Range LCM Queries

Given an array arr[] of integers of size N and an array of Q queries, query[], where each query is of type [L, R] denoting the range from index L to index R, the task is to find the LCM of all the numbers of the range for all the queries.

Examples:

Input: arr[] = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 44}
query[] = {{2, 5}, {5, 10}, {0, 10}}
Output: 60,15708, 78540
Explanation: In the first query LCM(5, 2, 10, 12) = 60 
In the second query LCM(12, 11, 17, 14, 1, 44) = 15708
In the last query LCM(5, 7, 5, 2, 10, 12, 11, 17, 14, 1, 44) = 78540

Input: arr[] = {2, 4, 8, 16}, query[] = {{2, 3}, {0, 1}}
Output: 16, 4

Naive Approach: The approach is based on the following mathematical idea:

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

LCM(a, b) = (a*b) / GCD(a,b)

So traverse the array for every query and calculate the answer by using the above formula for LCM. 

Time Complexity: O(N * Q)
Auxiliary Space: O(1)

RangeLCM Queries using  Segment tree:

As the number of queries can be large, the naive solution would be impractical. This time can be reduced

There is no update operation in this problem. So we can initially build a segment tree and use that to answer the queries in logarithmic time.

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.

Follow the steps mentioned below to implement the idea:

• Build a segment tree from the given array.
• Traverse through the queries. For each query:
  • Find that particular range in the segment tree.
  • Use the above mentioned formula to combine the segments and calculate the LCM for that range.
  • Print the answer for that segment.

Below is the implementation of the above approach. 

C++

// 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;
}

Java

// LCM of given range queries
// using Segment Tree

class GFG {

    static final int MAX = 1000;

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

    // declaring the array globally
    static int arr[] = new int[MAX];

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

    // utility function to find lcm
    static 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
    static 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.
    static 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 code
    public static void main(String[] args)
    {

        // 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)
        System.out.println(query(1, 0, 10, 2, 5));

        // Print LCM of (5, 10)
        System.out.println(query(1, 0, 10, 5, 10));

        // Print LCM of (0, 10)
        System.out.println(query(1, 0, 10, 0, 10));
    }
}

// This code is contributed by 29AjayKumar

Python3

# LCM of given range queries using Segment Tree
MAX = 1000

# allocate space for tree
tree = [0] * (4 * MAX)

# declaring the array globally
arr = [0] * MAX

# Function to return gcd of a and b


def gcd(a: int, b: int):
    if a == 0:
        return b
    return gcd(b % a, a)

# utility function to find lcm


def lcm(a: int, b: int):
    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


def build(node: int, start: int, end: int):

    # If there is only one element
    # in current subarray
    if start == end:
        tree[node] = arr[start]
        return

    mid = (start + end) // 2

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

    # build the parent
    left_lcm = tree[2 * node]
    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.


def query(node: int, start: int,
          end: int, l: int, r: int):

    # Completely outside the segment,
    # returning 1 will not affect the lcm;
    if end < l or start > r:
        return 1

    # completely inside the segment
    if l <= start and r >= end:
        return tree[node]

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


# Driver Code
if __name__ == "__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)
    print(query(1, 0, 10, 2, 5))

    # Print LCM of (5, 10)
    print(query(1, 0, 10, 5, 10))

    # Print LCM of (0, 10)
    print(query(1, 0, 10, 0, 10))

# This code is contributed by
# sanjeev2552

C#

// LCM of given range queries
// using Segment Tree
using System;
using System.Collections.Generic;

class GFG {
    static readonly int MAX = 1000;

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

    // declaring the array globally
    static int[] arr = new int[MAX];

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

    // utility function to find lcm
    static 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
    static 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.
    static 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 code
    public static void Main(String[] args)
    {

        // 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)
        Console.WriteLine(query(1, 0, 10, 2, 5));

        // Print LCM of (5, 10)
        Console.WriteLine(query(1, 0, 10, 5, 10));

        // Print LCM of (0, 10)
        Console.WriteLine(query(1, 0, 10, 0, 10));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// LCM of given range queries using Segment Tree
const MAX = 1000

// allocate space for tree
var tree = new Array(4*MAX);

// declaring the array globally
var arr = new Array(MAX);

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

//utility function to find lcm
function lcm(a, b)
{
    return Math.floor(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
function build(node, start, end)
{
    // If there is only one element in current subarray
    if (start==end)
    {
        tree[node] = arr[start];
        return;
    }

    let mid = Math.floor((start+end)/2);

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

    // build the parent
    let left_lcm = tree[2*node];
    let 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.
function query(node, start, end, l, 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
    let mid = Math.floor((start+end)/2);
    let left_lcm = query(2*node, start, mid, l, r);
    let right_lcm = query(2*node+1, mid+1, end, l, r);
    return lcm(left_lcm, right_lcm);
}

//driver function to check the above program
    //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)
    document.write(query(1, 0, 10, 2, 5) +"<br>");

    // Print LCM of (5, 10)
    document.write(query(1, 0, 10, 5, 10) + "<br>");

    // Print LCM of (0, 10)
    document.write(query(1, 0, 10, 0, 10) + "<br>");

// This code is contributed by Manoj.
</script>

Output

60
15708
78540

Time Complexity: O(Log N * Log n) where N is the number of elements in the array. The other log n denotes the time required for finding the LCM. This time complexity is for each query. The total time complexity is O(N + Q*Log N*log n), this is because O(N) time is required to build the tree and then to answer the queries.
Auxiliary Space: O(N), where N is the number of elements in the array. This space is required for storing the segment tree.

Related Topic: Segment Tree

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