Java程序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。

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));
    }
}

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。

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));
    }
}

输出：

4

有关详细信息，请参阅有关总和小于给定值的计数三元组的完整文章！