Range LCM Queries

Software Development
#1

Hello Everyone,

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.

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.

Below is a 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