Skip to main content

Sliding Window

Sliding window is one common trick used in coding, which may not be counted as one "algorithm" by itself. Nevertheless, this trick is popular in tech code interviews, perhaps due to its relative "short" length (?).

The size of the window could be either fixed or non-fixed, depends on the questions.

When the size is non-fixed, two pointers are usually used to maintain the size of the window. Thus, the "two-pointer" method may be considered as a variant of the sliding window trick. The first point stands for the starting side of the "window" and the second one represents the ending side of the "window".

When does the sliding window trick work?

Usually string matching, or sub-array / sub-sequences may be considered to use the sliding window trick. The key point is that the question can be gradually solved by the evolution of the sliding window from the beginning of a string/array to the end.

To sum up, sliding window

1. is a common trick in coding interviews;

2. usually maintains two pointers: one for the left side of the window, the other is the right side of the "window";


Question List


Upper Layer

Comments

Post a Comment

Popular posts from this blog

Bit Manipulation - Medium Level

 Leetcode 416 Partition Equal Subset Sum Given a non-empty array nums containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100 Analysis: There are different ways to solve this problem, such as dp with a time complexity of O(N^2). Since N is small to this question, so it is Okay to pass OJ. Besides the small N, the value of each element is also very small. So this gives us some chance to use "space to trad off time". The data structure to be used is bitset. For bitset, each bit can be either 0 or 1. The index of that bit can be used as the corresponding sum. When the bit is 1, means there is a sum with the value of its index. When a new number comes, this number needs to be added to all the previous sums, to form new "previous" sums. Thus for each number, we need to go through all the previous sums, the time co...
Post a Comment
Read more

Priority_queue - Question 4

973 K Closest Points to Origin Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0). The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2). You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in). Constraints: 1 <= k <= points.length <= 10^4 -10^4 < xi, yi < 10^4 Analysis: It is quite straightforward to find the K points that have the closest distances to the original point. Since we need to find the smallest values, the root of the priority_queue (pq) is the largest one for all the elements inside the pq. Once the size of pq is larger than K, we need to pop out one. See the code below: class Solution { public: typedef pair<int, int> pp; vector<vector<int>> kClosest(vector<vector<int>>& points, int k) { vector...
Post a Comment
Read more