LeetCode 135: Candy — Python Solution

Solve LeetCode 135: Candy in Python with a two directional passes approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

Give each child one candy, scan left-to-right for increasing ratings, then right-to-left for decreasing ratings while taking the larger requirement.

Algorithm

1. Give each child one candy, scan left-to-right for increasing ratings, then right-to-left for decreasing ratings while taking the larger requirement.
2. Maintain this invariant: After both passes every higher-rated neighbor has strictly more candy while each allocation is minimal for its slope.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def candy(self, ratings):
        sweets = [1] * len(ratings)
        for i in range(1, len(ratings)):
            if ratings[i] > ratings[i - 1]:
                sweets[i] = sweets[i - 1] + 1
        for i in range(len(ratings) - 2, -1, -1):
            if ratings[i] > ratings[i + 1]:
                sweets[i] = max(sweets[i], sweets[i + 1] + 1)
        return sum(sweets)

Why this is correct

The proof follows the maintained state: After both passes every higher-rated neighbor has strictly more candy while each allocation is minimal for its slope. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(n) space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

Plateaus reset to one; peaks must satisfy both directions.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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