Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"
Output:
[0, 6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:
s: "abab" p: "ab"
Output:
[0, 1, 2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
// Brute Force
vector<int> findAnagrams(string s, string p) { // time: O(s_len * n_len); space: O(s_len)
vector<int> res, record(128, 0);
int s_len = s.size(), p_len = p.size(), i = 0;
for (const char& ch : p) ++record[ch];
while (i < s_len) {
bool success = true;
vector<int> tmp = record;
for (int j = i; j < i + p_len; ++j) {
if (--tmp[s[j]] < 0) {
success = false;
break;
}
}
if (success) {
res.push_back(i);
}
++i;
}
return res;
}
// Optimized Brute Force
vector<int> findAnagrams(string s, string p) { // time: O(s_len); space: O(s_len)
vector<int> res, cnt1(128, 0), cnt2(128, 0);
int s_len = s.length(), p_len = p.length();
for (int i = 0; i < p_len; ++i) {
++cnt1[s[i]];
++cnt2[p[i]];
}
if (cnt1 == cnt2) res.emplace_back(0);
for (int i = p_len; i < s_len; ++i) {
++cnt1[s[i]];
--cnt1[s[i - p_len]];
if (cnt1 == cnt2) res.emplace_back(i - p_len + 1);
}
return res;
}
// Sliding Window
vector<int> findAnagrams(string s, string p) { // time: O(s_len + p_len); space: O(s_len)
vector<int> res, record(128, 0);
for (const char& ch : p) ++record[ch];
int begin = 0, end = 0, counter = p.length(), s_len = s.length(), p_len = p.length();
while (end < s_len) {
if (record[s[end++]]-- >= 1) --counter;
if (counter == 0) res.emplace_back(begin);
if (end - begin == p_len && record[s[begin++]]++ >= 0) ++counter;
}
return res;
}