Given an array A of size N (> 2). The task is to find the maximum value of A[i] / A[j]
Note: A[i] ? 0.
Examples:
Input : A[] = {1, 2, 3, 4}
Output : 4
4 / 1 = 4 is maximum possible value.Input : A[] = {3, 7, 9, 3, 11}
Output : 3
Naive Approach: A naive approach is to run nested loops and find the maximum possible answer.
Time complexity : O(N2).
Efficient Approach: An efficient approach is to find maximum and minimum element in the array and print maximum / minimum
Proof:
Lets take an array A[] = {a, b, c, d} where a < b < c < d .
Now :
(b / a) < ( c / a ) < (d / a) (since b < c < d, and denominator is constant )
and since a is minimum, therefore
d / a is maximum
Below is the implementation of the above approach:
// CPP program to maximum value of // division of two numbers in an array #include <bits/stdc++.h> using namespace std;
// Function to maximum value of // division of two numbers in an array int Division( int a[], int n)
{ int maxi = INT_MIN, mini = INT_MAX;
// Traverse through the array
for ( int i = 0; i < n; i++)
{
maxi = max(a[i], maxi);
mini = min(a[i], mini);
}
// Return the required answer
return maxi / mini;
} // Driver code int main()
{ int a[] = {3, 7, 9, 3, 11};
int n = sizeof (a) / sizeof (a[0]);
cout << Division(a, n);
return 0;
} |
// Java program to maximum value of // division of two numbers in an array import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{ // Function to maximum value of // division of two numbers in an array static int Division( int a[], int n)
{ int maxi = Integer.MIN_VALUE,
mini = Integer.MAX_VALUE;
// Traverse through the array
for ( int i = 0 ; i < n; i++)
{
maxi = Math.max(a[i], maxi);
mini = Math.min(a[i], mini);
}
// Return the required answer
return maxi / mini;
} // Driver code public static void main (String[] args)
throws java.lang.Exception
{ int a[] = { 3 , 7 , 9 , 3 , 11 };
int n = a.length;
System.out.print(Division(a, n));
} } // This code is contributed by Nidhiva |
# Python3 program to maximum value of # division of two numbers in an array # Function to maximum value of # division of two numbers in an array def Division(a, n):
maxi = - 10 * * 9
mini = 10 * * 9
# Traverse through the array
for i in a:
maxi = max (i, maxi)
mini = min (i, mini)
# Return the required answer
return maxi / / mini
# Driver code a = [ 3 , 7 , 9 , 3 , 11 ]
n = len (a)
print (Division(a, n))
# This code is contributed by Mohit Kumar |
// C# program to maximum value of // division of two numbers in an array using System;
class GFG
{ // Function to maximum value of // division of two numbers in an array static int Division( int []a, int n)
{ int maxi = int .MinValue,
mini = int .MaxValue;
// Traverse through the array
for ( int i = 0; i < n; i++)
{
maxi = Math.Max(a[i], maxi);
mini = Math.Min(a[i], mini);
}
// Return the required answer
return maxi / mini;
} // Driver code public static void Main (String[] args)
{ int []a = {3, 7, 9, 3, 11};
int n = a.Length;
Console.WriteLine(Division(a, n));
} } // This code is contributed by Rajput-Ji |
<script> // Javascript program to maximum value of // division of two numbers in an array // Function to maximum value of // division of two numbers in an array function Division(a, n)
{ let maxi = Number.MIN_VALUE, mini = Number.MAX_VALUE;
// Traverse through the array
for (let i = 0; i < n; i++)
{
maxi = Math.max(a[i], maxi);
mini = Math.min(a[i], mini);
}
// Return the required answer
return parseInt(maxi / mini);
} // Driver code let a = [3, 7, 9, 3, 11];
let n = a.length;
document.write(Division(a, n));
</script> |
Output :
3
Time complexity : O(N), since the loop runs only once from 0 to (n – 1).
Auxiliary space : O(1), since no extra space has been taken.
Note: Above solution works only for positive integers