Skip to content
Related Articles

Related Articles

Online Queries for GCD of array after divide operations

View Discussion
Improve Article
Save Article
Like Article
  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021

You are given an array of N integer values and M update operations. An update consists of choosing an element of the array and dividing it by a given value. It is guaranteed that the element is divisible by the chosen value. After each update, you should compute the greatest common divisor of all the elements of the array.
Examples: 
 

Input : 3 3
        36 24 72
        1 3
        3 12
        2 4
Output :12
        6
        6
After each operation the array values will be:
1. 12, 24, 72
2. 12, 24, 6
3. 12, 6, 6

Input :5 6
       100 150 200 600 300
       4 6
       2 3
       4 4
       1 4
       2 5
       5 25
Output : 50
         50
         25
         25
         5
         1

 

Approach 
First, you should compute the Greatest Common Divisor(gcd) of all the initial numbers. Because the queries consist of dividing a number to one of its divisors it means the after each query the new gcd is a divisor of the old gcd. So for each query, you should simply compute the gcd between the updated value and the previous gcd. 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
void print_gcd_online(int n, int m,
                      int query[][2], int arr[])
{
    // stores the gcd of the initial array elements
    int max_gcd = 0;
    int i = 0;
 
    // calculates the gcd
    for (i = 0; i < n; i++)
        max_gcd = __gcd(max_gcd, arr[i]);
 
    // performing online queries
    for (i = 0; i < m; i++)
    {
        // index is 1 based
        query[i][0]--;
 
        // divide the array element
        arr[query[i][0]] /= query[i][1];
 
        // calculates the current gcd
        max_gcd = __gcd(arr[query[i][0]], max_gcd);
 
        // print the gcd after each step
        cout << max_gcd << endl;
    }
}
 
// Driver code
int main()
{
    int n = 3;
    int m = 3;
    int query[m][2];
    int arr[] = {36, 24, 72};
    query[0][0] = 1;
    query[0][1] = 3;
    query[1][0] = 3;
    query[1][1] = 12;
    query[2][0] = 2;
    query[2][1] = 4;
 
    print_gcd_online(n, m, query, arr);
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java




// Java implementation of the approach
 
class GFG {
 
    // returns the gcd after all updates
    // in the array
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
 
        return gcd(b % a, a);
    }
 
    static void print_gcd_online(int n, int m,
                    int[][] query, int[] arr)
    {
 
        // stores the gcd of the initial array elements
        int max_gcd = 0;
 
        int i = 0;
        for (i = 0; i < n; i++) // calculates the gcd
            max_gcd = gcd(max_gcd, arr[i]);
 
        // performing online queries
        for (i = 0; i < m; i++) {
 
            query[i][0]--; // index is 1 based
 
            // divide the array element
            arr[query[i][0]] /= query[i][1];
  
            // calculates the current gcd
            max_gcd = gcd(arr[query[i][0]], max_gcd);
 
            // print the gcd after each step
            System.out.println(max_gcd);
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 3;
        int m = 3;
        int[][] query = new int[m][2];
        int[] arr = new int[] { 36, 24, 72 };
        query[0][0] = 1;
        query[0][1] = 3;
        query[1][0] = 3;
        query[1][1] = 12;
        query[2][0] = 2;
        query[2][1] = 4;
 
        print_gcd_online(n, m, query, arr);
    }
}

Python3




# Python3 implementation of the
# above approach
 
# Returns the gcd after all
# updates in the array
def gcd(a, b):
     
    if a == 0:
        return b
 
    return gcd(b % a, a)
 
def print_gcd_online(n, m, query, arr):
 
    # Stores the gcd of the initial
    # array elements
    max_gcd = 0
 
    for i in range(0, n): # calculates the gcd
        max_gcd = gcd(max_gcd, arr[i])
 
    # performing online queries
    for i in range(0, m):
 
        query[i][0] -= 1 # index is 1 based
 
        # divide the array element
        arr[query[i][0]] //= query[i][1]
     
        # calculates the current gcd
        max_gcd = gcd(arr[query[i][0]], max_gcd)
 
        # Print the gcd after each step
        print(max_gcd)
 
# Driver code
if __name__ == "__main__":
     
    n, m = 3, 3
    query = [[1,3], [3,12], [2,4]]
    arr = [36, 24, 72]
         
    print_gcd_online(n, m, query, arr)
     
