Calculate GCD of all pairwise sum of given two Arrays
Last Updated :
04 May, 2022
Given two arrays A[] and B[] of size N and M calculate GCD of all pairwise sum of (A[i]+B[j]) 1<=i<=N and 1<=j<=M.
Examples:
Input: A[] = {1, 7, 25, 55}, B[] = {1, 3, 5}
Output: 2
Explanation: The GCD of all pairwise sum of (A[i]+B[j]) is equals to
GCD(1+1, 1+3, 1+5, 7+1, 7+3, 7+5, 25+1, 25+3, 25+5, 55+1, 55+3, 55+5)
GCD(2, 4, 6, 8, 10, 12, 26, 28, 30, 56, 58, 60) = 2
Input: A[] = {8, 16, 20}, B[] = {12, 24}
Output: 4
Naive Approach: The simple approach of this problem is to calculate all pairwise sums and then calculate their GCD.
Below is the implementation of this approach.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate gcd
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
int calculateGCD(vector<int>& a,
vector<int>& b,
int N, int M)
{
int ans = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
// Pairwise sum of all elements
int sum = a[i] + b[j];
// Finding gcd of the elements
ans = gcd(ans, sum);
}
}
return ans;
}
// Driver code
int main()
{
int N = 4, M = 3;
// Initialization of the vector
vector<int> A = { 1, 7, 25, 55 };
vector<int> B = { 1, 3, 5 };
// output
cout << calculateGCD(A, B, N, M);
return 0;
}
// Java code to implement the approach
import java.util.*;
class GFG {
// Function to calculate gcd
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
static int calculateGCD(int a[],
int b[],
int N, int M)
{
int ans = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
// Pairwise sum of all elements
int sum = a[i] + b[j];
// Finding gcd of the elements
ans = gcd(ans, sum);
}
}
return ans;
}
// Driver code
public static void main (String[] args) {
int N = 4, M = 3;
// Initialization of the vector
int A[] = { 1, 7, 25, 55 };
int B[] = { 1, 3, 5 };
// output
System.out.print(calculateGCD(A, B, N, M));
}
}
// This code is contributed by hrithikgarg03188.
# Python code to implement the approach
# Function to calculate gcd
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
# Function to calculate the GCD
# of all pairwise sums
def calculateGCD(a, b, N, M):
ans = 0
for i in range(N):
for j in range(M):
# Pairwise sum of all elements
sum = a[i]+b[j]
# Finding gcd of the elements
ans = gcd(ans, sum)
return ans
# Driver code
N = 4
M = 3
A = [1, 7, 25, 55]
B = [1, 3, 5]
print(calculateGCD(A, B, N, M))
'''This Code is contributed by Rajat Kumar'''
// C# code to implement the approach
using System;
using System.Numerics;
using System.Collections.Generic;
public class GFG {
// Function to calculate gcd
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
static int calculateGCD(int[] a, int[] b, int N, int M)
{
int ans = 0;
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
// Pairwise sum of all elements
int sum = a[i] + b[j];
// Finding gcd of the elements
ans = gcd(ans, sum);
}
}
return ans;
}
// Driver Code
public static void Main(string[] args)
{
int N = 4, M = 3;
// Initialization of the vector
int[] A = { 1, 7, 25, 55 };
int[] B = { 1, 3, 5 };
// output
Console.WriteLine(calculateGCD(A, B, N, M));
}
}
// This code is contributed by phasing17
<script>
// JavaScript code to implement the approach
// Function to calculate gcd
const gcd = (a, b) => {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
const calculateGCD = (a, b, N, M) => {
let ans = 0;
for (let i = 0; i < N; i++) {
for (let j = 0; j < M; j++) {
// Pairwise sum of all elements
let sum = a[i] + b[j];
// Finding gcd of the elements
ans = gcd(ans, sum);
}
}
return ans;
}
// Driver code
let N = 4, M = 3;
// Initialization of the vector
let A = [1, 7, 25, 55];
let B = [1, 3, 5];
// output
document.write(calculateGCD(A, B, N, M));
// This code is contributed by rakeshsahni
</script>
Time Complexity: O(N*M)
Auxiliary Space: O(N)
Efficient Approach: The problem can be solved efficiently based on the following mathematical observation:
Observation:
- By Euclidean algorithm it can be said that GCD( a, b) = GCD(a-b, b) where(a>b) .
- Then for any j:
GCD(A[0] + B[j], A[1] + B[j], . . ., A[N-1] + B[j]) = GCD(A[0]+B[j], A[1]-A[0], A[2]-A[0], . . ., A[N]-A[0]).
The term GCD(A[1]-A[0], A[2]-A[0], . . ., A[N]-A[0]) is common for all j from 0 to M-1 (say this is X) - This leaves us with only the elements of type (A[0] + B[j]) for which will calculate their GCD and will get the desired GCD upon calculating their GCD with X.
Follow the below steps to solve this problem:
- First sort both the arrays.
- Then iterate through i = 1 to N-1 and calculate GCD of all pairs of A[i] -A[0] (say X).
- Then iterate through i = 1 to M-1 and calculate GCD of all pairs of A[0] +B[i] (say Y).
