Pair with maximum GCD from two arrays
Given two arrays of n integers with values of array being small (values never exceed a small number say 100). Find the pair(x, y) which has maximum gcd. x and y cannot be of the same array. If multiple pairs have same gcd, then consider the pair which has the maximum sum.
Examples:
Input : a[] = {3, 1, 4, 2, 8}
b[] = {5, 2, 12, 8, 3}
Output : 8 8
Explanation: The maximum gcd is 8 which is
of pair(8, 8).
Input: a[] = {2, 3, 5}
b[] = {7, 11, 13}
Output: 5 13
Explanation: Every pair has a gcd of 1.
The maximum sum pair with GCD 1 is (5, 13)
A naive approach will be to iterate for every pair in both the arrays and find out the maximum gcd possible.
An efficient (only when elements are small) is to apply the sieve property and for that we need to pre calculate the following things.
- A cnt array to mark the presence of array elements.
- We check for all the numbers from 1 to N and for each its multiple we check that if the number exists then the max of the pre-existing number or the present existing multiple is stored.
- Step 1 and 2 is repeated for the other array also.
- At the end we check for the maximum multiple which is the common in both first and second array to get the maximum GCD, and in the position of is stored the element, in first the element of a array is stored, and in second the element of b array is stored, so we print the pair.
Below is the implementation of the above approach
C++
// CPP program to find maximum GCD pair// from two arrays#include <bits/stdc++.h>using namespace std; // Find the maximum GCD pair with maximum// sumvoid gcdMax(int a[], int b[], int n, int N){ // array to keep a count of existing elements int cnt[N] = { 0 }; // first[i] and second[i] are going to store // maximum multiples of i in a[] and b[] // respectively. int first[N] = { 0 }, second[N] = { 0 }; // traverse through the first array to // mark the elements in cnt for (int i = 0; i < n; ++i) cnt[a[i]] = 1; // Find maximum multiple of every number // in first array for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) if (cnt[j]) first[i] = max(first[i], j); // Find maximum multiple of every number // in second array // We re-initialise cnt[] and traverse // through the second array to mark the // elements in cnt memset(cnt, 0, sizeof(cnt)); for (int i = 0; i < n; ++i) cnt[b[i]] = true; for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) // if the multiple is present in the // second array then store the max // of number or the pre-existing // element if (cnt[j]) second[i] = max(second[i], j); // traverse for every elements and checks // the maximum N that is present in both // the arrays int i; for (i = N - 1; i >= 0; i--) if (first[i] && second[i]) break; cout << "Maximum GCD pair with maximum " "sum is " << first[i] << " " << second[i] << endl;} // driver program to test the above functionint main(){ int a[] = { 3, 1, 4, 2, 8 }; int b[] = { 5, 2, 12, 8, 3 }; int n = sizeof(a) / sizeof(a[0]); // Maximum possible value of elements // in both arrays. int N = 20; gcdMax(a, b, n, N); return 0;} |
Java
// Java program to find maximum // GCD pair from two arraysclass GFG{ // Find the maximum GCD// pair with maximum sumstatic void gcdMax(int[] a, int[] b, int n, int N){ // array to keep a count // of existing elements int[] cnt = new int[N]; // first[i] and second[i] // are going to store // maximum multiples of // i in a[] and b[] // respectively. int[] first = new int[N]; int[] second = new int[N]; // traverse through the // first array to mark // the elements in cnt for (int i = 0; i < n; ++i) cnt[a[i]] = 1; // Find maximum multiple // of every number in // first array for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) if (cnt[j] > 0) first[i] = Math.max(first[i], j); // Find maximum multiple // of every number in second // array. We re-initialise // cnt[] and traverse through // the second array to mark // the elements in cnt cnt = new int[N]; for (int i = 0; i < n; ++i) cnt[b[i]] = 1; for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) // if the multiple is present // in the second array then // store the max of number or // the pre-existing element if (cnt[j] > 0) second[i] = Math.max(second[i], j); // traverse for every // elements and checks // the maximum N that // is present in both // the arrays int x; for (x = N - 1; x >= 0; x--) if (first[x] > 0 && second[x] > 0) break; System.out.println(first[x] + " " + second[x]);} // Driver Codepublic static void main(String[] args){ int[] a = { 3, 1, 4, 2, 8 }; int[] b = { 5, 2, 12, 8, 3 }; int n = a.length; // Maximum possible // value of elements // in both arrays. int N = 20; gcdMax(a, b, n, N);}} // This code is contributed // by mits |
Python3
