466

466. Count The Repetitions (Hard)

Define S = [s,n] as the string S which consists of n connected strings s. For example, ["abc", 3] ="abcabcabc".

On the other hand, we define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1. For example, “abc” can be obtained from “abdbec” based on our definition, but it can not be obtained from “acbbe”.

You are given two non-empty strings s1 and s2 (each at most 100 characters long) and two integers 0 ≤ n1 ≤ 10⁶ and 1 ≤ n2 ≤ 10⁶. Now consider the strings S1 and S2, where S1=[s1,n1] and S2=[s2,n2]. Find the maximum integer M such that [S2,M] can be obtained from S1.

Example:

Input:
s1="acb", n1=4
s2="ab", n2=2
Return:
2

Related Topics:
Dynamic Programming

Solution 1.

See explanation here

// OJ: https://leetcode.com/problems/count-the-repetitions/
// Author: github.com/lzl124631x
// Time: O(|s1| * n1) where |s1| is the length of s1
// Space: O(n1)
class Solution {
public:
    int getMaxRepetitions(string s1, int n1, string s2, int n2) {
        vector<int> repeatCount(n1 + 1, 0);
        vector<int> nextIndex(n1 + 1, 0);
        int j = 0, cnt = 0;
        for (int k = 1; k <= n1; ++k) {
            for (int i = 0; i < s1.size(); ++i) {
                if (s1[i] == s2[j]) {
                    ++j;
                    if (j == s2.size()) {
                        j = 0;
                        ++cnt;
                    }
                }
            }
            repeatCount[k] = cnt;
            nextIndex[k] = j;
            for (int start = 0; start < k; ++start) {
                if (nextIndex[start] == j) {
                    int prefixCount = repeatCount[start];
                    int patternCount = (n1 - start) / (k - start) * (repeatCount[k] - prefixCount);
                    int suffixCount = repeatCount[start + (n1 - start) % (k - start)] - prefixCount;
                    return (prefixCount + patternCount + suffixCount) / n2;
                }
            }
        }
        return repeatCount[n1] / n2;
    }
};

Solution 2.

Same idea as Solution 1, but use unordered_map to save computation. Reduce runtime from ~80ms to ~4ms.

// OJ: https://leetcode.com/problems/count-the-repetitions/
// Author: github.com/lzl124631x
// Time: O(|s1| * n1) where |s1| is the length of s1
// Space: O(n1)
class Solution {
public:
    int getMaxRepetitions(string s1, int n1, string s2, int n2) {
        unordered_map<int, int> kToRepeatCount;
        unordered_map<int, int> nextIndexToK;
        kToRepeatCount[0] = 0;
        nextIndexToK[0] = 0;
        int j = 0, cnt = 0;
        for (int k = 1; k <= n1; ++k) {
            for (int i = 0; i < s1.size(); ++i) {
                if (s1[i] == s2[j]) {
                    ++j;
                    if (j == s2.size()) {
                        j = 0;
                        ++cnt;
                    }
                }
            }
            if (nextIndexToK.find(j) != nextIndexToK.end()) {
                int start = nextIndexToK[j];
                int prefixCount = kToRepeatCount[start];
                int patternCount = (n1 - start) / (k - start) * (cnt - prefixCount);
                int suffixCount = kToRepeatCount[start + (n1 - start) % (k - start)] - prefixCount;
                return (prefixCount + patternCount + suffixCount) / n2;
            }
            kToRepeatCount[k] = cnt;
            nextIndexToK[j] = k;
        }
        return kToRepeatCount[n1] / n2;
    }
};