Count elements which divide all numbers in range L-R

Given N numbers and Q queries, each query consists of L and R. Task is to write a program which prints the count of numbers which divides all numbers in the given range L-R. 

Examples :

Input : a = {3, 4, 2, 2, 4, 6} 
        Q = 2
        L = 1 R = 4  
        L = 2 R = 6
Output :  0
          2 
Explanation : The range 1-4 has {3, 4, 2, 2} 
which does not have any number that divides all the 
numbers in this range. 
The range 2-6 has {4, 2, 2, 4, 6} which has  2 numbers {2, 2} which divides 
all numbers in the given range. 

Input: a = {1, 2, 3, 5} 
       Q = 2 
       L = 1 R = 4 
       L = 2 R = 4 
Output: 1 
        0      

Naive approach : Iterate from range L-R for every query and check if the given element at index-i divide all the numbers in the range. We keep a count for of all the elements which divides all the numbers. The complexity of every query at worst case will be O(n²).

Below is the implementation of Naive Approach :

C++

// CPP program to Count elements which
// divides all numbers in range L-R
#include <bits/stdc++.h>
using namespace std;

// function to count element
// Time complexity O(n^2) worst case
int answerQuery(int a[], int n,
                int l, int r)
{
    // answer for query
    int count = 0;

    // 0 based index
    l = l - 1;

    // iterate for all elements
    for (int i = l; i < r; i++)
    {
        int element = a[i];
        int divisors = 0;

        // check if the element divides
        // all numbers in range
        for (int j = l; j < r; j++)
        {
            // no of elements
            if (a[j] % a[i] == 0)
                divisors++;
            else
                break;
        }

        // if all elements are divisible by a[i]
        if (divisors == (r - l))
            count++;
    }

    // answer for every query
    return count;
}

// Driver Code
int main()
{
    int a[] = { 1, 2, 3, 5 };
    int n = sizeof(a) / sizeof(a[0]);

    int l = 1, r = 4;
    cout << answerQuery(a, n, l, r) << endl;

    l = 2, r = 4;
    cout << answerQuery(a, n, l, r) << endl;
    return 0;
}

Java

// Java program to Count elements which
// divides all numbers in range L-R
import java.io.*;

class GFG
{

// function to count element
// Time complexity O(n^2) worst case
static int answerQuery(int a[], int n,
                       int l, int r)
{
    // answer for query
    int count = 0;

    // 0 based index
    l = l - 1;

    // iterate for all elements
    for (int i = l; i < r; i++)
    {
        int element = a[i];
        int divisors = 0;

        // check if the element divides
        // all numbers in range
        for (int j = l; j < r; j++)
        {
            // no of elements
            if (a[j] % a[i] == 0)
                divisors++;
            else
                break;
        }

        // if all elements are divisible by a[i]
        if (divisors == (r - l))
            count++;
    }

    // answer for every query
    return count;
}

// Driver Code
public static void main (String[] args)
{
    int a[] = { 1, 2, 3, 5 };
    int n = a.length;

    int l = 1, r = 4;
    System.out.println( answerQuery(a, n, l, r));

    l = 2; r = 4;
    System.out.println( answerQuery(a, n, l, r));
}
}

// This code is contributed by anuj_67..

Python3

# Python 3 program to Count elements which
# divides all numbers in range L-R

# function to count element
# Time complexity O(n^2) worst case
def answerQuery(a, n, l, r):

    # answer for query
    count = 0

    # 0 based index
    l = l - 1

    # iterate for all elements
    for i in range(l, r, 1):
        element = a[i]
        divisors = 0

        # check if the element divides
        # all numbers in range
        for j in range(l, r, 1):

            # no of elements
            if (a[j] % a[i] == 0):
                divisors += 1
            else:
                break

        # if all elements are divisible
        # by a[i]
        if (divisors == (r - l)):
            count += 1

    # answer for every query
    return count

# Driver Code
if __name__ =='__main__':
    a = [1, 2, 3, 5]
    n = len(a)

    l = 1
    r = 4
    print(answerQuery(a, n, l, r))

    l = 2
    r = 4
    print(answerQuery(a, n, l, r))

