Skip to main content

Segment Tree -Question 1

Most Beautiful Item for Each Query

You are given a 2D integer array items where items[i] = [pricei, beautyi] denotes the price and beauty of an item respectively.

You are also given a 0-indexed integer array queries. For each queries[j], you want to determine the maximum beauty of an item whose price is less than or equal to queries[j]. If no such item exists, then the answer to this query is 0.

Return an array answer of the same length as queries where answer[j] is the answer to the jth query.

Constraints:

1 <= items.length, queries.length <= 10^5

items[i].length == 2

1 <= pricei, beautyi, queries[j] <= 10^9


Analysis:

The first step can be to sort the items based on the price. Then we can narrow down the search range by a binary search.

But the beauty is not sorted yet. So if without further treatment, we need to scan through the query range, the time complexity is O(N). The query length could be as large as 1e5, thus the overall time complexity is O(N*M) ~ 1e10, which is too large to this question.

If use a segment tree, then the time complexity can decrease to O(M*log(N)).

The key step is to save the beauty values with the original order to a helper array with a double size, then build up the maximum values in a way like building a maximum-root heap (see the code below).

After having the segment-tree array, the time complexity is O(log(N)) for one query. The overall time complexity is O(M*log(N)) for M queries.


See the code below:

class Solution {
public:
    vector<int> maximumBeauty(vector<vector<int>>& items, vector<int>& queries) {
        vector<int> res;
        sort(items.begin(), items.end(), [](auto &a, auto &b) {
            return a[0] < b[0]; 
        });
        int n = items.size();
        vector<int> data(2*n, 0);
        for(int i=n; i<2*n; ++i) data[i] = items[i-n][1];
        build(data);
        for(auto &a : queries) {
            int id = findUpperBound(items, a);
            //cout<<id<<endl;
            res.push_back(query(data, 0, id-1));
        }
        return res;
    }
private:
    int findUpperBound(vector<vector<int>>& its, int val) {
        int left = 0, right = its.size();
        while(left < right) {
            int mid = left + (right - left) / 2;
            //cout<<mid<<" "<<left<<" "<<right<<endl;
            if(its[mid][0] <= val) left = mid + 1;
            else right = mid;
        }
        return left;
    }
    void build(vector<int>& data) {
        int n = data.size();
        n /= 2;
        for(int i=n-1; i>0; --i) data[i] = max(data[2*i], data[2*i+1]);
    }
    int query(vector<int>& data, int left, int right) {
        if(left > right) return 0;
        int res = 0, m = data.size(), n = m/2;
        for(int i = left + n, j = right + n; i <= j; i /= 2, j /= 2) {
            if(i&1) res = max(res, data[i++]);
            if(!(j&1)) res = max(res, data[j--]);
        }
        return res;
    }
};


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