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#LC438 - Find All Anagrams - Sliding Window Algorithm

Updated: Nov 23, 2021

In this topic we will cover the leetcode #438 Find all anagram programming question.

DirtyRoom - Dormitory (Anagrams)

Problem Statement:

Given two strings s and p, return an array of all the start indices of p's anagrams in s. You may return the answer in any order.


String from -- this is the main string from which we have to find all the anagrams.

String find -- this is the string for which we have to find if its anagrams are present in the from string.


We will be using Sliding Window Algorithm. The logic here is to prepare an int array of 26 length (called hash), which contains the 0 as the initial value. It will increment value at those indexes, for which the character at the specific index in the find string is present.

public int[] prepareHash(String str){
    int[] charCount = new int[26];
    for(char c : str.toCharArray()){
        charCount[c - 'a']++;
    return charCount;

The find string length defines the length of the window. We will keep the left and right variable to keep the windows start and end indexes.

int[] findHash = prepareHash(find);
int left=0, right=find.length();

Now, for the characters present in the window, we will find the hash again and compare this int array with the hash we computed from the find string. If both the arrays are equal, then we will add the left value in the result list. If both the arrays are not equal, we will slide the window one step ahead, means we will increment the left and right variable values.

while(right <= from.length()){
    String temp = from.substring(left, right);
    int[] tempCharCount = prepareHash(temp);
    if(Arrays.equals(findHash, tempCharCount)){

Sliding the window, we will get all the possible anagrams present in the input from string. The below code will list all the indexes where we can find the anagrams.


You can find the entire code base here. Github Link

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