TCS Digital Coding Question | August 17 2019

Question

This is the TCS Digital Coding Question asked on August 17, 2019. Given a range, find all the unique number within that range. Find the count of unique numbers and print the count.

SOLUTION 1:

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int a,b,tem2,rem2,tem,rem,count=0,flag,k,j,i,m;
    scanf("%d %d",&a,&b);
    int *arr = (int*)malloc(sizeof(int)*b);
    for(i=a; i<=b; i++)     //loop for the given range
    {
        flag = 1;
        tem = i;
        k=0;
        while(tem)          //seperate each digit in the number and store them in a array
        {
            rem = tem%10;
            arr[k++] = rem;
            tem = tem/10;
        }
        arr[k] = '\0';
        tem = i;
        j = 0;              //used keep track of index position
        while(tem&&flag)
        {
            rem = tem%10;

            for(m = 0; m<k; m++)
            {
                if(rem == arr[m]&&j!=m)     //element is found in the array and
                {                           // it not same index Ex: duplicate element
                    flag = 0;
                    break;
                }
            }

            j++;                    //increase index count
            tem = tem/10;

        }
        if(flag)        //if flag == 1
            count++;    //increase count of unique number

    }
  printf("%d",count);   //total count of unique number
  return 0;
}

SOLUTION 2 (using Hashing)

In this we store the count of each digit in a array using hashing. For example num = 133 then count[1] =1, count[3] =2.

#include<stdio.h>
int main()
{
    int i,a,flag,j,b,tem,rem,total=0;
    int count[10]={0};
    scanf("%d %d",&a,&b);

    for(i = a; i<=b; i++)        //loop for the given range
    {
        tem = i;
        flag =1;

        while(tem)
        {
            rem = tem %10;
            count[rem] +=1;         //increase the count of rem Ex: if rem=2 then count[2]++
            tem = tem/10;
        }
        for(j = 0;j<10; j++)        //loop from 0 to 9
        {

            //any digit have count more than 1 EX: 1222 then count[2]=>3 which is greater than 1
            if(count[j]>1)
            {
                flag = 0;
                break;
            }
        }
        if(flag)
          {
             total++;
             printf("%d ",i);
          }          //increase the count of unique number

          /*Reset the count of digits to 0  */
         for(j=0;j<10;j++)
         {
             count[j] = 0;
         }

    }
    printf("%d ",total);
}

SOLUTION 3: (Using Set)

In this, we use Set to check whether the duplicate element is present or not because Set can hold only unique elements.

import java.util.Scanner;
import java.util.Set;
import java.util.HashSet;

public class TCSdigital
{
  public static void main(String ar[])
  {
    Scanner sc = new Scanner(System.in);
    int a = sc.nextInt();
    int b = sc.nextInt();
    boolean flag = true;
    int count=0,tem,rem;
    /*Set is the check whether duplicate digit is present 
     * cuz set can store only unique elements
     */
    Set<Integer> s = new HashSet<Integer>();
    for(int i = a; i<=b; i++)
    { tem = i;
      flag = true;
      while(tem!= 0)
      {
        rem = tem%10;
        //s.add() returns false if it is a duplicate element 
        if(!(s.add(rem)))
        { flag = false;
          break;
        }
        tem = tem/10;
      }
      if(flag)
        count++;
      s.clear();         //clear the set so that the next number in the range can be stored
  }
    System.out.println(count);
 }
}

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