给定一个数组arr[] ,任务是找到数组中的最大比率对。
例子:
Input: arr[] = { 15, 10, 3 }
Output: 5
Explanation:
Maximum ratio pair will be –
Input: arr[] = { 15, 10, 3, 2 }
Output: 7.5
Explanation:
Maximum ratio pair will be –
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
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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