Check if Two Strings are Anagrams

Problem Statement❯

Given two strings s1 and s2, determine whether they are anagrams of each other.

Two strings are considered anagrams if they contain the same characters with the same frequency, but possibly in a different order.
The comparison should be case-insensitive, and spacing or punctuation is not ignored unless explicitly stated.

If either string is empty or null, handle appropriately.

Examples❯

s1	s2	Are Anagrams?	Description
"listen"	"silent"	true	Characters match in count and type
"Listen"	"Silent"	true	Case-insensitive comparison
"triangle"	"integral"	true	Same characters in different order
"abc"	"def"	false	Completely different characters
"aabbcc"	"abccba"	true	Same letters and same frequency
"aabbcc"	"abc"	false	Frequencies do not match
"hello"	"helloo"	false	Different lengths
""	""	true	Both strings are empty
""	"abc"	false	One string is empty
""	""	true	Empty strings are technically anagrams of each other

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Solution❯

To determine if two strings are anagrams, we need to check if they contain the exact same characters with the exact same frequency, regardless of order or case. For beginners, think of it like this: can one string be rearranged to form the other?

Let’s Break Down the Possibilities

Case 1: Different lengths
If the strings have different lengths, they cannot be anagrams. For example, 'hello' and 'helloo' cannot be rearranged into each other—one has an extra character.

Case 2: Same letters, same frequency
This is the classic anagram case. 'listen' and 'silent' both have the letters l, i, s, t, e, n exactly once. So they are anagrams.

Case 3: Same letters, different frequency
If 's1' = 'aabbcc' and 's2' = 'abc', the letters may match but the count is different. Two 'a's in one string and only one in the other. So, not an anagram.

Case 4: Completely different letters
Obvious case—'abc' and 'xyz' share no characters at all, so they are not anagrams.

Case 5: Case sensitivity
We usually want to ignore case. So 'Listen' and 'Silent' should be considered anagrams. This means we convert both strings to lowercase before comparing.

Case 6: Empty strings
If both strings are empty, we consider them trivially anagrams—they contain the same (no) characters. But if only one is empty and the other is not, they are not anagrams.

What’s the Best Way to Solve This?

We compare the frequency of each character in both strings using a data structure like a map or dictionary. After converting both to lowercase, we count how many times each character appears in each string. If all characters and their frequencies match, they’re anagrams.

This solution is efficient and works in linear time relative to the length of the strings, since we only pass through each character a couple of times.

Algorithm Steps❯

If the lengths of s1 and s2 are different, return false immediately.
Convert both strings to lowercase to handle case-insensitivity.
Initialize two hash maps: map1 and map2.
Loop through each character in s1 and count frequency in map1.
Loop through each character in s2 and count frequency in map2.
If the number of unique keys in map1 and map2 differ, return false.
Check each key in map1 and compare its count with map2. If any mismatch, return false.
If all checks pass, return true.

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

def is_anagram(s1, s2):
    if len(s1) != len(s2):
        return False  # If lengths differ, can't be anagram

    s1, s2 = s1.lower(), s2.lower()
    map1, map2 = {}, {}

    for ch in s1:
        map1[ch] = map1.get(ch, 0) + 1  # Count frequency in s1
    for ch in s2:
        map2[ch] = map2.get(ch, 0) + 1  # Count frequency in s2

    if len(map1) != len(map2):
        return False  # Different set of characters

    for key in map1:
        if map1[key] != map2.get(key):
            return False  # Frequency mismatch

    return True

# Example usage
print(is_anagram("listen", "silent"))  # True
print(is_anagram("Hello", "Olelh"))    # True
print(is_anagram("Hi", "Bye"))          # False

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(n)	We scan both strings once to populate the maps and compare them.
Average Case	O(n)	Each character is processed once in both strings and looked up in hash maps.
Worst Case	O(n)	All characters are processed and map comparisons done linearly.

Space Complexity❯

O(1)

Explanation: At most 26 characters are stored in the maps (for alphabetic strings), making space constant for English letters.

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