📜  在数组中查找具有最大比率的对

📅  最后修改于: 2021-09-08 12:38:45             🧑  作者: Mango

给定一个数组arr[] ,任务是找到数组中的最大比率对。
例子:

方法:这个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。对于任何对,可以使用以下方法获得最大比率\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


输出:
7.5

时间复杂度: O(N 2 )

辅助空间: O(1)

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