LeetCode #135 — HARD

Candy

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.
Children with a higher rating get more candies than their neighbors.

Return the minimum number of candies you need to have to distribute the candies to the children.

Example 1:

Input: ratings = [1,0,2]
Output: 5
Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: ratings = [1,2,2]
Output: 4
Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
The third child gets 1 candy because it satisfies the above two conditions.

Constraints:

n == ratings.length
1 <= n <= 2 * 10⁴
0 <= ratings[i] <= 2 * 10⁴

Step 01

Brute Force Baseline

Problem summary: There are n children standing in a line. Each child is assigned a rating value given in the integer array ratings. You are giving candies to these children subjected to the following requirements: Each child must have at least one candy. Children with a higher rating get more candies than their neighbors. Return the minimum number of candies you need to have to distribute the candies to the children.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,0,2]

Example 2

[1,2,2]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #135: Candy
class Solution {
    public int candy(int[] ratings) {
        int n = ratings.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, 1);
        Arrays.fill(right, 1);
        for (int i = 1; i < n; ++i) {
            if (ratings[i] > ratings[i - 1]) {
                left[i] = left[i - 1] + 1;
            }
        }
        for (int i = n - 2; i >= 0; --i) {
            if (ratings[i] > ratings[i + 1]) {
                right[i] = right[i + 1] + 1;
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += Math.max(left[i], right[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #135: Candy
func candy(ratings []int) int {
	n := len(ratings)
	left := make([]int, n)
	right := make([]int, n)
	for i := range left {
		left[i] = 1
		right[i] = 1
	}
	for i := 1; i < n; i++ {
		if ratings[i] > ratings[i-1] {
			left[i] = left[i-1] + 1
		}
	}
	for i := n - 2; i >= 0; i-- {
		if ratings[i] > ratings[i+1] {
			right[i] = right[i+1] + 1
		}
	}
	ans := 0
	for i, a := range left {
		b := right[i]
		ans += max(a, b)
	}
	return ans
}

# Accepted solution for LeetCode #135: Candy
class Solution:
    def candy(self, ratings: List[int]) -> int:
        n = len(ratings)
        left = [1] * n
        right = [1] * n
        for i in range(1, n):
            if ratings[i] > ratings[i - 1]:
                left[i] = left[i - 1] + 1
        for i in range(n - 2, -1, -1):
            if ratings[i] > ratings[i + 1]:
                right[i] = right[i + 1] + 1
        return sum(max(a, b) for a, b in zip(left, right))

// Accepted solution for LeetCode #135: Candy
impl Solution {
    pub fn candy(ratings: Vec<i32>) -> i32 {
        let n = ratings.len();
        let mut left = vec![1; n];
        let mut right = vec![1; n];

        for i in 1..n {
            if ratings[i] > ratings[i - 1] {
                left[i] = left[i - 1] + 1;
            }
        }

        for i in (0..n - 1).rev() {
            if ratings[i] > ratings[i + 1] {
                right[i] = right[i + 1] + 1;
            }
        }

        ratings
            .iter()
            .enumerate()
            .map(|(i, _)| left[i].max(right[i]) as i32)
            .sum()
    }
}

// Accepted solution for LeetCode #135: Candy
function candy(ratings: number[]): number {
    const n = ratings.length;
    const left = new Array(n).fill(1);
    const right = new Array(n).fill(1);
    for (let i = 1; i < n; ++i) {
        if (ratings[i] > ratings[i - 1]) {
            left[i] = left[i - 1] + 1;
        }
    }
    for (let i = n - 2; i >= 0; --i) {
        if (ratings[i] > ratings[i + 1]) {
            right[i] = right[i + 1] + 1;
        }
    }
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        ans += Math.max(left[i], right[i]);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.