    // Otherwise, return answer
    else
    {
        return MinimumSteps[N % 10];
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 42, K = 7;
   
    System.out.println(minCount(N, K));
}
}
 
// This code is contributed by hritikrommie

Python3




# Python3 program for the above approach
import sys
 
def minCount(N, K):
   
    # Stores the smallest number that
    # ends with digit i (0, 9)
    SmallestNumber = [0 for i in range(10)]
 
    # Stores the minimum number of
    # steps to create a number ending
    # with digit i
    MinimumSteps = [0 for i in range(10)]
 
    # Initialize elements as infinity
    for i in range(10):
        SmallestNumber[i] = sys.maxsize;
        MinimumSteps[i] = sys.maxsize
 
    for i in range(1,11,1):
        num = K * i
 
        # Minimum number
        # ending with digit i
        SmallestNumber[num % 10] = min(SmallestNumber[num % 10],num)
 
        # Minimum steps to create a
        # number ending with digit i
        MinimumSteps[num % 10] = min(MinimumSteps[num % 10], i)
 
    # If N < SmallestNumber then,
    # return -1
    if (N < SmallestNumber[N % 10]):
        return -1
 
    # Otherwise, return answer
    else:
        return MinimumSteps[N % 10]
 
# Driver Code
if __name__ == '__main__':
    N = 42
    K = 7
    print(minCount(N, K))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for the above approach
using System;
 
public class GFG{
     
    static int minCount(int N, int K)
    {
         
        // Stores the smallest number that
        // ends with digit i (0, 9)
        int[] SmallestNumber = new int[10];
     
        // Stores the minimum number of
        // steps to create a number ending
        // with digit i
        int[] MinimumSteps = new int[10];
     
        // Initialize elements as infinity
        for(int i = 0; i <= 9; i++)
        {
            SmallestNumber[i] = Int32.MaxValue;
            MinimumSteps[i] = Int32.MaxValue;
        }
     
        for(int i = 1; i <= 10; i++)
        {
            int num = K * i;
             
            // Minimum number
            // ending with digit i
            SmallestNumber[num % 10] = Math.Min(
                SmallestNumber[num % 10], num);
     
            // Minimum steps to create a
            // number ending with digit i
            MinimumSteps[num % 10] = Math.Min(
                MinimumSteps[num % 10], i);
        }
     
        // If N < SmallestNumber then,
        // return -1
        if (N < SmallestNumber[N % 10])
        {
            return -1;
        }
     
        // Otherwise, return answer
        else
        {
            return MinimumSteps[N % 10];
        }
    }
     
    // Driver Code
    static public void Main ()
    {
        int N = 42, K = 7;
       
        Console.Write(minCount(N, K));
    }
}
 
// This code is contributed by shubhamsingh10

Javascript




<script>
 
// JavaScript program for the above approach
 
function minCount(N, K)
{
    // Stores the smallest number that
    // ends with digit i (0, 9)
    let SmallestNumber = new Array(10);
 
    // Stores the minimum number of
    // steps to create a number ending
    // with digit i
    let MinimumSteps = new Array(10);
 
    // Initialize elements as infinity
    for (let i = 0; i <= 9; i++) {
        SmallestNumber[i] = Number.MAX_VALUE;
        MinimumSteps[i] = Number.MAX_VALUE;
    }
 
    for (let i = 1; i <= 10; i++) {
 
        let num = K * i;
 
        // Minimum number
        // ending with digit i
        SmallestNumber[num % 10]
            = Math.min(
                SmallestNumber[num % 10],
                num);
 
        // Minimum steps to create a
        // number ending with digit i
        MinimumSteps[num % 10]
            = Math.min(
                MinimumSteps[num % 10], i);
    }
 
    // If N < SmallestNumber then,
    // return -1
    if (N < SmallestNumber[N % 10]) {
        return -1;
    }
 
    // Otherwise, return answer
    else {
        return MinimumSteps[N % 10];
    }
}
 
// Driver Code
    let N = 42, K = 7;
    document.write(minCount(N, K));
 
</script>
Output: 
6

 

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

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