Given an array arr[] of size N > 1. The task is to find the maximum possible GCD of the array by replacing at most one element.
Examples:
Input: arr[] = {6, 7, 8}
Output: 2
Replace 7 with 2 and gcd(6, 2, 8) = 2
which is maximum possible.
Input: arr[] = {12, 18, 30}
Output: 6
Approach:
- Idea is to find the GCD value of all the sub-sequences of length (N – 1) and removing the element which has to be replaced in the sub-sequence as it can be replaced with any other element already in the subsequence. The maximum GCD found would be the answer.
- To find the GCD of the sub-sequences optimally, maintain a prefixGCD[] and a suffixGCD[] array using single state dynamic programming.
- The maximum value of GCD(prefixGCD[i – 1], suffixGCD[i + 1]) is the required answer. Also note that suffixGCD[1] and prefixGCD[N – 2] also need to be compared in case the first or the last element has to be replaced.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum possible gcd after // replacing a single element int MaxGCD( int a[], int n)
{ // Prefix and Suffix arrays
int Prefix[n + 2];
int Suffix[n + 2];
// Single state dynamic programming relation for storing
// gcd of first i elements from the left in Prefix[i]
Prefix[1] = a[0];
for ( int i = 2; i <= n; i += 1)
Prefix[i] = __gcd(Prefix[i - 1], a[i - 1]);
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing gcd of all the elements having
// index greater than or equal to i in Suffix[i]
for ( int i = n - 1; i >= 1; i -= 1)
Suffix[i] = __gcd(Suffix[i + 1], a[i - 1]);
// If first or last element of the array has to be
// replaced
int ans = max(Suffix[2], Prefix[n - 1]);
// If any other element is replaced
for ( int i = 2; i < n; i += 1)
ans = max(ans, __gcd(Prefix[i - 1], Suffix[i + 1]));
// Return the maximized gcd
return ans;
} // Driver code int main()
{ int a[] = { 6, 7, 8 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d" , MaxGCD(a, n));
return 0;
} |
C
// C implementation of the approach #include <stdio.h> // Find maximum between two numbers. int max( int num1, int num2)
{ return (num1 > num2) ? num1 : num2;
} // Find minimum between two numbers. int min( int num1, int num2)
{ return (num1 > num2) ? num2 : num1;
} // Function to return gcd of a and b int gcd( int a, int b)
{ int result = min(a, b); // Finding minimum of a nd b
while (result > 0) {
if (a % result == 0 && b % result == 0) {
break ;
}
result--;
}
return result; // return gcd of a nd b
} // Function to return the maximum possible gcd after // replacing a single element int MaxGCD( int a[], int n)
{ // Prefix and Suffix arrays
int Prefix[n + 2];
int Suffix[n + 2];
// Single state dynamic programming relation for storing
// gcd of first i elements from the left in Prefix[i]
Prefix[1] = a[0];
for ( int i = 2; i <= n; i += 1)
Prefix[i] = gcd(Prefix[i - 1], a[i - 1]);
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing gcd of all the elements having
// index greater than or equal to i in Suffix[i]
for ( int i = n - 1; i >= 1; i -= 1)
Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
// If first or last element of the array has to be
// replaced
int ans = max(Suffix[2], Prefix[n - 1]);
// If any other element is replaced
for ( int i = 2; i < n; i += 1)
ans = max(ans, gcd(Prefix[i - 1], Suffix[i + 1]));
// Return the maximized gcd
return ans;
} // Driver code int main()
{ int a[] = { 6, 7, 8 };
int n = sizeof (a) / sizeof (a[0]);
printf ( "%d" , MaxGCD(a, n));
return 0;
} // This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of the approach class GFG
{ // Function to return the maximum // possible gcd after replacing // a single element static int MaxGCD( int a[], int n)
{ // Prefix and Suffix arrays
int []Prefix = new int [n + 2 ];
int []Suffix = new int [n + 2 ];
// Single state dynamic programming relation
// for storing gcd of first i elements
// from the left in Prefix[i]
Prefix[ 1 ] = a[ 0 ];
for ( int i = 2 ; i <= n; i += 1 )
{
Prefix[i] = __gcd(Prefix[i - 1 ],
a[i - 1 ]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1 ];
// Single state dynamic programming relation
// for storing gcd of all the elements having
