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Problem Statement:


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:


Prerequisite: Sliding Window

At any point of time we would have to maintain a window of length = p.length and check if the substring in the window is anagram of p.

To achieve this we would keep a frequency map which would contain all the character counts for p.

For any window we would keep similar frequency map which would contain all character counts for the current window.

If the window character frequency map looks same as the character frequency map for p, we would know that the substring in the current window is an anagram of p.

Java Solution:



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Python Solution:



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Related Other Problems:




The above content is written by:

Abhishek Dey

Abhishek Dey

A Visionary Software Engineer With A Mission To Empower Every Person & Every Organization On The Planet To Achieve More

Microsoft | University of Florida

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