Get free ebooK with 50 must do coding Question for Product Based Companies solved
Fill the details & get ebook over email
Thank You!
We have sent the Ebook on 50 Must Do Coding Questions for Product Based Companies Solved over your email. All the best!
Sorting interview programming

Maximum Divisors in a Range

Prepbytes

Concepts Used

Sorting

Difficulty Level

Hard, Mathematics

Problem Statement (Simplified):

For a given range p and q, find the highest common factor of two given numbers, i.e. a and b. If no such number exists, return -1.

Test Case

Input:
8 12
4
2 10
3 7
2 2
5 6

Output:
4
4
2
-1

Explanation:
For 8 and 12, common divisors are [1,2,4].

In the range [2,10] we have 4 which is the maximum divisor in range.

In the range [3,7] we have 4 which is the maximum divisor in range.

In the range [2,2] we have 2 which is the maximum divisor in range.

In the range [5,6], there lies no divisor so we print -1.

Solving Approach :

1) The Highest common factor of any both numbers will be gcd(a,b) and there exists no factor above this.
2) So, to calculate all the factors we check all number which divide a and b both from 1 to gcd(a,b) and save them in an array.
3) We sort the array containing all common factors so that all factors are organized and searching factors in the range are easier.
4) Highest common factor in the given range [p,q] can be found by searching lower bound of p and upper bound of q. Where,

  • Upper Bound: Upper bound of any number K returns the position of immediate next number which exists in the array and is just greater than K.
  • Lower Bound: Lower bound of any number K returns the position of immediate previous number which exists in the array and is just smaller than K.
    5) If the lower bound of p and upper bound of q is the same that means, there is no common factor in the range [p,q].
    6) If they are not same, the Highest Common factor in range will be the previous element to element at upper bound of q, as an element at the upper bound is greater than q, and element previous to it will be smaller than q and will be a common factor. Hence it is our answer, so we return the element.

Example

Let given numbers be a=8 and b=12

We find gcd(a,b) as no factor lies beyond gcd(a,b) which divided both numbers, hence,

gcd(a,b) = 4

We check all numbers between 1 and gcd(a,b) which divides both of our numbers a and b, we save such numbers in a array,

We also sort this array so that factors are arranged in increasing order,

For given query p and q, we check lower bound of p and upper bound of q, let’s assume two queries

  • Query-1:

p=5 and q=6

Here, lowerBound(p) = 4 and upperBound(q)=4, as both values are same that means our query is beyond the range of factors, hence we return -1.

  • Query-2:
    p=2 and q=3

Here, lowerBound(p) = 2 and upperBound(q)=4, as both values are not same that means our range coincides with range of factors, hence our answer lies at factor less than upperBound(q) i.e. 3.

Solutions

#include <stdio.h>

int gcd(int a, int b){
    while(a!=0 && b!=0){
        if(b%a==0)
            return a;
        int temp = a;
        a = b%a;
        b = temp;
    }
    return a;
}

int lowerBound(int a[], int start, int end, int key){
  if (start > end)
    return end;
  int mid = (start + end) / 2;
  if (a[mid] == key)
    return mid;
  else if (a[mid] > key)
    return lowerBound(a, start, mid-1, key);
  else
    return lowerBound(a, mid+1, end, key);
}

int upperBound(int a[], int start, int end, int key){
  if (start > end)
    return start;
  int mid = (start + end) / 2;
  if (a[mid] == key)
    return upperBound(a, mid+1, end, key);
  else if (a[mid] > key)
    return upperBound(a, start, mid-1, key);
  else
    return upperBound(a, mid+1, end, key);
}

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    int g = (a<b)?gcd(a,b):gcd(b,a);

    int divisors[9999999]={0}, count=0;
    for(int i=1; i<=g; i++)
        if(g%i==0)
            divisors[count++] = i;

    int n;
    scanf("%d",&n);
    while(n--){

      int low,high;
      scanf("%d%d",&low,&high);

      int id1 = lowerBound(divisors,0,count-1, low);
      int id2 = upperBound(divisors,0,count-1, high);
      if(low>divisors[id1] || count==0)
        id1++;
      int diff = id2 - id1;
      if(diff == 0)
        printf("-1\n");
      else
        printf("%d\n", divisors[id2-1]);
    }  
}

