# Range LCM Queries

Given an array of integers, evaluate queries of the form LCM(l, r). There might be many queries, hence evaluate the queries efficiently.

```LCM (l, r) denotes the LCM of array elements
that lie between the index l and r
(inclusive of both indices)

Mathematically,
LCM(l, r) = LCM(arr[l],  arr[l+1] , ......... ,
arr[r-1], arr[r])
```

Examples:

```Inputs : Array = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 44}
Queries: LCM(2, 5), LCM(5, 10), LCM(0, 10)
Outputs: 60 15708 78540
Explanation : In the first query LCM(5, 2, 10, 12) = 60,
similarly in other queries.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A naive solution would be to traverse the array for every query and calculate the answer by using,
LCM(a, b) = (a*b) / GCD(a,b)

However as the number of queries can be large, this solution would be impractical.

An efficient solution would be to use segment tree. Recall that in this case, where no update is required, we can build the tree once and can use that repeatedly to answer the queries. Each node in the tree should store the LCM value for that particular segment and we can use the same formula as above to combine the segments. Hence we can answer each query efficiently!

Below is a C++ solution for the same.

```// LCM of given range queries using Segment Tree
#include <bits/stdc++.h>
using namespace std;

#define MAX 1000

// allocate space for tree
int tree[4*MAX];

// declaring the array globally
int arr[MAX];

// Function to return gcd of a and b
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b%a, a);
}

//utility function to find lcm
int lcm(int a, int b)
{
return a*b/gcd(a,b);
}

// Function to build the segment tree
// Node starts beginning index of current subtree.
// start and end are indexes in arr[] which is global
void build(int node, int start, int end)
{
// If there is only one element in current subarray
if (start==end)
{
tree[node] = arr[start];
return;
}

int mid = (start+end)/2;

// build left and right segments
build(2*node, start, mid);
build(2*node+1, mid+1, end);

// build the parent
int left_lcm = tree[2*node];
int right_lcm = tree[2*node+1];

tree[node] = lcm(left_lcm, right_lcm);
}

// Function to make queries for array range )l, r).
// Node is index of root of current segment in segment
// tree (Note that indexes in segment tree begin with 1
// for simplicity).
// start and end are indexes of subarray covered by root
// of current segment.
int query(int node, int start, int end, int l, int r)
{
// Completely outside the segment, returning
// 1 will not affect the lcm;
if (end<l || start>r)
return 1;

// completely inside the segment
if (l<=start && r>=end)
return tree[node];

// partially inside
int mid = (start+end)/2;
int left_lcm = query(2*node, start, mid, l, r);
int right_lcm = query(2*node+1, mid+1, end, l, r);
return lcm(left_lcm, right_lcm);
}

//driver function to check the above program
int main()
{
//initialize the array
arr[0] = 5;
arr[1] = 7;
arr[2] = 5;
arr[3] = 2;
arr[4] = 10;
arr[5] = 12;
arr[6] = 11;
arr[7] = 17;
arr[8] = 14;
arr[9] = 1;
arr[10] = 44;

// build the segment tree
build(1, 0, 10);

// Now we can answer each query efficiently

// Print LCM of (2, 5)
cout << query(1, 0, 10, 2, 5) << endl;

// Print LCM of (5, 10)
cout << query(1, 0, 10, 5, 10) << endl;

// Print LCM of (0, 10)
cout << query(1, 0, 10, 0, 10) << endl;

return 0;
}
```

Output:

```60
15708
78540
```

This article is contributed by Ashutosh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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