LeetCode 0466. Count The Repetitions Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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

Solution 1: Preprocessing + Iteration

We preprocess the string s_2 such that for each starting position i, we calculate the next position j and the count of s_2 after matching a complete s_1, i.e., d[i] = (cnt, j), where cnt represents the count of s_2, and j represents the next position in the string s_2.

Next, we initialize j=0, and then loop n1 times. Each time, we add d[j][0] to the answer, and then update j=d[j][1].

The final answer is the count of s_2 that can be matched by n1 s_1, divided by n2.

The time complexity is O(m × n + n_1), and the space complexity is O(n). Where m and n are the lengths of s_1 and s_2 respectively.

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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); }(code-box)