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# Python 3 Program For Range LCM Queries

Ranging a query is a common database current interest operation present in data structure to restore all the records where the output value lies between an upper and lower boundary. This process works with some input data, to structure them in an efficient manner on any subset of a particular input. The range function, denoted as range() is used to iterate a for loop over a series. We need to declare the start as 0 at the beginning of a process. If somehow we miss this step, the process will run and iterate the loop until the end (-1).

A range is a minimum and maximum value which can be stored inside a variable. They are generated with an operator and used to call arrays of the variables. They return an output list present in a range between x to y-1.

### Example

Let’s assume, we have an array set of integers and we need to evaluate the query in the form of LCM(a,r). Hence, we have to evaluate the queries in an efficient manner.

LCM(a,r) denotes the least common multiple of an array that present in between the index a and r. Here both indicates are inclusive in manner.

Mathematically we all know; L.C.M. = L.C.M Of Numerator/H.C.F of Denominator.

So, by using this logic we can follow the rule to range LCM query written below:

LCM(a, r) = LCM(arr[a], arr[a+1] , ......... ,arr[r-1], arr[r])

Applying This Logic:

Input –

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(1, 0, 10); cout << query(1, 0, 10, 2, 5) << endl; cout << query(1, 0, 10, 5, 10) << endl; cout << query(1, 0, 10, 0, 10) << endl;

Output –

60 15708 78540

The time complexity of this particular process denoted as O(Log N * Log n). Here N is the total numbers of elements present in an array. We need to declare the Log n as the time requirement finder for the LCM operation in a particular coding environment and O(N) time is required to build a tree to get an output from the written program. It also indicates the space requirement for the process.

### Algorithm to range LCM queries

Step 1 − Start

Step 2 − Initialize the two numbers of variables for tow numbers.

Step 3 − Find the storage value for each number.

Step 4 − Separate the variable with a 'max' function.

Step 5 − If, the max is divisible by the first number and also with the second.

Step 6 − Print the max as LCM.

Step 7 − Else, it is not divisible then increase it by 1.

Step 8 − Then again go for step five, until a number has been printed.

Step 9 − Repeat the process until max value is found which satisfies the condition.

Step 10 − Terminate.

### Syntax to range LCM query

int find_therangelcm(int a, int tl, int ts, int r) { if (r > t[a]) return -1; if (tl == ts) return tl; int tm = (tl + ts) / 2; if (t[a*2] >= r) return find_therangelcm(a*2, tl, tm, r); else return find_therangelcm(a*2+1, tm+1, ts, r - t[a*2]); }

In this syntax, we have explained how to range a LCM queries in a particular coding environment.

### Approaches to follow

Approach 1 − The naïve approach by using a segment tree.

Approach 2 − Find LCM of two in a general way.

## The naive approach by using a segment tree

There is no update operation for this problem but naive approach is not the right one alone. We need to implement a segment tree to get the possible result. Here we will use a logic -

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

Here is the implementation steps −

Built segment tree for the array.

Traverse the particular range of segment tree.

Calculate the LCM from that range.

Print the answer for that segment.

### Example 1

MAX = 1000 tree = [0] * (4 * MAX) arr = [0] * MAX def gcd(a: int, b: int): if a == 0: return b return gcd(b % a, a) def lcm(a: int, b: int): return (a * b) // gcd(a, b) def build(node: int, start: int, end: int): if start == end: tree[node] = arr[start] return mid = (start + end) // 2 build(2 * node, start, mid) build(2 * node + 1, mid + 1, end) left_lcm = tree[2 * node] right_lcm = tree[2 * node + 1] tree[node] = lcm(left_lcm, right_lcm) def query(node: int, start: int, end: int, l: int, r: int): if end < l or start > r: return 1 if l <= start and r >= end: return tree[node] 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) if __name__ == "__main__": # initialize the array arr[0] = 16 arr[1] = 7 arr[2] = 10 arr[3] = 2 arr[4] = 22 arr[5] = 31 arr[6] = 11 arr[7] = 17 arr[8] = 14 arr[9] = 1 arr[10] = 44 build(1, 0, 10) print(query(1, 0, 10, 7, 5)) print(query(1, 0, 10, 5, 10)) print(query(1, 0, 10, 0, 10))

### Output

1 162316 3246320

## Find LCM of two in a general way

Here in this program, we have two integers n1 and n2. And the larger of those two numbers is stored in max.

### Example 2

def compute_lcm(x, y): if x > y: greater = x else: greater = y while(True): if((greater % x == 0) and (greater % y == 0)): lcm = greater break greater += 1 return lcm num1 = 16 num2 = 7 print("The L.C.M. is of the given number", compute_lcm(num1, num2))

### Output

The L.C.M. is of the given number 112

## Conclusion

In this article today, we have found how to write a program by using a particular coding environment to find out the range of the given LCM queries.