从两个数组中获得对使得第一个数组的元素小于第二个数组的元素的概率

给定两个数组arr1[]和arr2[] 分别由N和M 个整数组成，任务是找到分别从arr1[]和arr2[]中随机选择这两个数字的概率，使得第一个选择的元素严格小于比第二个选定的元素。

例子：

Input: arr1[] = {3, 2, 1, 1}, arr2[] = {1, 2, 5, 2}
Output: 0.5
Explanation:
Following are the ways of selecting the array elements from both the arrays first number is less than the second number:

1. Selecting arr1[0], there are 1 way of selecting an element in arr2[].
2. Selecting arr1[1], there are 1 way of selecting an element in arr2[].
3. Selecting arr1[2], there are 3 way of selecting an element in arr2[].
4. Selecting arr1[3], there are 3 way of selecting an element in arr2[]

Therefore, there are totals of (3 + 3 + 1 + 1 = 8) ways of selecting the elements from both arrays satisfying the conditions. Hence, the probability is (8/(4*4)) = 0.5.

Input: arr1[] = {5, 2, 6, 1}, arr2[] = {1, 6, 10, 1}
Output: 0.4375

朴素的方法：可以根据以下观察解决给定的问题：

• 这个想法是使用条件概率的概念。从数组arr1[]中选择一个元素的概率是1/N。
• 现在假设X是arr2[]中元素的计数大于arr1[]的选定元素，那么从arr2[]中选择一个这样的元素的概率是X/M 。
• 因此，选择两个元素使得第一个元素小于第二个所选元素的概率是arr1[] 中每个元素的(1/N)*(X/M) 之和。

请按照以下步骤解决问题：

• 初始化一个变量，比如res为0来存储结果概率。
• 遍历给定的数组arr1[]并执行以下步骤：
  • 通过遍历数组arr2[]找到arr2[] 中大于arr1[i]的元素计数，然后通过它增加res 。
  • 将res的值更新为res = res/N*M 。
• 完成上述步骤后，打印res的值作为结果概率。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find  probability
// such that x < y and X belongs to
// arr1[] and Y belongs to arr2[]
double probability(vector arr1,vector arr2)
{
    // Stores the length of arr1
    int N = arr1.size();
 
    // Stores the length of arr2
    int M = arr2.size();
 
    // Stores the result
    double res = 0;
 
    // Traverse the arr1[]
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // elements in arr2 that
        // are greater than arr[i]
        int y = 0;
 
        // Traverse the arr2[]
        for (int j = 0; j < M; j++) {
 
            // If arr2[j] is greater
            // than arr1[i]
            if (arr2[j] > arr1[i])
                y++;
        }
 
        // Increment res by y
        res += y;
    }
 
    // Update the value of res
    res = (double)res / (double)(N * M);
 
    // Return resultant probability
    return res;
}
 
// Driver Code
int main()
{
    vector arr1 = { 5, 2, 6, 1 };
    vector arr2 = { 1, 6, 10, 1 };
    cout<

---

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Function to find  probability
    // such that x < y and X belongs to
    // arr1[] and Y belongs to arr2[]
    static double probability(int[] arr1,
                              int[] arr2)
    {
        // Stores the length of arr1
        int N = arr1.length;
 
        // Stores the length of arr2
        int M = arr2.length;
 
        // Stores the result
        double res = 0;
 
        // Traverse the arr1[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count of
            // elements in arr2 that
            // are greater than arr[i]
            int y = 0;
 
            // Traverse the arr2[]
            for (int j = 0; j < M; j++) {
 
                // If arr2[j] is greater
                // than arr1[i]
                if (arr2[j] > arr1[i])
                    y++;
            }
 
            // Increment res by y
            res += y;
        }
 
        // Update the value of res
        res = (double)res / (double)(N * M);
 
        // Return resultant probability
        return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 5, 2, 6, 1 };
        int[] arr2 = { 1, 6, 10, 1 };
        System.out.println(
            probability(arr1, arr2));
    }
}

Python3

# Python 3 program for the above approach
 
# Function to find  probability
# such that x < y and X belongs to
# arr1[] and Y belongs to arr2[]
def probability(arr1, arr2):
 
