数组 (i, j, k) 中三元组的计数，使得 i < j < k 和 a[k] < a[i] < a[j]

给定一个由N 个整数组成的数组arr[] ，任务是计算数组中三元组(i, j, k)的数量，使得a[k] < a[i] < a[j]且i < j < k .
例子：

Input: arr[] = {2, 5, 1, 3, 0} 
Output: 4 
Explanation: 
Below are the triplets (i, j, k) such that i < j < k and a[k] < a[i] < a[j]: 
1. (0, 1, 2) and arr[2] < arr[0] 1 < 2 < 5.
2. (0, 1, 4) and arr[4] < arr[0] 0 < 2 < 5.
3. (0, 3, 4) and arr[4] < arr[0] 0 < 2 < 3.
4. (2, 3, 4) and arr[4] < arr[2] 0 < 1 < 3.
Input: arr[] = {2, 5, 1, 2, 0, 3, 10, 1, 5, 0 } 
Output: 25 
 

朴素的方法：这个想法是迭代 3 个循环并检查每个三元组(i, j, k) 是否满足给定条件。如果是，则增加该三元组并在检查所有三元组后打印最终计数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count triplets with the
// given conditions
int CountTriplets(int arr[], int n)
{
    int cnt = 0;
 
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            for (int k = j + 1; k < n; k++)
 
                // If it satisfy the
                // given conditions
                if (arr[k] < arr[i]
                    && arr[i] < arr[j]) {
 
                    cnt += 1;
                }
 
    // Return the final count
    return cnt;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 5, 1, 3, 0 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << CountTriplets(arr, n)
         << endl;
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to count triplets
// with the given conditions
static int CountTriplets(int arr[], int n)
{
    int cnt = 0;
 
    for(int i = 0; i < n; i++)
       for(int j = i + 1; j < n; j++)
          for(int k = j + 1; k < n; k++)
              
             // If it satisfy the
             // given conditions
             if (arr[k] < arr[i] &&
                 arr[i] < arr[j])
             {
                 cnt += 1;
             }
              
    // Return the final count
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = new int[]{ 2, 5, 1, 3, 0 };
 
    int n = arr.length;
     
    System.out.print(CountTriplets(arr, n));
}
}
 
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
 
# Function to count triplets with the
# given conditions
def CountTriplets(arr, n):
 
    cnt = 0;
 
    for i in range(0, n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
 
                # If it satisfy the
                # given conditions
                if (arr[k] < arr[i] and arr[i] < arr[j]):
                    cnt += 1;
                 
    # Return the final count
    return cnt;
 
# Driver Code
 
# Given array arr[]
arr = [ 2, 5, 1, 3, 0 ];
 
n = len(arr);
 
# Function Call
print(CountTriplets(arr, n))
 
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to count triplets
// with the given conditions
static int CountTriplets(int []arr, int n)
{
    int cnt = 0;
 
    for(int i = 0; i < n; i++)
       for(int j = i + 1; j < n; j++)
          for(int k = j + 1; k < n; k++)
              
             // If it satisfy the
             // given conditions
             if (arr[k] < arr[i] &&
                 arr[i] < arr[j])
             {
                 cnt += 1;
             }
             
    // Return the final count
    return cnt;
}
 
// Driver Code
public static void Main(string[] args)
{
     
    // Given array arr[]
    int []arr = new int[]{ 2, 5, 1, 3, 0 };
 
    int n = arr.Length;
     
    Console.Write(CountTriplets(arr, n));
}
}
 
// This code is contributed by Ritik Bansal

Javascript

 
    // Javascript program for the above approach
     
    // Function to count triplets with the
    // given conditions
    function CountTriplets(arr, n)
    {
        let cnt = 0;
 
        for (let i = 0; i < n; i++)
            for (let j = i + 1; j < n; j++)
                for (let k = j + 1; k < n; k++)
 
