Maximum value of division of two numbers in an Array

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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:

C++

// 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

// 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

#  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#

// 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

Javascript

<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 :

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

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