# This code is contributed by
# Shashank_Sharma

C#

// C# program to Count elements which
// divides all numbers in range L-R
using System;

class GFG
{

// function to count element
// Time complexity O(n^2) worst case
static int answerQuery(int []a, int n,
                       int l, int r)
{
    // answer for query
    int count = 0;

    // 0 based index
    l = l - 1;

    // iterate for all elements
    for (int i = l; i < r; i++)
    {
        //int element = a[i];
        int divisors = 0;

        // check if the element divides
        // all numbers in range
        for (int j = l; j < r; j++)
        {
            // no of elements
            if (a[j] % a[i] == 0)
                divisors++;
            else
                break;
        }

        // if all elements are divisible by a[i]
        if (divisors == (r - l))
            count++;
    }

    // answer for every query
    return count;
}

// Driver Code
public static void Main ()
{
    int []a = { 1, 2, 3, 5 };
    int n = a.Length;

    int l = 1, r = 4;
    Console.WriteLine(answerQuery(a, n, l, r));

    l = 2; r = 4;
    Console.WriteLine(answerQuery(a, n, l, r));
}
}

// This code is contributed by anuj_67..

PHP

<?php
// PHP program to Count elements which
// divides all numbers in range L-R

// function to count element
// Time complexity O(n^2) worst case
function answerQuery($a, $n, $l, $r)
{
    // answer for query
    $count = 0;

    // 0 based index
    $l = $l - 1;

    // iterate for all elements
    for ($i = $l; $i < $r; $i++)
    {
        $element = $a[$i];
        $divisors = 0;

        // check if the element divides
        // all numbers in range
        for ($j = $l; $j < $r; $j++)
        {
            // no of elements
            if ($a[$j] % $a[$i] == 0)
                $divisors++;
            else
                break;
        }

        // if all elements are divisible by a[i]
        if ($divisors == ($r - $l))
            $count++;
    }

    // answer for every query
    return $count;
}

// Driver Code
$a = array(1, 2, 3, 5);
$n = sizeof($a);

$l = 1; $r = 4;
echo answerQuery($a, $n, $l, $r) . "\n";

$l = 2; $r = 4;
echo answerQuery($a, $n, $l, $r) . "\n";

// This code is contributed
// by Akanksha Rai

Javascript

<script>
// javascript program to Count elements which
// divides all numbers in range L-R

    // function to count element
    // Time complexity O(n^2) worst case
    function answerQuery(a , n , l , r)
    {

        // answer for query
        var count = 0;

        // 0 based index
        l = l - 1;

        // iterate for all elements
        for (i = l; i < r; i++) {
            var element = a[i];
            var divisors = 0;

            // check if the element divides
            // all numbers in range
            for (j = l; j < r; j++) {
                // no of elements
                if (a[j] % a[i] == 0)
                    divisors++;
                else
                    break;
            }

            // if all elements are divisible by a[i]
            if (divisors == (r - l))
                count++;
        }

        // answer for every query
        return count;
    }

    // Driver Code
        var a = [ 1, 2, 3, 5 ];
        var n = a.length;

        var l = 1, r = 4;
        document.write(answerQuery(a, n, l, r)+"<br/>");

        l = 2;
        r = 4;
        document.write(answerQuery(a, n, l, r));

// This code is contributed by gauravrajput1
</script>

Output

1
0

Efficient approach : 

Use Segment Trees to solve this problem. If an element divides all the numbers in a given range, then the element is the minimum number in that range and it is the gcd of all elements in the given range L-R. So the count of the number of minimums in range L-R, given that minimum is equal to the gcd of that range will be our answer to every query. The problem boils down to finding the GCD, MINIMUM and countMINIMUM for every range using Segment trees. On every node of the tree, three values are stored. 

On querying for a given range, if the gcd and minimum of the given range are equal, countMINIMUM is returned as the answer. If they are unequal, 0 is returned as the answer.

