Understanding the Problem
We are given a string, and our task is to sort the characters in the string based on how frequently they appear. Characters that appear more often should come first in the result. If multiple characters have the same frequency, their relative order doesn't matter.
For example, if the input is "tree"
, the character 'e' appears twice, while 't' and 'r' appear once. A valid output would be "eetr"
or "eert"
, as both start with the most frequent character.
Step-by-Step Solution with Example
Step 1: Count character frequencies
We first count how many times each character appears in the string. We can use a hash map (or dictionary) for this.
For the input "tree"
, we get:
t → 1
r → 1
e → 2
Step 2: Organize characters by frequency
Now that we have the frequencies, we need a way to sort characters from highest to lowest frequency. A max heap (priority queue) is ideal for this because it allows us to always extract the character with the highest frequency.
Step 3: Build the output string
We extract characters from the max heap one by one. For each character, we repeat it as many times as its frequency and append it to the result string.
Continuing the example:
Take 'e' → frequency 2 → append "ee"
Then 't' or 'r' → frequency 1 → append "t" and "r"
Final output could be "eert"
or "eetr"
.
Step 4: Return the final string
Once all characters are extracted and appended, return the final string as the answer.
Edge Cases
- Empty string: If the input is
""
, the output should also be an empty string.
- Single character: For input like
"aaaa"
, the result is the same as input, "aaaa"
.
- All characters same frequency: For input
"abcde"
, any order is acceptable: "abcde"
, "edcba"
, etc.
- Mixed characters: For input
"a1a1b!"
, treat all characters equally. Count and sort based on frequency, not type.
- Very large input: Ensure the solution handles large strings efficiently using heap or bucket sort.
Finally
This problem tests your understanding of frequency counting and using data structures like hash maps and heaps. The logic is straightforward:
count → organize → build result. By carefully handling different input types and edge cases, we ensure the solution is both correct and robust.
Whether you’re a beginner or experienced coder, this step-by-step approach helps break the problem into manageable parts and makes the logic easy to follow.
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