Skip to main content

Rolling Hash - Question 1

2156. Find Substring With Given Hash Value

The hash of a 0-indexed string s of length k, given integers p and m, is computed using the following function:

hash(s, p, m) = (val(s[0]) * p0 + val(s[1]) * p1 + ... + val(s[k-1]) * pk-1) mod m.
Where val(s[i]) represents the index of s[i] in the alphabet from val('a') = 1 to val('z') = 26.

You are given a string s and the integers power, modulo, k, and hashValue. Return sub, the first substring of s of length k such that hash(sub, power, modulo) == hashValue.

The test cases will be generated such that an answer always exists.

A substring is a contiguous non-empty sequence of characters within a string.


Constraints:

1 <= k <= s.length <= 2 * 104
1 <= power, modulo <= 109
0 <= hashValue < modulo
s consists of lowercase English letters only.
The test cases are generated such that an answer always exists.


Analysis


This is question is about rolling hash.

The formula is given, but we cannot use it directly. Or if we scan the string from left to right, the hashValue needs to divide the power, which may give incorrectly hashing result.

To bypass this, we can just scan from right to left. Then no division is needed.

See the code below,

 

class Solution {
public:
    string subStrHash(string s, int power, int modulo, int k, int hashValue) {
        int n = s.size(), id = n;
        long sum = 0, p = 1;
        for(int i=n-1; i>=0; --i) {
            if(i>=n-k) p = p*power%modulo;  // p = power^k%modulo
            sum = (sum*power + (s[i]-'a'+1))%modulo;
            if(i<=n-k-1) {
                sum = (sum - (s[i+k]-'a'+1)*p%modulo + modulo)%modulo;
            }
            if(i<=n-k && sum == hashValue) {
                cout<<i<<endl;
                id = i;
            }
        }
        return id<n ? s.substr(id, k) : "";
    }
};



Comments

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...

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...