- Return the final answer which is GCD of (X and Y).
Below is the implementation of this approach.
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function for calculating gcd
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the required GCD
int calculateGCD(vector<int>& a,
vector<int>& b, int N, int M)
{
int ans = 0;
// Sorting the arrays
sort(a.begin(), a.end());
sort(b.begin(), b.end());
// Calculating the gcd of all the elements
// of type (i, j) i>0 and j>=0
// Using the property
// gcd(a[0]+b[j], a[i]+b[j])
// =gcd(a[0]+b[j], a[i]-a[0])
for (int i = 1; i < N; i++) {
ans = gcd(ans, a[i] - a[0]);
}
// Calculating the gcd of the remaining
// elements of the type (a[0]+b[j])
for (int i = 0; i < M; i++) {
ans = gcd(ans, a[0] + b[i]);
}
return ans;
}
// Driver code
int main()
{
int N = 4, M = 3;
// Initialization of the array
vector<int> A = { 1, 7, 25, 55 };
vector<int> B = { 1, 3, 5 };
// Function call
cout << calculateGCD(A, B, N, M);
return 0;
}
// JAVA code to implement the above approach
import java.util.*;
class GFG {
// Function to calculate gcd
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
static int calculateGCD(int a[],
int b[],
int N, int M)
{
int ans = 0;
// Sorting the arrays
Arrays.sort(a);
Arrays.sort(b);
// Calculating the gcd of all the elements
// of type (i, j) i>0 and j>=0
// Using the property
// gcd(a[0]+b[j], a[i]+b[j])
// =gcd(a[0]+b[j], a[i]-a[0])
for (int i = 1; i < N; i++) {
ans = gcd(ans, a[i] - a[0]);
}
// Calculating the gcd of the remaining
// elements of the type (a[0]+b[j])
for (int i = 0; i < M; i++) {
ans = gcd(ans, a[0] + b[i]);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 4, M = 3;
// Initialization of the vector
int A[] = { 1, 7, 25, 55 };
int B[] = { 1, 3, 5 };
// output
System.out.print(calculateGCD(A, B, N, M));
}
}
// This code is contributed by sanjoy_62.
# Python3 code to implement the approach
# Function for calculating gcd
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
# Function to calculate the required GCD
def calculateGCD(a, b, N, M):
ans = 0
# sorting the arrays
a.sort()
b.sort()
# calculating the gcd of all the elements
# of type (i, j) i>0 and j>=0
# Using the property
# gcd(a[0]+b[j], a[i]+b[j])
# =gcd(a[0]+b[j], a[i]-a[0])
for i in range(1, N):
ans = gcd(ans, a[i] - a[0])
# Calculating the gcd of the remaining
# elements of the type (a[0]+b[j])
for i in range(M):
ans = gcd(ans, a[0] + b[i])
return ans
# Driver code
N, M = 4, 3
A = [1, 7, 25, 55]
B = [1, 3, 5]
# function all
print(calculateGCD(A, B, N, M))
# This code is contributed by phasing17.
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to calculate gcd
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the GCD
// of all pairwise sums
static int calculateGCD(int[] a,
int[] b,
int N, int M)
{
int ans = 0;
// Sorting the arrays
Array.Sort(a);
Array.Sort(b);
// Calculating the gcd of all the elements
// of type (i, j) i>0 and j>=0
// Using the property
// gcd(a[0]+b[j], a[i]+b[j])
// =gcd(a[0]+b[j], a[i]-a[0])
for (int i = 1; i < N; i++) {
ans = gcd(ans, a[i] - a[0]);
}
// Calculating the gcd of the remaining
// elements of the type (a[0]+b[j])
for (int i = 0; i < M; i++) {
ans = gcd(ans, a[0] + b[i]);
}
return ans;
}
// Driver Code
public static void Main()
{
int N = 4, M = 3;
// Initialization of the vector
int[] A = { 1, 7, 25, 55 };
int[] B = { 1, 3, 5 };
// output
Console.Write(calculateGCD(A, B, N, M));
}
}
// This code is contributed by sanjoy_62.
<script>
// JavaScript code to implement the approach
// Function for calculating gcd
function gcd(a, b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the required GCD
function calculateGCD(a, b, N, M)
{
var ans = 0;
// sorting the arrays
a.sort()
b.sort()
// calculating the gcd of all the elements
// of type (i, j) i>0 and j>=0
// Using the property
// gcd(a[0]+b[j], a[i]+b[j])
// =gcd(a[0]+b[j], a[i]-a[0])
for (var i = 1; i < N; i++)
ans = gcd(ans, a[i] - a[0]);
// Calculating the gcd of the remaining
// elements of the type (a[0]+b[j])
for (var i = 0; i < M; i++)
ans = gcd(ans, a[0] + b[i]);
return ans;
}
// Driver code
var N = 4;
var M = 3;
var A = [1, 7, 25, 55];
var B = [1, 3, 5];
// function all
document.write(calculateGCD(A, B, N, M))
// This code is contributed by phasing17.
</script>
Time Complexity : O(N*logN + M*logM)
Auxiliary Space: O(1)
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