计算数组中的三元组，使得 i <j<k and a[j] – a[i] = a[k] – a[j] = D

给定一个数组arr和一个整数D ，任务是计算数组中三元组(arr[i], arr[j], arr[k]) 的数量，使得：

1. i < j < k
2. arr[j] – arr[i] = arr[k] – arr[j] = D

例子：

Input: arr = {1, 6, 7, 7, 8, 10, 12, 13, 14}, D = 3 
Output: 2 
Explanation: 
There are two triplets in the array that satisfies the given criteria. 
-> 1st triplet(7, 10, 13) such that i = 3, j = 6 and k = 8, such that (i < j < k) and (10 – 7 = 13 – 10 = D (=3)) 
-> 2nd triplet(7, 10, 13) such that i = 4, j = 6 and k = 8, such that (i < j < k) and (10 – 7 = 13 – 10 = D (=3))

Input: arr = {6, 3, 4, 5}, D = 1 
Output: 0 
 

朴素的方法：最简单的方法是使用三个嵌套的 for 循环并找到满足给定标准的三元组。下面是这个方法的实现。

C++

// C++ program to count the triplets
#include
using namespace std;
 
// Function to count the triplets
int CountTriplets(int arr[], int d, int n)
{
    int count = 0;
 
    // Three nested for loops to count the
    // triplets that satisfy the given criteria
    for(int i = 0; i < n - 2; i++)
    {
       for(int j = i + 1; j < n - 1; j++)
       {
          for(int k = j + 1; k < n; k++)
          {
             if ((arr[j] - arr[i] == d) &&
                 (arr[k] - arr[j] == d))
             {
                 count++;
             }
          }
       }
    }
    return count;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
    int D = 3;
    int n = sizeof(A) / sizeof(A[0]);
     
    cout << (CountTriplets(A, D, n));
}
 
// This code is contributed by chitranayal

Java

// Java program to count the triplets
 
class GFG {
 
    // Function to count the triplets
    static int CountTriplets(int[] arr, int d)
    {
        int count = 0;
        int n = arr.length;
 
        // Three nested for loops to count the
        // triplets that satisfy the given criteria
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
 
                    if ((arr[j] - arr[i] == d)
                        && (arr[k] - arr[j] == d)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int A[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int D = 3;
        System.out.println(CountTriplets(A, D));
    }
}

Python3

# Python3 program to count the triplets
 
# Function to count the triplets
def CountTriplets(arr, d):
     
    count = 0;
    n = len(arr);
 
    # Three nested for loops to count the
    # triplets that satisfy the given criteria
    for i in range(n - 1):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
 
                if ((arr[j] - arr[i] == d) and
                    (arr[k] - arr[j] == d)):
                    count += 1;
    return count;
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 1, 6, 7, 7, 8, 10, 12, 13, 14 ];
    D = 3;
     
    print(CountTriplets(A, D));
 
# This code is contributed by Rajput-Ji

C#

// C# program to count the triplets
using System;
 
class GFG {
  
    // Function to count the triplets
    static int CountTriplets(int[] arr, int d)
    {
        int count = 0;
        int n = arr.Length;
  
        // Three nested for loops to count the
        // triplets that satisfy the given criteria
        for (int i = 0; i < n - 2; i++) {
            for (int j = i + 1; j < n - 1; j++) {
                for (int k = j + 1; k < n; k++) {
  
                    if ((arr[j] - arr[i] == d)
                        && (arr[k] - arr[j] == d)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
  
    // Driver Code
    public static void Main(String []args)
    {
        int []A = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int D = 3;
        Console.WriteLine(CountTriplets(A, D));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript program to count the triplets
 
// Function to count the triplets
    function CountTriplets(arr, d)
    {
        let count = 0;
        let n = arr.length;
  
        // Three nested for loops to count the
        // triplets that satisfy the given criteria
        for (let i = 0; i < n - 2; i++) {
            for (let j = i + 1; j < n - 1; j++) {
                for (let k = j + 1; k < n; k++) {
  
                    if ((arr[j] - arr[i] == d)
                        && (arr[k] - arr[j] == d)) {
                        count++;
                    }
                }
            }
        }
        return count;
    }
  
