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All Anagrams in a String

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a C program to find all the start indices of anagrams of a pattern string p in a text string s. An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once. Return the result as an array of starting indices.

Example 1
    
  
  • Input: s = "abab", p = "ab"
    • 
  
  • Output: [0, 2]
    • 
  
  • Explanation: 
    • The substring starting at index 0 is "ab", which is an anagram of "ab". 
    • The substring starting at index 2 is "ab", which is an anagram of "ab".
    • Both substrings have the same character frequency as pattern "ab". 
    • Therefore, return indices [0, 2].
    •
Example 2
    
  
  • Input: s = "abcdefg", p = "abc"
    • 
  
  • Output: [0]
    • 
  
  • Explanation: 
    • The substring starting at index 0 is "abc", which is an anagram of "abc". 
    • No other substring of length 3 forms an anagram of "abc". 
    • Checking all positions: "bcd", "cde", "def", "efg" - none match. 
    • Therefore, return [0].
    •
Constraints
    
  
  • 1 ≤ s.length, p.length ≤ 3 * 10^4
    • 
  
  • s and p consist of lowercase English letters only
    • 
  
  • Time Complexity: O(n) where n is length of string s
    • 
  
  • Space Complexity: O(1) as we use fixed size arrays for character counts
    • 
  
  • Use sliding window technique for optimal solution
    •
StringsHCL TechnologiesSamsung
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Hint
Solution

Solution Hints

    
  
  • Use sliding window technique with two frequency arrays to track character counts
    • 
  
  • Create frequency arrays for pattern p and current window in string s
    • 
  
  • Initialize the first window of size equal to pattern length
    • 
  
  • Compare frequency arrays to check if current window is an anagram
    • 
  
  • Slide the window by removing leftmost character and adding new rightmost character
    • 
  
  • Continue until the window reaches the end of string s
    •

Steps to solve by this approach:

 Step 1: Check if the string s is shorter than pattern p; if so, return empty result.
 Step 2: Create frequency arrays to count characters in pattern p and current sliding window.
 Step 3: Initialize the pattern frequency array by counting all characters in p.
 Step 4: Initialize the first window of size equal to pattern length and count its character frequencies.
 Step 5: Compare the two frequency arrays; if equal, add starting index 0 to results.
 Step 6: Slide the window by adding the new character and removing the character that goes out of window.
 Step 7: After each slide, compare frequency arrays and add valid starting indices to results.

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