计算总和小于给定值的三元组

给定一个不同整数的数组和一个总和值。查找总和小于给定总和值的三元组计数。预期的时间复杂度为 O(n ² )。
例子：

Input : arr[] = {-2, 0, 1, 3}
        sum = 2.
Output : 2
Explanation :  Below are triplets with sum less than 2
               (-2, 0, 1) and (-2, 0, 3) 

Input : arr[] = {5, 1, 3, 4, 7}
        sum = 12.
Output : 4
Explanation :  Below are triplets with sum less than 12
               (1, 3, 4), (1, 3, 5), (1, 3, 7) and 
               (1, 4, 5)

一个简单的解决方案是运行三个循环以逐个考虑所有三元组。对于每个三元组，如果三元组总和小于给定总和，则比较总和并增加计数。

C++

// A Simple C++ program to count triplets with sum smaller
// than a given value
#include
using namespace std;
 
int countTriplets(int arr[], int n, int sum)
{
    // Initialize result
    int ans = 0;
 
    // Fix the first element as A[i]
    for (int i = 0; i < n-2; i++)
    {
       // Fix the second element as A[j]
       for (int j = i+1; j < n-1; j++)
       {
           // Now look for the third number
           for (int k = j+1; k < n; k++)
               if (arr[i] + arr[j] + arr[k] < sum)
                   ans++;
       }
    }
 
    return ans;
}
 
// Driver program
int main()
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    cout << countTriplets(arr, n, sum) << endl;
    return 0;
}

Java

// A Simple Java program to count triplets with sum smaller
// than a given value
 
class Test
{
    static int arr[] = new int[]{5, 1, 3, 4, 7};
     
    static int countTriplets(int n, int sum)
    {
        // Initialize result
        int ans = 0;
      
        // Fix the first element as A[i]
        for (int i = 0; i < n-2; i++)
        {
           // Fix the second element as A[j]
           for (int j = i+1; j < n-1; j++)
           {
               // Now look for the third number
               for (int k = j+1; k < n; k++)
                   if (arr[i] + arr[j] + arr[k] < sum)
                       ans++;
           }
        }
      
        return ans;
    }
     
    // Driver method to test the above function
    public static void main(String[] args)
    {
        int sum = 12;
        System.out.println(countTriplets(arr.length, sum));
    }
}

Python 3

# A Simple Python 3 program to count triplets with sum smaller
# than a given value
#include
def countTriplets(arr, n, sum):
 
    # Initialize result
    ans = 0
 
    # Fix the first element as A[i]
    for i in range( 0 ,n-2):
     
        # Fix the second element as A[j]
        for j in range( i+1 ,n-1):
     
            # Now look for the third number
            for k in range( j+1, n):
                if (arr[i] + arr[j] + arr[k] < sum):
                    ans+=1
     
    return ans
 
# Driver program
arr = [5, 1, 3, 4, 7]
n = len(arr)
sum = 12
print(countTriplets(arr, n, sum))
 
#Contributed by Smitha

C#

// A Simple C# program to count triplets with sum smaller
// than a given value
  
using System;
class Test
{
    static int[] arr = new int[]{5, 1, 3, 4, 7};
      
    static int countTriplets(int n, int sum)
    {
        // Initialize result
        int ans = 0;
       
        // Fix the first element as A[i]
        for (int i = 0; i < n-2; i++)
        {
           // Fix the second element as A[j]
           for (int j = i+1; j < n-1; j++)
           {
               // Now look for the third number
               for (int k = j+1; k < n; k++)
                   if (arr[i] + arr[j] + arr[k] < sum)
                       ans++;
           }
        }
       
        return ans;
    }
      
    // Driver method to test the above function
    public static void Main()
    {
        int sum = 12;
        Console.Write(countTriplets(arr.Length, sum));
    }
}

Javascript

// A Simple Javascript program to count triplets with sum smaller
// than a given value
    let arr = [5, 1, 3, 4, 7];
     
    function countTriplets(n,sum)
    {
     
        // Initialize result
        let ans = 0;
        
        // Fix the first element as A[i]
        for (let i = 0; i < n-2; i++)
        {
           // Fix the second element as A[j]
           for (let j = i + 1; j < n-1; j++)
           {
               // Now look for the third number
               for (let k = j + 1; k < n; k++)
                   if (arr[i] + arr[j] + arr[k] < sum)
                       ans++;
           }
        }
        return ans;
    }
     
    // Driver method to test the above function
    let sum = 12;
    document.write(countTriplets(arr.length, sum));
     
    // This code is contributed by avanitrachhadiya2155  

输出：

4

上述解决方案的时间复杂度为 O(n ³ )。一个有效的解决方案可以通过首先对数组进行排序，然后在循环中使用本文的方法 1 来计算 O(n ² ) 中的三元组。

1) Sort the input array in increasing order.
2) Initialize result as 0.
3) Run a loop from i = 0 to n-2.  An iteration of this loop finds all
   triplets with arr[i] as first element.
     a) Initialize other two elements as corner elements of subarray
        arr[i+1..n-1], i.e., j = i+1 and k = n-1
     b) Move j and k toward each other until they meet, i.e., while (j= sum
                then k--
            // Else for current i and j, there can (k-j) possible third elements
            // that satisfy the constraint.
            (ii) Else Do ans += (k - j) followed by j++ 

下面是上述思想的实现。

C++

// C++ program to count triplets with sum smaller than a given value
#include
using namespace std;
 
int countTriplets(int arr[], int n, int sum)
{
    // Sort input array
    sort(arr, arr+n);
 
