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✏️Fill the BlankpythonEasyArrays & Hashing

Valid Anagram - Fill the Blank (Python)

Complete the Python solution

Problem Brief

Given two strings s and t, return true if t is an anagram of s, and false otherwise. 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.

Puzzle Hints

Read the blank inside this exact statement: "return ___".
Python dict.get() for safe access with default 0.
Check the variable that is read immediately after the blank; that usually tells you what the blank must produce.

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Valid Anagram — Frequency Count

hashmap

Input: s="anagram", t="nagaram". Count char frequencies

dict & set in Pythonref

Python dict is a hash map with O(1) average access. set stores unique hashable elements. Both are built-in and highly optimized. Use collections.Counter for frequency counting and collections.defaultdict to avoid KeyError.

-dict — {key: value}, O(1) access via d[key]

-set() — unique elements, O(1) in/add/discard

-dict.get(key, default) avoids KeyError

-Counter(iterable) counts element frequencies

from collections import Counter, defaultdict

d = {}
d['a'] = 1
d.get('b', 0)       # 0 (default)

s = set([1, 2, 2])  # {1, 2}
2 in s               # True — O(1)

Counter("aab")       # Counter({'a': 2, 'b': 1})

Official docs →

dict & set in Pythonref

-dict — {key: value}, O(1) access via d[key]

-set() — unique elements, O(1) in/add/discard

-dict.get(key, default) avoids KeyError

-Counter(iterable) counts element frequencies

from collections import Counter, defaultdict

d = {}
d['a'] = 1
d.get('b', 0)       # 0 (default)

s = set([1, 2, 2])  # {1, 2}
2 in s               # True — O(1)

Counter("aab")       # Counter({'a': 2, 'b': 1})

Official docs →

How to think: Hash Map / Setguide

You need O(1) lookups — checking if something exists, counting frequencies, or finding pairs.

1.Ask: "Am I searching for something repeatedly?" → Hash Map

2.Ask: "Do I need to check existence?" → Set

3.Ask: "Do I need to count occurrences?" → Map with value = count

4.Ask: "Do I need to find a pair that satisfies a condition?" → Store complement in Map

5.The tradeoff: O(n) extra space buys you O(1) per lookup

vs Nested loops (O(n²)): You're comparing every element against every other — a Map does it in one pass

vs Sorting (O(n log n)): You just need existence/frequency checks, not order

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

How to think: Hash Map / Setguide

You need O(1) lookups — checking if something exists, counting frequencies, or finding pairs.

1.Ask: "Am I searching for something repeatedly?" → Hash Map

2.Ask: "Do I need to check existence?" → Set

3.Ask: "Do I need to count occurrences?" → Map with value = count

4.Ask: "Do I need to find a pair that satisfies a condition?" → Store complement in Map

5.The tradeoff: O(n) extra space buys you O(1) per lookup

vs Nested loops (O(n²)): You're comparing every element against every other — a Map does it in one pass

vs Sorting (O(n log n)): You just need existence/frequency checks, not order

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

Drag tokens into the blanks

False

True

1def isAnagram(s, t):
2    if len(s) != len(t):
3        return ___
4    count = {}
5    for c in s:
6        count[c] = count.get(c, 0) + ___
7    for c in t:
8        count[c] = count.get(c, 0) - 1
9        if count[c] < ___:
10            return False
11    return True

from collections import Counter, defaultdict d = {} d['a'] = 1 d.get('b', 0) # 0 (default) s = set([1, 2, 2]) # {1, 2} 2 in s # True — O(1) Counter("aab") # Counter({'a': 2, 'b': 1})

1def isAnagram(s, t):

2 if len(s) != len(t):

3 return ___

4 count = {}

5 for c in s:

6 count[c] = count.get(c, 0) + ___

7 for c in t:

8 count[c] = count.get(c, 0) - 1

9 if count[c] < ___:

10 return False

11 return True