Maximum candies you can get from boxes

Array Breadth-First Search Graph Theory

67.9% acceptance

Mar 1, 2026

772

225

You have n boxes labeled from 0 to n - 1. You are given four arrays: status, candies, keys, and containedBoxes where: status[i] is 1 if the ith box is open and 0 if the ith box is closed, candies[i] is the number of candies in the ith box, keys[i] is a list of the labels of the boxes you can open after opening the ith box. containedBoxes[i] is a list of the boxes you found inside the ith box. You are given an integer array initialBoxes that contains the labels of the boxes you initially have. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it. Return the maximum number of candies you can get following the rules above.

Solution

Rust

Time O(n²)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn max_candies(
    status: Vec<i32>,
    candies: Vec<i32>,
    keys: Vec<Vec<i32>>,
    contained_boxes: Vec<Vec<i32>>,
    initial_boxes: Vec<i32>,
  ) -> i32 {
    let n = status.len();
    let mut has_box = vec![false; n];
    let mut has_key = vec![false; n];
    let mut queued = vec![false; n];
    let mut total = 0;
    let mut queue = std::collections::VecDeque::new();

    for b in initial_boxes {
      let b = b as usize;
      has_box[b] = true;
      if status[b] == 1 && !queued[b] {
        queued[b] = true;
        queue.push_back(b);
      }
    }

    while let Some(b) = queue.pop_front() {
      total += candies[b];
      for &k in &keys[b] {
        let k = k as usize;
        has_key[k] = true;
        if has_box[k] && !queued[k] {
          queued[k] = true;
          queue.push_back(k);
        }
      }
      for &cb in &contained_boxes[b] {
        let cb = cb as usize;
        has_box[cb] = true;
        if (status[cb] == 1 || has_key[cb]) && !queued[cb] {
          queued[cb] = true;
          queue.push_back(cb);
        }
      }
    }
    total
  }
}