Find the GCD of LCM of all unique pairs in an Array
Last Updated :
10 Apr, 2023
Given an integer array arr[] of size N, the task is to find the GCD of LCM of all unique pair (i, j) of the array, such that i < j.
Examples:
Input: arr[] = {10, 24, 40, 80}
Output: 40
Explanation:
LCM of all unique pairs following given conditions are:
LCM(10, 24) = 120
LCM(10, 40) = 40
LCM(10, 80) = 80
LCM(24, 40) = 120
LCM(24, 80) = 240
LCM(40, 80) = 80
Therefore, GCD of these LCM = GCD(120, 40, 80, 120, 240, 80) = 40
Input: arr[] = {1, 1}
Output: 1
Naive Approach: The simplest approach is the find all unique pairs in the array. Then find their LCM. Then find the GCD of all the LCM.
Efficient Approach: The above naive approach can be optimised with the help of Suffix array. We can use the Suffix array to find the LCM of each element paired with other elements, efficiently. Then we can simply find and return the GCD of this LCM array.
- For every element A[i], we need to calculate LCM(a[i], a[j]), where j belong to [i+1, N-1].
- LCM of all pairs where starting element is A[i] can be written as
LCM(A[i], GCD(all j in range i+1 to n-1))
- For that we build a suffix array. Let say suffix[] that stores the gcd of elements that belong to the range [i+1, N-1].
- Then create a LCM array to store the LCM of A[i] and GCD of all elements after it, i.e.
LCM[i] = LCM(A[i], suffix[i+1])
Where suffix[i+1] stores the GCD of elements [i+1, n-1]
- Finally compute the GCD of all elements in LCM array.
C++
// C++ code to find the GCD of LCM
// of all unique pairs in an Array
#include <bits/stdc++.h>
using namespace std;
// Find lcm of element x and y
int LCM(int x, int y)
{
return x * y / __gcd(x, y);
}
// Function that finds gcd of lcm
// of all pairs of elements.
void gcd_of_lcm(int n, int arr[])
{
// n is the size of the array.
// arr is the array.
// Suffix array that stores
// the gcd of the array elements.
int suff[n];
// Initialize suffix array.
for(int x = 0; x < n; x++)
{
suff[x] = 1;
}
// Loop that make the suffix gcd array
suff[n - 1] = arr[n - 1];
for(int i = n - 2; i >= 0; i--)
{
suff[i] = __gcd(arr[i], suff[i + 1]);
}
// lcm array that store the lcm
// of ith elements for all j
// that satisfy given condition.
vector<int> lcm;
for(int i = 0; i < n - 1; i++)
{
// we find lcm[i] for lcm
// of ith elements for all j
// using below formula.
int y = LCM(arr[i], suff[i + 1]);
// Add lcm of ith elements
// for all j in lcm array.
lcm.push_back(y);
}
// Now we find gcd of all ith elements.
// where i = 0, 1, 2, 3.....n-2.
int ans = lcm[0];
for(int i = 1; i < n - 1; i++)
{
ans = __gcd(ans, lcm[i]);
}
cout << ans << endl;
}
// Driver code
int main()
{
int n = 4;
int a[] = { 10, 24, 40, 80 };
// Function call for input 1
gcd_of_lcm(n, a);
n = 10;
int b[] = { 540, 648, 810, 648, 720,
540, 594, 864, 972, 648 };
// Function call for input 2
gcd_of_lcm(n, b);
}
// This code is contributed by shobhitgupta907
// Java code to find the GCD of LCM
// of all unique pairs in an array
class GFG{
// Function to evaluate GCD of x and y
static int gcd(int x, int y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
// Function that finds gcd of lcm
// of all pairs of elements.
static void gcd_of_lcm(int n, int arr[])
{
// n = size of array i.e. arr[]
// Suffix array that stores
// the GCD of the array elements.
int suff[] = new int[n];
// Initialise the suffix array
for(int i = 0; i < n; i++)
suff[i] = 1;
// Loop that make suffix GCD array
suff[n - 1] = arr[n - 1];
for(int i = n - 2; i >= 0; i--)
suff[i] = gcd(arr[i], suff[i + 1]);
// lcm array that store lcm for pairwise
// consecutive elements
int lcm[] = new int[n - 1];
for(int i = 0; i < n - 1; i++)
// Find LCM using standard known formula
lcm[i] = (arr[i] * suff[i + 1]) /
gcd(arr[i], suff[i + 1]);
// Now we find gcd of all ith elements.
