📜  Javascript程序计算总和小于给定值的三元组

📅  最后修改于: 2022-05-13 01:57:48.463000             🧑  作者: Mango

Javascript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。

Javascript


Javascript


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。

Javascript


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!