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// returns the gcd after all
// updates in the array
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
static void print_gcd_online(int n, int m,
                             int[,] query,
                             int[] arr)
{
 
    // stores the gcd of the
    // initial array elements
    int max_gcd = 0;
 
    int i = 0;
    for (i = 0; i < n; i++) // calculates the gcd
        max_gcd = gcd(max_gcd, arr[i]);
 
    // performing online queries
    for (i = 0; i < m; i++)
    {
 
        query[i,0]--; // index is 1 based
 
        // divide the array element
        arr[query[i, 0]] /= query[i, 1];
 
        // calculates the current gcd
        max_gcd = gcd(arr[query[i, 0]], max_gcd);
 
        // print the gcd after each step
        Console.WriteLine(max_gcd);
    }
}
 
// Driver code
public static void Main()
{
    int n = 3;
    int m = 3;
    int[,] query = new int[m, 2];
    int[] arr = new int[] { 36, 24, 72 };
    query[0, 0] = 1;
    query[0, 1] = 3;
    query[1, 0] = 3;
    query[1, 1] = 12;
    query[2, 0] = 2;
    query[2, 1] = 4;
 
    print_gcd_online(n, m, query, arr);
}
}
 
// This code is contributed
// by Subhadeep Gupta

PHP




<?php
// PHP implementation of the approach
// returns the gcd after all updates
// in the array
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
 
    return gcd($b % $a, $a);
}
 
function print_gcd_online($n, $m,
                          $query, $arr)
{
 
    // stores the gcd of the
    // initial array elements
    $max_gcd = 0;
 
    $i = 0;
     
    // calculates the gcd
    for ($i = 0; $i < $n; $i++)
        $max_gcd = gcd($max_gcd,
                       $arr[$i]);
 
    // performing online queries
    for ($i = 0; $i < $m; $i++)
    {
 
        $query[$i][0]--; // index is 1 based
 
        // divide the array element
        $arr[$query[$i][0]] /= $query[$i][1];
 
        // calculates the current gcd
        $max_gcd = gcd($arr[$query[$i][0]],
                            $max_gcd);
 
        // print the gcd after each step
        echo ($max_gcd),"\n";
    }
}
 
// Driver code
$n = 3; $m = 3; $query;
$arr = array( 36, 24, 72 );
$query[0][0] = 1; $query[0][1] = 3;
$query[1][0] = 3; $query[1][1] = 12;
$query[2][0] = 2; $query[2][1] = 4;
 
print_gcd_online($n, $m, $query, $arr);
 
// This code is contributed by Sach_Code
?>

Javascript




<script>
 
    // JavaScript implementation of the approach
     
    // returns the gcd after all updates
    // in the array
    function gcd(a, b)
    {
        if (a == 0)
            return b;
   
        return gcd(b % a, a);
    }
     
    function print_gcd_online(n,m,query,arr)
    {
        // stores the gcd of the initial array elements
        let max_gcd = 0;
        let i = 0;
 
        // calculates the gcd
        for (i = 0; i < n; i++)
            max_gcd = gcd(max_gcd, arr[i]);
 
        // performing online queries
        for (i = 0; i < m; i++)
        {
            // index is 1 based
            query[i][0]--;
 
            // divide the array element
            arr[query[i][0]] /= query[i][1];
 
            // calculates the current gcd
            max_gcd = gcd(arr[query[i][0]], max_gcd);
 
            // print the gcd after each step
            document.write(max_gcd + "</br>");
        }
    }
 
    // Driver code
 
    let n = 3;
    let m = 3;
    let query = new Array(m);
    for(let i = 0; i < m; i++)
    {
        query[i] = new Array(2);
        for(let j = 0; j < 2; j++)
        {
            query[i][j] = 0;
        }
    }
    let arr = [36, 24, 72];
    query[0][0] = 1;
    query[0][1] = 3;
    query[1][0] = 3;
    query[1][1] = 12;
    query[2][0] = 2;
    query[2][1] = 4;
 
    print_gcd_online(n, m, query, arr);
     
</script>

Output: 

12
6
6

 

Time Complexity : O(m + n)
 


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!