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!