Skip to main content

Double Pointers

Double Pointers is one of the common tricks used in algorithm interviews. For some circumstances, for example, one or two sorted arrays, we can use two pointers to track the loop process. The final answer can be updated during the looping, and obtained when the loop is done.

The logic behind the double pointers is that for every step, we just need to move one of the two pointers, or two of them. The optimal or final answer can be converged by doing this. Thus the time complexity is linear, and the space complexity is constant.

One of the key step is that we have to make sure (or proof) by moving the two pointers, we will not miss the possible local optimal solutions, which then guarantees that we will obtain the global optimal solutions after finishing the looping.


Question List

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