Below is the implementation of efficient approach :

C++

// CPP program to Count elements
// which divides all numbers in
// range L-R efficient approach
#include <bits/stdc++.h>
using namespace std;

#define N 100005

// predefines the tree with nodes
// storing gcd, min and count
struct node
{
    int gcd;
    int min;
    int cnt;
} tree[5 * N];

// function to construct the tree
void buildtree(int low, int high,
               int pos, int a[])
{
    // base condition
    if (low == high)
    {
        // initially always gcd and min
        // are same at leaf node
        tree[pos].min =  tree[pos].gcd = a[low];
        tree[pos].cnt = 1;

        return;
    }

    int mid = (low + high) >> 1;

    // left-subtree
    buildtree(low, mid, 2 * pos + 1, a);

    // right-subtree
    buildtree(mid + 1, high, 2 * pos + 2, a);

    // finds gcd of left and right subtree
    tree[pos].gcd = __gcd(tree[2 * pos + 1].gcd,
                      tree[2 * pos + 2].gcd);

    // left subtree has the minimum element
    if (tree[2 * pos + 1].min < tree[2 * pos + 2].min)
    {
        tree[pos].min = tree[2 * pos + 1].min;
        tree[pos].cnt = tree[2 * pos + 1].cnt;
    }

    // right subtree has the minimum element
    else
    if (tree[2 * pos + 1].min > tree[2 * pos + 2].min)
    {
        tree[pos].min = tree[2 * pos + 2].min;
        tree[pos].cnt = tree[2 * pos + 2].cnt;
    }

    // both subtree has the same minimum element
    else
    {
        tree[pos].min = tree[2 * pos + 1].min;
        tree[pos].cnt = tree[2 * pos + 1].cnt +
                        tree[2 * pos + 2].cnt;
    }
}

// function that answers every query
node query(int s, int e, int low, int high, int pos)
{
    node dummy;

    // out of range
    if (e < low or s > high)
    {
        dummy.gcd = dummy.min = dummy.cnt = 0;
        return dummy;
    }

    // in range
    if (s >= low and e <= high)
    {
        node dummy;
        dummy.gcd = tree[pos].gcd;
        dummy.min = tree[pos].min;
        if (dummy.gcd != dummy.min)
            dummy.cnt = 0;
        else
            dummy.cnt = tree[pos].cnt;

        return dummy;
    }

    int mid = (s + e) >> 1;

    // left-subtree
    node ans1 = query(s, mid, low,
                high, 2 * pos + 1);

    // right-subtree
    node ans2 = query(mid + 1, e, low,
                   high, 2 * pos + 2);

    node ans;

    // when both left subtree and
    // right subtree is in range
    if (ans1.gcd and ans2.gcd)
    {
        // merge two trees
        ans.gcd = __gcd(ans1.gcd, ans2.gcd);
        ans.min = min(ans1.min, ans2.min);

        // when gcd is not equal to min
        if (ans.gcd != ans.min)
            ans.cnt = 0;
        else
        {
            // add count when min is
            // same of both subtree
            if (ans1.min == ans2.min)
                ans.cnt = ans2.cnt + ans1.cnt;

            // store the minimal's count
            else
            if (ans1.min < ans2.min)
                ans.cnt = ans1.cnt;
            else
                ans.cnt = ans2.cnt;
        }

        return ans;
    }

    // only left subtree is in range
    else if (ans1.gcd)
        return ans1;

    // only right subtree is in range
    else if (ans2.gcd)
        return ans2;
}

// function to answer query in range l-r
int answerQuery(int a[], int n, int l, int r)
{
    // calls the function which returns
    // a node this function returns the
    // count which will be the answer
    return query(0, n - 1, l - 1, r - 1, 0).cnt;
}

// Driver Code
int main()
{
    int a[] = { 3, 4, 2, 2, 4, 6 };

    int n = sizeof(a) / sizeof(a[0]);
    buildtree(0, n - 1, 0, a);
    int l = 1, r = 4;

    // answers 1-st query
    cout << answerQuery(a, n, l, r) << endl;

    l = 2, r = 6;
    // answers 2nd query
    cout << answerQuery(a, n, l, r) << endl;
    return 0;
}

Output

0
2

Time Complexity: Time Complexity for tree construction is O(n logn) since tree construction takes O(n) and finding out gcd takes O(log n). The time taken for every query in worst case will be O(log n * log n) since the inbuilt function __gcd takes O(log n)