Sort Characters by Frequency

1. Step 1: Initialize Solution.frequencySort

   Replace the empty starter with the first real state owned by Solution.frequencySort. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Assemble the primary transition

   Extend the initialized state with the next contiguous part of the popular solution. The transition explains how one input element or operation changes the state; boundaries are easier to reason about after this invariant is visible.

3. Step 3: Pass the Single case

   Complete the readable core algorithm for one representative Interview case. Count characters and use frequency-indexed buckets to emit them in descending count order.

4. Step 4: Harden the Three One boundary

   Repair the reviewed boundary and pass the complete submission contract. Each character is placed in exactly the bucket equal to its frequency. Visiting bucket indices in descending order emits every higher-frequency character first, and repeating by the index preserves exactly all input occurrences.

• Emit each character count times; sorting only the distinct characters loses repeated occurrences.
• hashing and sets
• strings

• LeetCode 451 — Sort Characters By Frequency

• Rank Top Words(opens in a new tab)

  Ranking frequency entries by a deterministic compound key prepares the frequency-first ordering required when sorting characters.

• Update a Bounded Heap(opens in a new tab)

  Update a Bounded Heap isolates after each value, the heap contains the largest min(k, processed_count) values seen so far. That focused state discipline is required when implementing sort characters by frequency as a complete Interview Problem.

Count each character, place characters into a bucket indexed by count, then traverse buckets from the maximum count down and repeat each character by its count.

Why it works: Each character is placed in exactly the bucket equal to its frequency. Visiting bucket indices in descending order emits every higher-frequency character first, and repeating by the index preserves exactly all input occurrences.

class Solution:
    def frequencySort(self, s):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        counts = {}
        for character in s:
            counts[character] = counts.get(character, 0) + 1
        buckets = [[] for _ in range(len(s) + 1)]
        for character, count in counts.items():
            buckets[count].append(character)
        output = []
        for count in range(len(s), 0, -1):
            for character in buckets[count]:
                output.append(character * count)
        return ''.join(output)

• Group Words by Anagram Signature(opens in a new tab)