    // Initialize result
    int ans = 0;
 
    // Every iteration of loop counts triplet with
    // first element as arr[i].
    for (int i = 0; i < n - 2; i++)
    {
        // Initialize other two elements as corner elements
        // of subarray arr[j+1..k]
        int j = i + 1, k = n - 1;
 
        // Use Meet in the Middle concept
        while (j < k)
        {
            // If sum of current triplet is more or equal,
            // move right corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum)
                k--;
 
            // Else move left corner
            else
            {
                // This is important. For current i and j, there
                // can be total k-j third elements.
                ans += (k - j);
                j++;
            }
        }
    }
    return ans;
}
 
// Driver program
int main()
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    cout << countTriplets(arr, n, sum) << endl;
    return 0;
}

Java

// A Simple Java program to count triplets with sum smaller
// than a given value
 
import java.util.Arrays;
 
class Test
{
    static int arr[] = new int[]{5, 1, 3, 4, 7};
     
    static int countTriplets(int n, int sum)
    {
        // Sort input array
        Arrays.sort(arr);
      
        // Initialize result
        int ans = 0;
      
        // Every iteration of loop counts triplet with
        // first element as arr[i].
        for (int i = 0; i < n - 2; i++)
        {
            // Initialize other two elements as corner elements
            // of subarray arr[j+1..k]
            int j = i + 1, k = n - 1;
      
            // Use Meet in the Middle concept
            while (j < k)
            {
                // If sum of current triplet is more or equal,
                // move right corner to look for smaller values
                if (arr[i] + arr[j] + arr[k] >= sum)
                    k--;
      
                // Else move left corner
                else
                {
                    // This is important. For current i and j, there
                    // can be total k-j third elements.
                    ans += (k - j);
                    j++;
                }
            }
        }
        return ans;
    }
     
    // Driver method to test the above function
    public static void main(String[] args)
    {
        int sum = 12;
        System.out.println(countTriplets(arr.length, sum));
    }
}

Python3

# Python3 program to count triplets with
# sum smaller than a given value
 
 
# Function to count triplets with sum smaller
# than a given value        
def countTriplets(arr,n,sum):
     
    # Sort input array
    arr.sort()
     
    # Initialize result
    ans = 0
     
    # Every iteration of loop counts triplet with
    # first element as arr[i].
    for i in range(0,n-2):
         
        # Initialize other two elements as corner elements
        # of subarray arr[j+1..k]
        j = i + 1
        k = n-1
 
        # Use Meet in the Middle concept
        while(j < k):
             
            # If sum of current triplet is more or equal,
            # move right corner to look for smaller values
            if (arr[i]+arr[j]+arr[k] >=sum):
                k = k-1
             
            # Else move left corner
            else:
                 
                # This is important. For current i and j, there
                # can be total k-j third elements.
                ans += (k - j)
                j = j+1
     
    return ans
 
# Driver program
if __name__=='__main__':
    arr = [5, 1, 3, 4, 7]
    n = len(arr)
    sum = 12
    print(countTriplets(arr, n, sum))
     
# This code is contributed by
# Yatin Gupta

C＃

// A Simple C# program to count
// triplets with sum smaller
// than a given value
using System;
 
class GFG
{
    static int []arr = new int[]{5, 1, 3, 4, 7};
     
    static int countTriplets(int n, int sum)
    {
        // Sort input array
        Array.Sort(arr);
     
        // Initialize result
        int ans = 0;
     
        // Every iteration of loop
        // counts triplet with
        // first element as arr[i].
        for (int i = 0; i < n - 2; i++)
        {
            // Initialize other two
            // elements as corner elements
            // of subarray arr[j+1..k]
            int j = i + 1, k = n - 1;
     
            // Use Meet in the Middle concept
            while (j < k)
            {
                // If sum of current triplet
                // is more or equal, move right
                // corner to look for smaller values
                if (arr[i] + arr[j] + arr[k] >= sum)
                    k--;
     
                // Else move left corner
                else
                {
                    // This is important. For
                    // current i and j, there
                    // can be total k-j third elements.
                    ans += (k - j);
                    j++;
                }
            }
        }
        return ans;
    }
     
    // Driver Code
    public static void Main()
    {
        int sum = 12;
        Console.Write(countTriplets(arr.Length, sum));
    }
}
 
// This code is contributed by Smitha

Javascript

// A Simple Javascript program to count triplets with sum smaller
// than a given value
    let arr = [5, 1, 3, 4, 7];
     
    function countTriplets(n,sum)
    {
     
        // Sort input array
        arr.sort(function(a,b){return b-a});
         
        // Initialize result
        let ans = 0;
         
        // Every iteration of loop counts triplet with
        // first element as arr[i].
        for (let i = 0; i < n - 2; i++)
        {
         
            // Initialize other two elements as corner elements
            // of subarray arr[j+1..k]
            let j = i + 1, k = n - 1;
       
            // Use Meet in the Middle concept
            while (j < k)
            {
                // If sum of current triplet is more or equal,
                // move right corner to look for smaller values
                if (arr[i] + arr[j] + arr[k] >= sum)
                    k--;
       
                // Else move left corner
                else
                {
                 
                    // This is important. For current i and j, there
                    // can be total k-j third elements.
                    ans += (k - j);
                    j++;
                }
            }
        }
        return ans;
    }
     
    // Driver method to test the above function
    let sum = 12;
    document.write(countTriplets(arr.length, sum));
     
    // This code is contributed by rag2127

输出：

4