    # Stores the length of arr1
    N = len(arr1)
 
    # Stores the length of arr2
    M = len(arr2)
 
    # Stores the result
    res = 0
 
    # Traverse the arr1[]
    for i in range(N):
 
        # Stores the count of
        # elements in arr2 that
        # are greater than arr[i]
        y = 0
 
        # Traverse the arr2[]
        for j in range(M):
 
            # If arr2[j] is greater
            # than arr1[i]
            if (arr2[j] > arr1[i]):
                y += 1
 
        # Increment res by y
        res += y
 
    # Update the value of res
    res = res / (N * M)
 
    # Return resultant probability
    return res
 
 
# Driver Code
if __name__ == "__main__":
 
    arr1 = [5, 2, 6, 1]
    arr2 = [1, 6, 10, 1]
    print(probability(arr1, arr2))
 
    # This code is contributed by ukasp.

C#

//C# program for the above approach
using System;
 
class GFG {
 
    // Function to find  probability
    // such that x < y and X belongs to
    // arr1[] and Y belongs to arr2[]
    static double probability(int[] arr1, int[] arr2)
    {
        // Stores the length of arr1
        int N = arr1.Length;
 
        // Stores the length of arr2
        int M = arr2.Length;
 
        // Stores the result
        double res = 0;
 
        // Traverse the arr1[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count of
            // elements in arr2 that
            // are greater than arr[i]
            int y = 0;
 
            // Traverse the arr2[]
            for (int j = 0; j < M; j++) {
 
                // If arr2[j] is greater
                // than arr1[i]
                if (arr2[j] > arr1[i])
                    y++;
            }
 
            // Increment res by y
            res += y;
        }
 
        // Update the value of res
        res = (double)res / (double)(N * M);
 
        // Return resultant probability
        return res;
    }
 
    // Driver Code
    static void Main()
    {
        int[] arr1 = { 5, 2, 6, 1 };
        int[] arr2 = { 1, 6, 10, 1 };
        Console.WriteLine(probability(arr1, arr2));
    }
}
 
 
// This code is contributed by SoumikMondal.

Javascript

 
// Javascript program for the above approach
 
    // Function to find  probability
    // such that x < y and X belongs to
    // arr1[] and Y belongs to arr2[]
    function probability(arr1, arr2)
    {
        // Stores the length of arr1
        let N = arr1.length;
 
        // Stores the length of arr2
        let M = arr2.length;
 
        // Stores the result
        let res = 0;
 
        // Traverse the arr1[]
        for (let i = 0; i < N; i++) {
 
            // Stores the count of
            // elements in arr2 that
            // are greater than arr[i]
            let y = 0;
 
            // Traverse the arr2[]
            for (let j = 0; j < M; j++) {
 
                // If arr2[j] is greater
                // than arr1[i]
                if (arr2[j] > arr1[i])
                    y++;
            }
 
            // Increment res by y
            res += y;
        }
 
        // Update the value of res
        res = (res / (N * M));
 
        // Return resultant probability
        return res;
    }
 
 
// Driver Code
 
     let arr1 = [ 5, 2, 6, 1 ];
     let arr2 = [ 1, 6, 10, 1 ];
     document.write(
            probability(arr1, arr2));
 

输出：

0.4375

时间复杂度： O(N * M)
辅助空间： O(1)

高效方法：上述方法可以通过使用二分搜索进行优化。请按照以下步骤解决问题：

• 初始化一个变量，比如res为0来存储结果概率。
• 按升序对数组进行排序。
• 遍历给定的数组arr1[]并执行以下步骤：
  • 通过使用二分搜索找到arr2[] 中大于arr1[i]的元素计数，然后通过它增加res 。
  • 将res的值更新为res = res/N*M。
• 完成上述步骤后，打印res作为结果概率。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
int countGreater(int* arr, int k);
 
// Function to find  probability
// such that x < y and X belongs
// to arr1[] & Y belongs to arr2[]
float probability(int* arr1,
                          int* arr2)
{
    // Stores the length of arr1
    int N = 4;
 
    // Stores the length of arr2
    int M = 4;
 
    // Stores the result
    float res = 0;
 
    // Sort the arr2[] in the
    // ascending order
    sort(arr2, arr2 + M);
     
    // Traverse the arr1[]
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // elements in arr2 that
        // are greater than arr[i]
        int y = countGreater(
            arr2, arr1[i]);
         
        // Increment res by y
        res += y;
    }
 
    // Update the resultant
    // probability
    res = res / (N * M);
 