// Driver code
     
      let A = [ 1, 6, 7, 7, 8, 10, 12, 13, 14 ];
        let D = 3;
        document.write(CountTriplets(A, D));
                                                                                       

输出：

2

时间复杂度： O(N^3)。

有效的方法：

• 解决此问题的一种有效方法是使用映射来存储 (key, values) 对，其中 value 将是 key 的计数。
• 这个想法是从0到N遍历数组并执行以下操作：
  • 检查A[i] – D和A[i] – 2*D是否存在于地图中。
  • 如果它在地图中，那么我们将简单地乘以它们各自的值并以此增加答案。
  • 现在，我们只需要将 A[i] 放入地图并更新地图。

下面是上述方法的实现。

C++14

// C++14 program to count the number
// of triplets from an array.
#include
using namespace std;
 
// Function to count the triplets
int countTriplets (int arr[], int d, int n)
{
    int count = -1;
     
    // Create a map to store (key, values) pair.
    map set;
 
    // Traverse the array and check that we
    // have already put (a[i]-d and a[i]-2*d)
    // into map or not. If yes we have to get
    // the values of both the keys and
    // multiply them, else put a[i] into the map.
    for(int i = 0; i < n; i++)
    {
        if ((set.find(arr[i] - d) != set.end()) &&
            (set.find(arr[i] - 2 * d) != set.end()))
        {
            count += (set[arr[i] - d] *
                      set[arr[i] - 2 * d]);
        }
 
        // Update the map
        if (set.find(arr[i]) == set.end())
            set[arr[i]] = 1;
        else
            set[arr[i]] += 1;
    }
    return count;
}
 
// Driver Code
int main()
{
     
    // Test Case 1:
    int a1[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
    int d1 = 3;
    cout << countTriplets(a1, d1, 9) << endl;
 
    // Test Case 2:
    int a2[] = { 6, 3, 4, 5 };
    int d2 = 1;
    cout << countTriplets(a2, d2, 4) << endl;
 
    // Test Case 3:
    int a3[] = { 1, 2, 4, 5, 7, 8, 10 };
    int d3 = 3;
    cout << countTriplets(a3, d3, 7) << endl;
 
    return 0;
}
 
// This code is contributed by himanshu77

Java

// Java program to count the number
// of triplets from an array.
 
import java.util.*;
 
class GFG {
 
    // Function to count the triplets
    static int countTriplets(int[] arr, int d)
    {
        int count = -1;
 
        // Create a map to store (key, values) pair.
        Map set = new HashMap<>();
 
        // Traverse the array and check that we
        // have already put (a[i]-d and a[i]-2*d)
        // into map or not. If yes we have to get
        // the values of both the keys and
        // multiply them, else put a[i] into the map.
        for (int i = 0; i < arr.length; i++) {
 
            if (set.get(arr[i] - d) != null
                && set.get(arr[i] - 2 * d) != null) {
                count += (set.get(arr[i] - d)
                          * set.get(arr[i] - 2 * d));
            }
 
            // Update the map.
            if (set.get(arr[i]) == null) {
                set.put(arr[i], 1);
            }
            else {
                set.put(arr[i], set.get(arr[i]) + 1);
            }
        }
 
        return count;
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        // Test Case 1:
        int a1[] = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int d1 = 3;
        System.out.println(countTriplets(a1, d1));
 
        // Test Case 2:
        int a2[] = { 6, 3, 4, 5 };
        int d2 = 1;
        System.out.println(countTriplets(a2, d2));
 
        // Test Case 3:
        int a3[] = { 1, 2, 4, 5, 7, 8, 10 };
        int d3 = 3;
        System.out.println(countTriplets(a3, d3));
    }
}

蟒蛇3

# Python3 program to count the number
# of triplets from an array.
 
# Function to count the triplets
def countTriplets (arr, d, n):
     
    count = -1
     
    # Create a map to store (key, values) pair.
    set = {}
     
    # Traverse the array and check that we
    # have already put (a[i]-d and a[i]-2*d)
    # into map or not. If yes we have to get
    # the values of both the keys and
    # multiply them, else put a[i] into the map.
    for i in range (0, n):
        if ((arr[i] - d) in set.keys() and
            (arr[i] - 2 * d) in set.keys()):
            count += (set[arr[i] - d] *
                      set[arr[i] - 2 * d])
                       