// index greater than or equal to i in Suffix[i]
for ( int i = n - 1 ; i >= 1 ; i -= 1 )
{
Suffix[i] = __gcd(Suffix[i + 1 ],
a[i - 1 ]);
}
// If first or last element of
// the array has to be replaced
int ans = Math.max(Suffix[ 2 ], Prefix[n - 1 ]);
// If any other element is replaced
for ( int i = 2 ; i < n; i += 1 )
{
ans = Math.max(ans, __gcd(Prefix[i - 1 ],
Suffix[i + 1 ]));
}
// Return the maximized gcd
return ans;
} static int __gcd( int a, int b)
{ return b == 0 ? a : __gcd(b, a % b);
} // Driver code public static void main(String[] args)
{ int a[] = { 6 , 7 , 8 };
int n = a.length;
System.out.println(MaxGCD(a, n));
} } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach from math import gcd as __gcd
# Function to return the maximum # possible gcd after replacing # a single element def MaxGCD(a, n) :
# Prefix and Suffix arrays
Prefix = [ 0 ] * (n + 2 );
Suffix = [ 0 ] * (n + 2 );
# Single state dynamic programming relation
# for storing gcd of first i elements
# from the left in Prefix[i]
Prefix[ 1 ] = a[ 0 ];
for i in range ( 2 , n + 1 ) :
Prefix[i] = __gcd(Prefix[i - 1 ], a[i - 1 ]);
# Initializing Suffix array
Suffix[n] = a[n - 1 ];
# Single state dynamic programming relation
# for storing gcd of all the elements having
# index greater than or equal to i in Suffix[i]
for i in range (n - 1 , 0 , - 1 ) :
Suffix[i] = __gcd(Suffix[i + 1 ], a[i - 1 ]);
# If first or last element of
# the array has to be replaced
ans = max (Suffix[ 2 ], Prefix[n - 1 ]);
# If any other element is replaced
for i in range ( 2 , n) :
ans = max (ans, __gcd(Prefix[i - 1 ],
Suffix[i + 1 ]));
# Return the maximized gcd
return ans;
# Driver code if __name__ = = "__main__" :
a = [ 6 , 7 , 8 ];
n = len (a);
print (MaxGCD(a, n));
# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the maximum // possible gcd after replacing // a single element static int MaxGCD( int []a, int n)
{ // Prefix and Suffix arrays
int []Prefix = new int [n + 2];
int []Suffix = new int [n + 2];
// Single state dynamic programming relation
// for storing gcd of first i elements
// from the left in Prefix[i]
Prefix[1] = a[0];
for ( int i = 2; i <= n; i += 1)
{
Prefix[i] = __gcd(Prefix[i - 1],
a[i - 1]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing gcd of all the elements having
// index greater than or equal to i in Suffix[i]
for ( int i = n - 1; i >= 1; i -= 1)
{
Suffix[i] = __gcd(Suffix[i + 1],
a[i - 1]);
}
// If first or last element of
// the array has to be replaced
int ans = Math.Max(Suffix[2], Prefix[n - 1]);
// If any other element is replaced
for ( int i = 2; i < n; i += 1)
{
ans = Math.Max(ans, __gcd(Prefix[i - 1],
Suffix[i + 1]));
}
// Return the maximized gcd
return ans;
} static int __gcd( int a, int b)
{ return b == 0 ? a : __gcd(b, a % b);
} // Driver code public static void Main(String[] args)
{ int []a = { 6, 7, 8 };
int n = a.Length;
Console.WriteLine(MaxGCD(a, n));
} } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach function gcd(a, b) {
if (b==0) {
return a;
}
return gcd(b, a % b);
} // Function to return the maximum // possible gcd after replacing // a single element function MaxGCD(a, n)
{ // Prefix and Suffix arrays
var Prefix = Array(n + 2).fill(0);
var Suffix = Array(n + 2).fill(0);
var i;
// Single state dynamic programming relation
// for storing gcd of first i elements
// from the left in Prefix[i]
Prefix[1] = a[0];
for (i = 2; i <= n; i++) {
Prefix[i] = gcd(Prefix[i - 1], a[i - 1]);
}
// Initializing Suffix array
Suffix[n] = a[n - 1];
// Single state dynamic programming relation
// for storing gcd of all the elements having
// index greater than or equal to i in Suffix[i]
for (i = n - 1; i >= 1; i--) {
Suffix[i] = gcd(Suffix[i + 1], a[i - 1]);
}
// If first or last element of
// the array has to be replaced
var ans = Math.max(Suffix[2], Prefix[n - 1]);
// If any other element is replaced
for (i = 2; i < n; i++) {
ans = Math.max(ans, gcd(Prefix[i - 1],
Suffix[i + 1]));
}
// Return the maximized gcd
return ans;
} // Driver code var a = [6, 7, 8];
n = a.length;
document.write(MaxGCD(a, n));
</script> |
Output:
2
Time Complexity: O(n * max(a, b))
Auxiliary Space: O(n)