Count The Repetitions

This page explains Java solution to problem Count The Repetitions using String data structure.

Problem Statement

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 1:

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


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import java.util.*;

class CountTheRepetitions {
    public int getMaxRepetitions(String s1, int n1, String s2, int n2) {
        if(s1 == null || s2 == null || n1 <= 0 && n2 <= 0) {
            return 0;

        //Key: Position of S2
        //Value: Number of times S1 is used
        HashMap<Integer, Integer> loopMap = new HashMap<>();

        //Key: Number of times S1 is used
        //Value: Number of times S2 appears
        HashMap<Integer, Integer> repMap = new HashMap<>();

        int s2Count = 0, s1Count = 1, s2Index = 0;
        while (s1Count <= n1) {
            for(int s1Index = 0; s1Index < s1.length(); s1Index++) {
                if(s1.charAt(s1Index) == s2.charAt(s2Index)) {
                    if(s2Index == s2.length()) {
                        s2Index = 0;
            repMap.put(s1Count, s2Count);
            if(loopMap.containsKey(s2Index)) {
            loopMap.put(s2Index, s1Count);

        if(s1Count >= n1) return s2Count / n2;

        int s1CountStartOfLoop = loopMap.get(s2Index);
        int s1CountInLoop = s1Count - s1CountStartOfLoop;
        int s2CountInLoop = repMap.get(s1Count) - repMap.get(s1CountStartOfLoop);

        //S2 Count Within a loop
        int repeatCount = (n1 - s1CountStartOfLoop) / s1CountInLoop; //How many loops of s1 are possible with n1 as higher limit
        int res = repeatCount * s2CountInLoop;

        //S2 Count Outside of a loop
        res += repMap.get(s1CountStartOfLoop + (n1 - s1CountStartOfLoop) % s1CountInLoop);
        return res / n2;

Time Complexity

O(len1 * len2) Where
len1 is length of input string s1
len2 is length of input string s2

Space Complexity

O(len2 + N1) Where
N1 number of times input string s1 can be used
len2 is length of input string s2