    // Return the result
    return res;
}
 
// Function to return the count
// of elements from the array
// which are greater than k
int countGreater(int* arr,
                        int k)
{
    int n = 4;
    int l = 0;
    int r = n - 1;
 
    // Stores the index of the
    // leftmost element from the
    // array which is at least k
    int leftGreater = n;
 
    // Finds number of elements
    // greater than k
    while (l <= r) {
        int m = l + (r - l) / 2;
 
        // If mid element is at least
        // K, then update the value
        // of leftGreater and r
        if (arr[m] > k) {
 
            // Update leftGreater
            leftGreater = m;
 
            // Update r
            r = m - 1;
        }
 
        // If mid element is
        // at most K, then
        // update the value of l
        else
            l = m + 1;
    }
 
    // Return the count of
    // elements greater than k
    return (n - leftGreater);
}
 
// Driver Code
int main()
{
    int arr1[] = { 5, 2, 6, 1 };
    int arr2[] = { 1, 6, 10, 1 };
    cout << probability(arr1, arr2);
    return 0;
}
 
// This code is contributed by Shubhamsingh10

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Function to find  probability
    // such that x < y and X belongs
    // to arr1[] & Y belongs to arr2[]
    static double probability(int[] arr1,
                              int[] arr2)
    {
        // Stores the length of arr1
        int N = arr1.length;
 
        // Stores the length of arr2
        int M = arr2.length;
 
        // Stores the result
        double res = 0;
 
        // Sort the arr2[] in the
        // ascending order
        Arrays.sort(arr2);
 
        // Traverse the arr1[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count of
            // elements in arr2 that
            // are greater than arr[i]
            int y = countGreater(
                arr2, arr1[i]);
 
            // Increment res by y
            res += y;
        }
 
        // Update the resultant
        // probability
        res = (double)res / (double)(N * M);
 
        // Return the result
        return res;
    }
 
    // Function to return the count
    // of elements from the array
    // which are greater than k
    static int countGreater(int[] arr,
                            int k)
    {
        int n = arr.length;
        int l = 0;
        int r = n - 1;
 
        // Stores the index of the
        // leftmost element from the
        // array which is at least k
        int leftGreater = n;
 
        // Finds number of elements
        // greater than k
        while (l <= r) {
            int m = l + (r - l) / 2;
 
            // If mid element is at least
            // K, then update the value
            // of leftGreater and r
            if (arr[m] > k) {
 
                // Update leftGreater
                leftGreater = m;
 
                // Update r
                r = m - 1;
            }
 
            // If mid element is
            // at most K, then
            // update the value of l
            else
                l = m + 1;
        }
 
        // Return the count of
        // elements greater than k
        return (n - leftGreater);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr1 = { 5, 2, 6, 1 };
        int[] arr2 = { 1, 6, 10, 1 };
        System.out.println(
            probability(arr1, arr2));
    }
}

蟒蛇3

# Python3 program for the above approach
 
# Function to find  probability
# such that x < y and X belongs
# to arr1[] & Y belongs to arr2[]
def probability(arr1, arr2):
   
    # Stores the length of arr1
    n = len(arr1)
     
    # Stores the length of arr2
    m = len(arr2)
     
    # Stores the result
    res = 0
     
    # Sort the arr2[] in the
    # ascending order
    arr2.sort()
     
    # Traverse the arr1[]
    for i in range(n):
         
        # Stores the count of
        # elements in arr2 that
        # are greater than arr[i]
        y = countGreater(arr2, arr1[i])
         
        # Increment res by y
        res += y
         
    # Update the resultant
    # probability
    res /= (n * m)
     
    # Return the result
    return res
 
# Function to return the count
# of elements from the array
# which are greater than k
def countGreater(arr, k):
 
    n = len(arr)
    l = 0
    r = n - 1
     
    # Stores the index of the
    # leftmost element from the
    # array which is at least k
    leftGreater = n
     
    # Finds number of elements
    # greater than k
    while l <= r:
        m = (l + r) // 2
         
        # If mid element is at least
        # K, then update the value
        # of leftGreater and r
        if (arr[m] > k):
             
            # Update leftGreater
            leftGreater = m
             
            # Update r
            r = m - 1
             
        # If mid element is
        # at most K, then
        # update the value of l
        else:
            l = m + 1
             
    # Return the count of
    # elements greater than k
    return n - leftGreater
 
# Driver code
if __name__ == '__main__':
     
    arr1 = [ 5, 2, 6, 1 ]
    arr2 = [ 1, 6, 10, 1 ]
     
    print(probability(arr1, arr2))
 