# Python 3 program to find maximum GCD pair# from two arrays # Find the maximum GCD pair with maximum# sumdef gcdMax(a, b, n, N): # array to keep a count of existing elements cnt = [0]*N # first[i] and second[i] are going to store # maximum multiples of i in a[] and b[] # respectively. first = [0]*N second = [0]*N # traverse through the first array to # mark the elements in cnt for i in range(n): cnt[a[i]] = 1 # Find maximum multiple of every number # in first array for i in range(1,N): for j in range(i,N,i): if (cnt[j]): first[i] = max(first[i], j) # Find maximum multiple of every number # in second array # We re-initialise cnt[] and traverse # through the second array to mark the # elements in cnt cnt = [0]*N for i in range(n): cnt[b[i]] = 1 for i in range(1,N): for j in range(i,N,i): # if the multiple is present in the # second array then store the max # of number or the pre-existing # element if (cnt[j]>0): second[i] = max(second[i], j) # traverse for every elements and checks # the maximum N that is present in both # the arrays i = N-1 while i>= 0: if (first[i]>0 and second[i]>0): break i -= 1 print( str(first[i]) + " " + str(second[i])) # driver program to test the above functionif __name__ == "__main__": a = [ 3, 1, 4, 2, 8 ] b = [ 5, 2, 12, 8, 3 ] n = len(a) # Maximum possible value of elements # in both arrays. N = 20 gcdMax(a, b, n, N) # this code is contributed by ChitraNayal |
C#
// C# program to find // maximum GCD pair // from two arraysusing System; class GFG{// Find the maximum GCD// pair with maximum sumstatic void gcdMax(int[] a, int[] b, int n, int N){ // array to keep a count // of existing elements int[] cnt = new int[N]; // first[i] and second[i] // are going to store // maximum multiples of // i in a[] and b[] // respectively. int[] first = new int[N]; int[] second = new int[N]; // traverse through the // first array to mark // the elements in cnt for (int i = 0; i < n; ++i) cnt[a[i]] = 1; // Find maximum multiple // of every number in // first array for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) if (cnt[j] > 0) first[i] = Math.Max(first[i], j); // Find maximum multiple // of every number in second // array. We re-initialise // cnt[] and traverse through // the second array to mark // the elements in cnt cnt = new int[N]; for (int i = 0; i < n; ++i) cnt[b[i]] = 1; for (int i = 1; i < N; ++i) for (int j = i; j < N; j += i) // if the multiple is present // in the second array then // store the max of number or // the pre-existing element if (cnt[j] > 0) second[i] = Math.Max(second[i], j); // traverse for every // elements and checks // the maximum N that // is present in both // the arrays int x; for (x = N - 1; x >= 0; x--) if (first[x] > 0 && second[x] > 0) break; Console.WriteLine(first[x] + " " + second[x]);} // Driver Codestatic int Main(){ int[] a = { 3, 1, 4, 2, 8 }; int[] b = { 5, 2, 12, 8, 3 }; int n = a.Length; // Maximum possible // value of elements // in both arrays. int N = 20; gcdMax(a, b, n, N); return 0;}} // This code is contributed // by mits |
PHP
<?php// PHP program to find // maximum GCD pair// from two arrays // Find the maximum GCD // pair with maximum// sumfunction gcdMax($a, $b, $n, $N){ // array to keep a count // of existing elements $cnt = array_fill(0, $N, 0); // first[i] and second[i] are // going to store maximum // multiples of i in a[] and // b[] respectively. $first = array_fill(0, $N, 0); $second = array_fill(0, $N, 0); // traverse through the first // array to mark the elements // in cnt for ($i = 0; $i < $n; ++$i) $cnt[$a[$i]] = 1; // Find maximum multiple // of every number in // first array for ($i = 1; $i < $N; ++$i) for ($j = $i; $j < $N; $j += $i) if ($cnt[$j]) $first[$i] = max($first[$i], $j); // Find maximum multiple of every // number in second array // We re-initialise cnt[] and // traverse through the second // array to mark the elements in cnt $cnt = array_fill(0, $N, 0); for ($i = 0; $i < $n; $i++) $cnt[$b[$i]] = 1; for ($i = 1; $i < $N; $i++) for ($j = $i; $j < $N; $j += $i) // if the multiple is present // in the second array then // store the max of number or // the pre-existing element if ($cnt[$j]) $second[$i] = max($second[$i], $j); // traverse for every elements // and checks the maximum N // that is present in both // the arrays $x = $N - 1; for (; $x >= 0; $x--) if ($first[$x] && $second[$x]) break; echo $first[$x] . " " . $second[$x] . "\n";} // Driver code$a = array(3, 1, 4, 2, 8);$b = array(5, 2, 12, 8, 3);$n = sizeof($a); // Maximum possible value // of elements in both arrays.$N = 20; gcdMax($a, $b, $n, $N); // This code is contributed // by mits?> |
Output :
8 8
Time complexity : O(N Log N + n). Note that N + (N/2) + (N/3) + ….. + 1 = N log N.
Auxiliary Space : O(N)
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