在数组中查找具有最大比率的对

给定数组arr [] ，任务是在数组中找到最大比率对。
例子：

Input: arr[] = { 15, 10, 3 } 
Output: 5 
Explanation: 
Maximum ratio pair will be – $\frac{15}{3} = 5$
Input: arr[] = { 15, 10, 3, 2 } 
Output: 7.5 
Explanation: 
Maximum ratio pair will be – $\frac{15}{2} = 7.5$

方法：想法是使用两个嵌套循环遍历数组的每个可能对，并找到可能的最大比率对。对于任何一对，可以使用$\frac{max(arr_i, arr_j)}{min(arr_i, arr_j)}$
下面是上述方法的实现：

C++

// C++ implementation to find
// the maximum pair in the array
 
#include 
 
using namespace std;
 
// Function to find the maximum pair
// possible for the array
float computeMaxValue(float arr[], int n)
{
    float ans = 0;
 
    // Loop to iterate over every
    // possible pair in the array
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Check pair (x, y) as well as
            // (y, x) for maximum value
            float val = max(arr[i] / arr[j],
                            arr[j] / arr[i]);
 
            // Update the answer
            ans = max(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    float arr[] = { 15, 10, 3, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << computeMaxValue(arr, n);
    return 0;
}

Java

// Java implementation to find
// the maximum pair in the array
import java.io.*;
import java.util.*;
 
class GFG {
     
// Function to find the maximum pair
// possible for the array
static float computeMaxValue(float arr[], int n)
{
    float ans = 0;
     
    // Loop to iterate over every
    // possible pair in the array
    for(int i = 0; i < n - 1; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
           
          // Check pair (x, y) as well as
          // (y, x) for maximum value
          float val = Math.max(arr[i] / arr[j],
                               arr[j] / arr[i]);
 
          // Update the answer
          ans = Math.max(ans, val);
       }
    }
     
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    float arr[] = { 15, 10, 3, 2 };
    int N = arr.length;
     
    System.out.println(computeMaxValue(arr, N));
}
}
 
// This code is contributed by coder001

Python3

# Python3 implementation to find
# the maximum pair in the array
 
# Function to find the maximum pair
# possible for the array
def computeMaxValue(arr, n):
 
    ans = 0
 
    # Loop to iterate over every
    # possible pair in the array
    for i in range(n - 1):
        for j in range(i + 1, n):
 
            # Check pair (x, y) as well as
            # (y, x) for maximum value
            val = max(arr[i] / arr[j],
                      arr[j] / arr[i])
 
            # Update the answer
            ans = max(ans, val)
             
    return ans
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 15, 10, 3, 2 ]
    n = len(arr)
 
    print(computeMaxValue(arr, n))
     
# This code is contributed by chitranayal

C#

// C# implementation to find
// the maximum pair in the array
using System;
 
class GFG {
     
// Function to find the maximum pair
// possible for the array
static float computeMaxValue(float []arr, int n)
{
    float ans = 0;
     
    // Loop to iterate over every
    // possible pair in the array
    for(int i = 0; i < n - 1; i++)
    {
       for(int j = i + 1; j < n; j++)
       {
           
          // Check pair (x, y) as well as
          // (y, x) for maximum value
          float val = Math.Max(arr[i] / arr[j],
                               arr[j] / arr[i]);
           
          // Update the answer
          ans = Math.Max(ans, val);
       }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    float []arr = { 15, 10, 3, 2 };
    int N = arr.Length;
     
    Console.WriteLine(computeMaxValue(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation to find
// the maximum pair in the array
 
// Function to find the maximum pair
// possible for the array
function computeMaxValue(arr, n)
{
    var ans = 0;
 
    // Loop to iterate over every
    // possible pair in the array
    for (var i = 0; i < n - 1; i++) {
        for (var j = i + 1; j < n; j++) {
 
            // Check pair (x, y) as well as
            // (y, x) for maximum value
            var val = Math.max(arr[i] / arr[j],
                            arr[j] / arr[i]);
 
            // Update the answer
            ans = Math.max(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
var arr = [ 15, 10, 3, 2 ];
var n = arr.length;
document.write( computeMaxValue(arr, n));
 

输出：

7.5

时间复杂度： O(N ² )

辅助空间： O(1)