Medium Hash Table Two Pointers
Amazon
Microsoft
Oracle
Uber
2020-05-17
438. Find All Anagrams in a String
Similar Questions: LeetCode Question 567
Question:
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".
Solution:
Using hashmap to store the character appears in the p string. The integer or value means how many times of certain character left to use.
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<Integer>();
if (s.length() < p.length()) return result;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
char[] sarr = s.toCharArray();
for (char c: p.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
for (int i = 0; i < s.length(); i++) {
char c = sarr[i];
map.put(c, map.getOrDefault(c, 0) - 1);
// Have found all c, remove it from the map
if (map.get(c) == 0) {
map.remove(c);
}
if (i >= p.length()) {
char c_remove = sarr[i - p.length()];
map.put(c_remove, map.getOrDefault(c_remove, 0) + 1);
if (map.get(c_remove) == 0) {
map.remove(c_remove);
}
}
// Have found all character
if (map.size() == 0) {
result.add(i - p.length() + 1);
}
}
return result;
}
}
The following two solutions use the same idea of two pointers which run faster than hash map.
// Borrow from https://leetcode.com/problems/find-all-anagrams-in-a-string/discuss/92015/ShortestConcise-JAVA-O(n)-Sliding-Window-Solution
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> list = new ArrayList<>();
if (s == null || s.length() == 0 || p == null || p.length() == 0) return list;
int[] hash = new int[256];
for (char c : p.toCharArray()) {
hash[c]++;
}
int left = 0, right = 0, count = p.length();
while (right < s.length()) {
//move right everytime, if the character exists in p's hash, decrease the count
//current hash value >= 1 means the character is existing in p
if (hash[s.charAt(right++)]-- >= 1) count--;
//when the count is down to 0, means we found the right anagram
//then add window's left to result list
if (count == 0) list.add(left);
//if we find the window's size equals to p, then we have to move left (narrow the window) to find the new match window
//++ to reset the hash because we kicked out the left
//only increase the count if the character is in p
//the count >= 0 indicate it was original in the hash, cuz it won't go below 0
if (right - left == p.length() && hash[s.charAt(left++)]++ >= 0) count++;
}
return list;
}
}
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> list = new ArrayList<Integer>();
if (s == null || s.length() == 0 || p == null || p.length() == 0 || s.length() < p.length()) return list;
int[] hash = new int[256];
for (char c : p.toCharArray()) {
hash[c]++;
}
int left = 0, right = 0, count = p.length();
while (right < s.length()) {
hash[s.charAt(right)]--;
if (hash[s.charAt(right)] >= 0) {
count--;
}
if (count == 0) {
list.add(left);
}
if (right - left == p.length() - 1) {
if (hash[s.charAt(left)] >= 0) {
count++;
}
hash[s.charAt(left)]++;
left++;
}
right++;
}
return list;
}
}