// where i = 0, 1, 2, 3.....n-2.
int ans = lcm[0];
for(int i = 1; i < n - 1; i++)
{
ans = gcd(ans, lcm[i]);
}
// Print the answer
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
// 1st input case
int n = 4;
int a[] = { 10, 24, 40, 80 };
// Function call for input 1
gcd_of_lcm(n, a);
// 2nd input case
n = 10;
int b[] = { 540, 648, 810, 648, 720,
540, 594, 864, 972, 648 };
// Function call for input 2
gcd_of_lcm(n, b);
}
}
// This code is contributed by Soumitri Chattopadhyay
# Python3 code to find the GCD of LCM
# of all unique pairs in an Array
from math import gcd
# find lcm of element x and y
def LCM(x, y):
return (x * y)//gcd(x, y)
# Function that finds gcd of lcm
# of all pairs of elements.
def gcd_of_lcm(n, arr):
# n is the size of the array.
# arr is the array.
# suffix array that stores
# the gcd of the array elements.
suff = [1]*n
# initialize suffix array.
# loop that make the suffix gcd array.
suff[n-1] = arr[n-1]
for i in range(n-2, -1, -1):
suff[i] = gcd(arr[i], suff[i + 1])
# lcm array that store the lcm
# of ith elements for all j
# that satisfy given condition.
lcm = []
for i in range(n-1):
# we find lcm[i] for lcm
# of ith elements for all j
# using below formula.
y = LCM( arr[i], suff[i + 1])
# add lcm of ith elements
# for all j in lcm array.
lcm.append(y)
# now we find gcd of all ith elements.
# where i = 0, 1, 2, 3.....n-2.
ans = lcm[0]
for i in range(1, n-1):
ans = gcd(ans, lcm[i])
print(ans)
if __name__== "__main__":
n = 4
a =[10, 24, 40, 80]
# function call for input 1
gcd_of_lcm(n, a)
n = 10
a =[540, 648, 810, 648, 720,
540, 594, 864, 972, 648]
# function call for input 2
gcd_of_lcm(n, a)
// C# code to find the GCD of LCM
// of all unique pairs in an array
using System;
class GFG{
// Function to evaluate GCD of x and y
static int gcd(int x, int y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
// Function that finds gcd of lcm
// of all pairs of elements.
static void gcd_of_lcm(int n, int[] arr)
{
// n = size of array i.e. arr[]
// Suffix array that stores
// the GCD of the array elements.
int[] suff = new int[n];
// Initialise the suffix array
for(int i = 0; i < n; i++)
suff[i] = 1;
// Loop that make suffix GCD array
suff[n - 1] = arr[n - 1];
for(int i = n - 2; i >= 0; i--)
suff[i] = gcd(arr[i], suff[i + 1]);
// lcm array that store lcm for pairwise
// consecutive elements
int[] lcm = new int[n - 1];
for(int i = 0; i < n - 1; i++)
// Find LCM using standard known formula
lcm[i] = (arr[i] * suff[i + 1]) /
gcd(arr[i], suff[i + 1]);
// Now we find gcd of all ith elements.
// where i = 0, 1, 2, 3.....n-2.
int ans = lcm[0];
for(int i = 1; i < n - 1; i++)
{
ans = gcd(ans, lcm[i]);
}
// Print the answer
Console.WriteLine(ans);
}
// Driver code
public static void Main()
{
// 1st input case
int n = 4;
int[] a = { 10, 24, 40, 80 };
// Function call for input 1
gcd_of_lcm(n, a);
// 2nd input case
n = 10;
int[] b = { 540, 648, 810, 648, 720,
540, 594, 864, 972, 648 };
// Function call for input 2
gcd_of_lcm(n, b);
}
}
// This code is contributed by sanjoy_62
<script>
// Javascript code to find the GCD of LCM
// of all unique pairs in an array
// Function to evaluate GCD of x and y
function gcd(x, y)
{
if (y == 0)
return x;
else
return gcd(y, x % y);
}
// Function that finds gcd of lcm
// of all pairs of elements.
function gcd_of_lcm(n, arr)
{
// n = size of array i.e. arr[]
// Suffix array that stores
// the GCD of the array elements.
let suff = new Array(n);
// Initialise the suffix array
for(let i = 0; i < n; i++)
suff[i] = 1;
// Loop that make suffix GCD array
suff[n - 1] = arr[n - 1];
for(let i = n - 2; i >= 0; i--)
suff[i] = gcd(arr[i], suff[i + 1]);
// lcm array that store lcm for pairwise
// consecutive elements
let lcm = new Array(n - 1);
for(let i = 0; i < n - 1; i++)
// Find LCM using standard known formula
lcm[i] = parseInt((arr[i] * suff[i + 1]) /
gcd(arr[i], suff[i + 1]), 10);
// Now we find gcd of all ith elements.
// where i = 0, 1, 2, 3.....n-2.
let ans = lcm[0];
for(let i = 1; i < n - 1; i++)
{
ans = gcd(ans, lcm[i]);
}
// Print the answer
document.write(ans + "</br>");
}
// Driver code
// 1st input case
let n = 4;
let a = [ 10, 24, 40, 80 ];
// Function call for input 1
gcd_of_lcm(n, a);
// 2nd input case
n = 10;
let b = [ 540, 648, 810, 648, 720,
540, 594, 864, 972, 648 ];
// Function call for input 2
gcd_of_lcm(n, b);
// This code is contributed by rameshtravel07
</script>
Time Complexity: O(N * log M), where M is the maximum element in the array.
Space Complexity: O(N)
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