        # Update the map
        if (arr[i]) in set.keys():
            set[arr[i]] += 1
        else:
            set[arr[i]] = 1
             
    print(count)
     
# Driver Code 
 
# Test Case 1:
a1 = [ 1, 6, 7, 7, 8, 10, 12, 13, 14 ]
d1 = 3
countTriplets(a1, d1, 9)
 
# Test Case 2:
a2 = [ 6, 3, 4, 5 ]
d2 = 1
countTriplets(a2, d2, 4)
 
# Test Case 3:
a3 = [ 1, 2, 4, 5, 7, 8, 10 ]
d3 = 3
countTriplets(a3, d3, 7)
 
# This code is contributed by Stream_Cipher

C＃

// C# program to count the number
// of triplets from an array.
using System;
using System.Collections.Generic;
 
class GFG {
  
    // Function to count the triplets
    static int countTriplets(int[] arr, int d)
    {
        int count = -1;
  
        // Create a map to store (key, values) pair.
        Dictionary set = new Dictionary();
  
        // Traverse the array and check that we
        // have already put (a[i]-d and a[i]-2*d)
        // into map or not. If yes we have to get
        // the values of both the keys and
        // multiply them, else put a[i] into the map.
        for (int i = 0; i < arr.Length; i++) {
  
            if (set.ContainsKey(arr[i] - d)
                && set.ContainsKey(arr[i] - 2 * d)) {
                count += (set[arr[i] - d]
                          * set[arr[i] - 2 * d]);
            }
  
            // Update the map.
            if (!set.ContainsKey(arr[i])) {
                set.Add(arr[i], 1);
            }
            else {
                set[arr[i]] = set[arr[i]] + 1;
            }
        }
  
        return count;
    }
  
    // Driver Code
    public static void Main(String []args)
    {
  
        // Test Case 1:
        int []a1 = { 1, 6, 7, 7, 8, 10, 12, 13, 14 };
        int d1 = 3;
        Console.WriteLine(countTriplets(a1, d1));
  
        // Test Case 2:
        int []a2 = { 6, 3, 4, 5 };
        int d2 = 1;
        Console.WriteLine(countTriplets(a2, d2));
  
        // Test Case 3:
        int []a3 = { 1, 2, 4, 5, 7, 8, 10 };
        int d3 = 3;
        Console.WriteLine(countTriplets(a3, d3));
    }
}
 
// This code is contributed by Princi Singh

Javascript

// Javascript program to count the number
// of triplets from an array.
 
// Function to count the triplets
    function countTriplets(arr,d)
    {
        let count = -1;
  
        // Create a map to store (key, values) pair.
        let set = new Map();
  
        // Traverse the array and check that we
        // have already put (a[i]-d and a[i]-2*d)
        // into map or not. If yes we have to get
        // the values of both the keys and
        // multiply them, else put a[i] into the map.
        for (let i = 0; i < arr.length; i++) {
  
            if (set.get(arr[i] - d) != null
                && set.get(arr[i] - 2 * d) != null) {
                count += (set.get(arr[i] - d)
                          * set.get(arr[i] - 2 * d));
            }
  
            // Update the map.
            if (set.get(arr[i]) == null) {
                set.set(arr[i], 1);
            }
            else {
                set.set(arr[i], set.get(arr[i]) + 1);
            }
        }
  
        return count;
    }
     
      // Driver Code
     
    // Test Case 1:
    let a1=[1, 6, 7, 7, 8, 10, 12, 13, 14];
    let d1 = 3;
    document.write(countTriplets(a1, d1)+"<br>");
 
    // Test Case 2:
    let a2=[6, 3, 4, 5 ];
    let d2 = 1;
    document.write(countTriplets(a2, d2)+"<br>");
     
    // Test Case 3:
    let a3=[1, 2, 4, 5, 7, 8, 10];
    let d3 = 3;
    document.write(countTriplets(a3, d3)+"<br>");
     
// This code is contributed by avanitrachhadiya2155

输出：

1
0
2

时间复杂度： O(NlogN)

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