# This code is contributed by MuskanKalra1

C＃

// C# program for the above approach
using System;
class GFG {
 
   // Function to find  probability
    // such that x < y and X belongs
    // to arr1[] & Y belongs to arr2[]
    static double probability(int[] arr1,
                              int[] arr2)
    {
       
        // Stores the length of arr1
        int N = arr1.Length;
 
        // Stores the length of arr2
        int M = arr2.Length;
 
        // Stores the result
        double res = 0;
 
        // Sort the arr2[] in the
        // ascending order
        Array.Sort(arr2);
 
        // Traverse the arr1[]
        for (int i = 0; i < N; i++) {
 
            // Stores the count of
            // elements in arr2 that
            // are greater than arr[i]
            int y = countGreater(
                arr2, arr1[i]);
 
            // Increment res by y
            res += y;
        }
 
        // Update the resultant
        // probability
        res = (double)res / (double)(N * M);
 
        // Return the result
        return res;
    }
 
    // Function to return the count
    // of elements from the array
    // which are greater than k
    static int countGreater(int[] arr,
                            int k)
    {
        int n = arr.Length;
        int l = 0;
        int r = n - 1;
 
        // Stores the index of the
        // leftmost element from the
        // array which is at least k
        int leftGreater = n;
 
        // Finds number of elements
        // greater than k
        while (l <= r) {
            int m = l + (r - l) / 2;
 
            // If mid element is at least
            // K, then update the value
            // of leftGreater and r
            if (arr[m] > k) {
 
                // Update leftGreater
                leftGreater = m;
 
                // Update r
                r = m - 1;
            }
 
            // If mid element is
            // at most K, then
            // update the value of l
            else
                l = m + 1;
        }
 
        // Return the count of
        // elements greater than k
        return (n - leftGreater);
    }
   
    // Driver code
    public static void Main()
    {
       int[] arr1 = { 5, 2, 6, 1 };
        int[] arr2 = { 1, 6, 10, 1 };
        Console.Write(
            probability(arr1, arr2));
    }
}
 
// This code is contributed by sanjoy_62.

Javascript

// Javascript program for the above approach
 
// Function to find  probability
// such that x < y and X belongs
// to arr1[] & Y belongs to arr2[]
function probability(arr1, arr2)
{
    // Stores the length of arr1
    var N = 4;
 
    // Stores the length of arr2
    var M = 4;
 
    // Stores the result
    var res = 0;
 
    // Sort the arr2[] in the
    // ascending order
    arr2.sort(function(a, b) {
        return a - b;
    });
     
    // Traverse the arr1[]
    for (var i = 0; i < N; i++) {
 
        // Stores the count of
        // elements in arr2 that
        // are greater than arr[i]
        var y = countGreater( arr2, arr1[i]);
             
        // Increment res by y
        res += y;
    }
 
    // Update the resultant
    // probability
    res = res / (N * M);
     
    // Return the result
    return res;
}
 
// Function to return the count
// of elements from the array
// which are greater than k
function countGreater(arr, k)
{
    var n = 4;
    var l = 0;
    var r = n - 1;
 
    // Stores the index of the
    // leftmost element from the
    // array which is at least k
    var leftGreater = n;
 
    // Finds number of elements
    // greater than k
    while (l <= r) {
        var m = Math.floor(l + (r - l) / 2);
 
        // If mid element is at least
        // K, then update the value
        // of leftGreater and r
        if (arr[m] > k) {
 
            // Update leftGreater
            leftGreater = m;
 
            // Update r
            r = m - 1;
             
        }
 
        // If mid element is
        // at most K, then
        // update the value of l
        else
            l = m + 1;
        
    }
 
    // Return the count of
    // elements greater than k
    return n - leftGreater;
}
 
// Driver Code
var arr1 = [ 5, 2, 6, 1 ];
var arr2 = [ 1, 6, 10, 1 ];
document.write(probability(arr1, arr2));
 
// This code is contributed by Shubhamsingh10

输出：

0.4375

时间复杂度： O(N * log M)
辅助空间： O(1)

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