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

Binary Search - Hard Level - Question 3

Binary Search - Hard Level - Question 3 878. Nth Magical Number A positive integer is magical if it is divisible by either a or b. Given the three integers n, a, and b, return the nth magical number. Since the answer may be very large, return it modulo 10^9 + 7. Analysis: Let us consider some examples first. Example 1, a = 4, b = 2. If b is dividable by a, then all the numbers which is dividable by a should be dividable by b as well. So the nth magical number should be n*b; Example 2, a = 3, b = 2. The multiples of 2 are: 2, 4, 6, 8, 10, 12, ... The multiple of 3 are: 3, 6, 9, 12, ... So the overlap is related to the minimum common multiple between a and b, and we need to remove the overlap which is double-counted. So now, we make some conclusions: 1. the upper bound of the nth magical number should be n*b, where a is the smaller one (or b <= a); 2. there are n*b/a magical numbers smaller than n*b; 3. there are n*b/(minimum common multiple) overlaps. Thus, the overall count is: n + ...
Post a Comment
Read more

Segment Tree

Segment tree can be viewed as an abstract data structure which using some more space to trade for speed. For example, for a typical question with O(N^2) time complexity, the segment tree method can decrease it to O(N*log(N)).  To make it understandable, let us consider one example. Say we have an integer array of N size, and what we want is to query the maximum with a query range [idx1, idx2], where idx1 is the left indexes, and idx2 is the right indexes inclusive. If we only do this kind of query once, then we just need to scan through the array from idx1 to idx2 once, and record the maximum, done. The time complexity is O(N), which is decent enough in most cases even though it is not the optimal one (for example, with a segment tree built, the time complexity can decrease down to O(log(N))). However, how about we need to query the array N times? If we continue to use the naïve way above, then the time complexity is O(N^2), since for each query we need to scan the query range once...
Post a Comment
Read more

Recursion - Example

Recursion - Example Leetcode 231  Power of Two Given an integer n, return true if it is a power of two. Otherwise, return false. An integer n is a power of two, if there exists an integer x such that n == 2^x Constraints: -2^31 <= n <= 2^31 - 1 Analysis: One way is to think about this question recursively: if n%2 == 1, then n must not be power of 2; if not, then we just need to consider whether (n/2) is a power of 2 or not. This is exactly the "same question with a smaller size"! It is trivial to figure out the base cases: if n == 0, return false; if n == 1, return true. See the code below: class Solution { public: bool isPowerOfTwo(int n) { // base cases if(n == 0) return false; if(n == 1) return true; // converging if(n%2 == 1) return false; return isPowerOfTwo(n/2); } }; If interested, there are some other ways to solve this problem. For example, using bit manipulation, we can have the following solution: class ...
Post a Comment
Read more