LintCode 1529: Triplet Subarray With Absolute Diff Less Than or Equal to Limit (同向双指针经典题)

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本文介绍了一种高效算法，用于在给定的整数数组中找出所有满足特定条件的三元子数组，即任意两个元素之间的绝对差小于等于预设限制。通过使用双指针技巧，该算法能够在O(n)的时间复杂度内解决问题。

1529. Triplet Subarray With Absolute Diff Less Than or Equal to Limit

Given an increasing array of integers nums and an integer limit, return the number of the triplet subarray in which the absolute difference between any two elements is less than or equal to limit.

In case there is no subarray satisfying the given condition return 0.

Example

Example 1:

Input：[1, 2, 3, 4], 3
Output：4
Explanation：We can choose (1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4). Therefore, the number of the triplet subarray is 4.

Example 2:

Input: [1, 10, 20, 30, 50], 19
Output：1
Explanation：The only triplet array is (1, 10, 20), so the answer is 1。

Challenge

Can you figure out a solution with time complexity of O(n)?

Notice

1 ≤ len(nums) ≤ 1e4，1 ≤ limit ≤ 1e6，0 ≤ nums[i] ≤ 1e6
Since the answer may be too large, return it modulo 99997867.

Input test data (one parameter per line)How to understand a testcase?

解法1：这题本来可以用DP，但DP可能要O(N^2)，而这里数据量太大，所以采用双指针。

思路，p2先行。当nums[p2]-nums[p1]>limit时，说明p1..p2这个range还不符合要求，所以p1必须++，当p1符合条件后跳出while循环，此时我们可以看出，p1, p1+1, ... p2这个range，p2固定，p1..p2-1里面任取2数都可以满足3数之中任意之差的绝对值<=limit，所以count+=((p2 - p1) * (p2 - p1 - 1) / 2)。这里采用组合C(x,2)的算法。

注意中间要取模。

class Solution {
public:
    /**
     * @param nums: the given array
     * @param limit: the limit on the absolute difference of the subarray
     * @return: Find the number of triplet subarray with the absolute difference less than or equal to limit
     */
    int tripletSubarray(vector<int> &nums, int limit) {
        int len = nums.size();
        long long count = 0;
        int p1 = 0, p2 = 0;
        
        for (int p2 = 2; p2 < len; ++p2) {
            while(nums[p2] - nums[p1] > limit) {
                p1++;
            }
            count += ((p2 - p1) * (p2 - p1 - 1) / 2) % 99997867;
        }
        return count;
    }
};