#include <bits/stdc++.h>
using namespace std;

int gcd(int a, int b){
    while(a!=0 && b!=0){
        if(b%a==0)
            return a;
        int temp = a;
        a = b%a;
        b = temp;
    }
    return a;
}

int lowerBound(int a[], int start, int end, int key){
  if (start > end)
    return end;
  int mid = (start + end) / 2;
  if (a[mid] == key)
    return mid;
  else if (a[mid] > key)
    return lowerBound(a, start, mid-1, key);
  else
    return lowerBound(a, mid+1, end, key);
}

int upperBound(int a[], int start, int end, int key){
  if (start > end)
    return start;
  int mid = (start + end) / 2;
  if (a[mid] == key)
    return upperBound(a, mid+1, end, key);
  else if (a[mid] > key)
    return upperBound(a, start, mid-1, key);
  else
    return upperBound(a, mid+1, end, key);
}

int main()
{
    int a,b;
    cin>>a>>b;
    int g = (a<b)?gcd(a,b):gcd(b,a);

    int divisors[9999999]={0}, count=0;
    for(int i=1; i<=g; i++)
        if(g%i==0)
            divisors[count++] = i;

    int n;
    cin>>n;
    while(n--){

      int low,high;
      cin>>low>>high;

      int id1 = lowerBound(divisors,0,count-1, low);
      int id2 = upperBound(divisors,0,count-1, high);
      if(low>divisors[id1] || count==0)
        id1++;
      int diff = id2 - id1;
      if(diff == 0)
        cout<<"-1\n";
      else
        cout<<divisors[id2-1]<<endl;
    }  
}

import java.util.*;
import java.io.*;

public class Main {

  static int gcd(int a, int b){
      while(a!=0 && b!=0){
          if(b%a==0)
              return a;
          int temp = a;
          a = b%a;
          b = temp;
      }
      return a;
  }

  static int lowerBound(int a[], int start, int end, int key){
    if (start > end)
      return end;
    int mid = (start + end) / 2;
    if (a[mid] == key)
      return mid;
    else if (a[mid] > key)
      return lowerBound(a, start, mid-1, key);
    else
      return lowerBound(a, mid+1, end, key);
  }

  static int upperBound(int a[], int start, int end, int key){
    if (start > end)
      return start;
    int mid = (start + end) / 2;
    if (a[mid] == key)
      return upperBound(a, mid+1, end, key);
    else if (a[mid] > key)
      return upperBound(a, start, mid-1, key);
    else
      return upperBound(a, mid+1, end, key);
  }

  public static void main(String args[]) throws IOException {

    Scanner sc = new Scanner(System.in);
    int a = sc.nextInt(),b = sc.nextInt();
    int g = (a<b)?gcd(a,b):gcd(b,a);

    int divisors[] = new int[9999999], count=0;
    for(int i=1; i<=g; i++)
        if(g%i==0)
            divisors[count++] = i;

    int n = sc.nextInt();
    while(n--!=0){

      int low = sc.nextInt(),high = sc.nextInt();

      int id1 = lowerBound(divisors,0,count-1, low);
      int id2 = upperBound(divisors,0,count-1, high);
      if(count==0 || low>divisors[id1])
        id1++;
      int diff = id2 - id1;
      if(diff == 0)
        System.out.println("-1");
      else
        System.out.println(divisors[id2-1]);
    }

  }
}

[forminator_quiz id="1237"]
Space Complexity is O(min(A,B)) to save all common factors.

This article tried to discuss the concept of Sorting. Hope this blog helps you understand the implementation. To practice more problems you can check out .

Leave a Reply

Close

Related Post