                    // If it satisfy the
                    // given conditions
                    if (arr[k] < arr[i]
                        && arr[i] < arr[j]) {
 
                        cnt += 1;
                    }
 
        // Return the final count
        return cnt;
    }
     
    // Given array arr[]
    let arr = [ 2, 5, 1, 3, 0 ];
  
    let n = arr.length;
  
    // Function Call
    document.write(CountTriplets(arr, n));
     

输出：

4

时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：我们可以使用以下步骤将复杂度从 N^3 降低到 N^2：

1. 运行两个循环以找到对(i, j)使得i < j和arr[j] > arr[i]并将这些对的计数保持为cnt 。
2. 在上面的循环中，如果存在任何元素，例如arr[j] < arr[i]然后通过cnt增加三元组的计数，因为当前元素是第K 个元素，使得a[k] < a[i] < a[ j]对于三元组i < j < k 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to count  triplets
int CountTriplets(int a[], int n)
{
 
    // To store count of total triplets
    int ans = 0;
 
    for (int i = 0; i < n; i++) {
 
        // Initialize count to zero
        int cnt = 0;
 
        for (int j = i + 1; j < n; j++) {
 
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
 
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 5, 1, 3, 0 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << CountTriplets(arr, n) << endl;
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to count triplets
static int CountTriplets(int a[], int n)
{
 
    // To store count of total triplets
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // Initialize count to zero
        int cnt = 0;
 
        for (int j = i + 1; j < n; j++)
        {
 
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
 
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 5, 1, 3, 0 };
 
    int n = arr.length;
 
    System.out.print(CountTriplets(arr, n));
}
}
 
// This code is contributed by shivanisinghss2110

蟒蛇3

# Python3 program for
# the above approach
 
# Function to count  triplets
def CountTriplets(a, n):
 
    # To store count
    # of total triplets
    ans = 0
 
    for i in range (n):
 
        # Initialize count to zero
        cnt = 0
 
        for j in range (i + 1 , n):
 
            # If a[j] > a[i] then,
            # increment cnt
            if (a[j] > a[i]):
                cnt += 1
 
            # If a[j] < a[i], then
            # it mean we have found a[k]
            # such that a[k] < a[i] < a[j]
            else:
                ans += cnt
      
    # Return the final count
    return ans
 
# Driver code
if __name__ == "__main__": 
    arr = [2, 5, 1, 3, 0]
    n = len(arr)
    print (CountTriplets(arr, n))
 
# This code is contributed by Chitranayal

C＃

// C# program for the above approach
using System;
class GFG{
 
// Function to count triplets
static int CountTriplets(int []a, int n)
{
 
    // To store count of total triplets
    int ans = 0;
 
    for (int i = 0; i < n; i++)
    {
 
        // Initialize count to zero
        int cnt = 0;
 
        for (int j = i + 1; j < n; j++)
        {
 
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
 
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver code
public static void Main()
{
    int []arr = { 2, 5, 1, 3, 0 };
 
    int n = arr.Length;
 
    Console.Write(CountTriplets(arr, n));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program for the above approach
 
// Function to count  triplets
function CountTriplets(a, n)
{
     
    // To store count of total triplets
    let ans = 0;
 
    for(let i = 0; i < n; i++)
    {
         
        // Initialize count to zero
        let cnt = 0;
 
        for(let j = i + 1; j < n; j++)
        {
             
            // If a[j] > a[i] then,
            // increment cnt
            if (a[j] > a[i])
                cnt++;
 
            // If a[j] < a[i], then
            // it mean we have found a[k]
            // such that a[k] < a[i] < a[j]
            else
                ans += cnt;
        }
    }
 
    // Return the final count
    return ans;
}
 
// Driver code
let arr = [ 2, 5, 1, 3, 0 ];
let n = arr.length;
 
document.write(CountTriplets(arr, n));
 
// This code is contributed by rameshtravel07
 

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)

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