Greedy Algorithm

Greedy Algorithm is one of the intuitive algorithms that appear not complicated to understand. As indicated by its name, greedy algorithm picks up the extreme values for each step since the beginning. Thus, greedy algorithm is only valid when the optimal of every step, or the optimal on the fly, is guarantee to be true, which is usually not.

The interesting part is that, when the greedy algorithm is valid to a question, it is usually not difficulty to "find it". However, it is not rare that the proving of the valid is not straightforward. One of the ways to prove the correctness is proof by contradiction.

To make a brief summary:

1. greedy algorithm requires the correctness to pick up the extreme value for each step;

2. the proof of the validity of greedy algorithm usually is not easy.

Question list

Upper Layer

Comments

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 + ...

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...

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 ...

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