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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 ``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

 ``

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

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

Approach#2: Using math

We first define a helper function lcm() to calculate the least common multiple of two numbers. Then, for each query, we iterate through the subarray of arr defined by the query range and calculate the LCM using the lcm() function. The LCM value is stored in a list, which is returned as the final result.

Algorithm

1. Define a helper function lcm(a, b) to calculate the least common multiple of two numbers.
2. Define a function range_lcm_queries(arr, queries) that takes an array arr and a list of query ranges queries as input.
3. Create an empty list results to store the LCM values for each query.
4. For each query in queries, extract the left and right indices l and r.
5. Set lcm_val to the value of arr[l].
6. For each index i in the range l+1 to r, update lcm_val to be the LCM of lcm_val and arr[i] using the lcm() function.
7. Append lcm_val to the results list.
8. Return the results list.

Python3

 `from` `math ``import` `gcd` `def` `lcm(a, b):``    ``return` `a``*``b ``/``/` `gcd(a, b)` `def` `range_lcm_queries(arr, queries):``    ``results ``=` `[]``    ``for` `query ``in` `queries:``        ``l, r ``=` `query``        ``lcm_val ``=` `arr[l]``        ``for` `i ``in` `range``(l``+``1``, r``+``1``):``            ``lcm_val ``=` `lcm(lcm_val, arr[i])``        ``results.append(lcm_val)``    ``return` `results` `# example usage``arr ``=` `[``5``, ``7``, ``5``, ``2``, ``10``, ``12` `,``11``, ``17``, ``14``, ``1``, ``44``]``queries ``=` `[(``2``, ``5``), (``5``, ``10``), (``0``, ``10``)]``print``(range_lcm_queries(arr, queries))  ``# output: [60, 15708, 78540]`

Output

`[60, 15708, 78540]`

Time Complexity: O(log(min(a,b))). For each query range, we iterate through a subarray of size O(n), where n is the length of arr. Therefore, the time complexity of the overall function is O(qn log(min(a_i))) where q is the number of queries and a_i is the i-th element of arr.
Space Complexity: O(1) since we are only storing a few integers at a time. The space used by the input arr and queries is not considered.

My Personal Notes arrow_drop_up