# 466. Count The Repetitions

## Description

We define str = [s, n] as the string str which consists of the string s concatenated n times.

• For example, str == ["abc", 3] =="abcabcabc".

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, s1 = "abc" can be obtained from s2 = "abdbec" based on our definition by removing the bolded underlined characters.

You are given two strings s1 and s2 and two integers n1 and n2. You have the two strings str1 = [s1, n1] and str2 = [s2, n2].

Return the maximum integer m such that str = [str2, m] can be obtained from str1.

Example 1:

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


Example 2:

Input: s1 = "acb", n1 = 1, s2 = "acb", n2 = 1
Output: 1


Constraints:

• 1 <= s1.length, s2.length <= 100
• s1 and s2 consist of lowercase English letters.
• 1 <= n1, n2 <= 106

## Solutions

• class Solution {
public int getMaxRepetitions(String s1, int n1, String s2, int n2) {
int m = s1.length(), n = s2.length();
int[][] d = new int[n][0];
for (int i = 0; i < n; ++i) {
int j = i;
int cnt = 0;
for (int k = 0; k < m; ++k) {
if (s1.charAt(k) == s2.charAt(j)) {
if (++j == n) {
j = 0;
++cnt;
}
}
}
d[i] = new int[] {cnt, j};
}
int ans = 0;
for (int j = 0; n1 > 0; --n1) {
ans += d[j][0];
j = d[j][1];
}
return ans / n2;
}
}

• class Solution {
public:
int getMaxRepetitions(string s1, int n1, string s2, int n2) {
int m = s1.size(), n = s2.size();
vector<pair<int, int>> d;
for (int i = 0; i < n; ++i) {
int j = i;
int cnt = 0;
for (int k = 0; k < m; ++k) {
if (s1[k] == s2[j]) {
if (++j == n) {
++cnt;
j = 0;
}
}
}
d.emplace_back(cnt, j);
}
int ans = 0;
for (int j = 0; n1; --n1) {
ans += d[j].first;
j = d[j].second;
}
return ans / n2;
}
};

• class Solution:
def getMaxRepetitions(self, s1: str, n1: int, s2: str, n2: int) -> int:
n = len(s2)
d = {}
for i in range(n):
cnt = 0
j = i
for c in s1:
if c == s2[j]:
j += 1
if j == n:
cnt += 1
j = 0
d[i] = (cnt, j)

ans = 0
j = 0
for _ in range(n1):
cnt, j = d[j]
ans += cnt
return ans // n2


• func getMaxRepetitions(s1 string, n1 int, s2 string, n2 int) (ans int) {
n := len(s2)
d := make([][2]int, n)
for i := 0; i < n; i++ {
j := i
cnt := 0
for k := range s1 {
if s1[k] == s2[j] {
j++
if j == n {
cnt++
j = 0
}
}
}
d[i] = [2]int{cnt, j}
}
for j := 0; n1 > 0; n1-- {
ans += d[j][0]
j = d[j][1]
}
ans /= n2
return
}

• function getMaxRepetitions(s1: string, n1: number, s2: string, n2: number): number {
const n = s2.length;
const d: number[][] = new Array(n).fill(0).map(() => new Array(2).fill(0));
for (let i = 0; i < n; ++i) {
let j = i;
let cnt = 0;
for (const c of s1) {
if (c === s2[j]) {
if (++j === n) {
j = 0;
++cnt;
}
}
}
d[i] = [cnt, j];
}
let ans = 0;
for (let j = 0; n1 > 0; --n1) {
ans += d[j][0];
j = d[j][1];
}
return Math.floor(ans / n2);
}