LeetCode #1054 — MEDIUM

Distant Barcodes

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

In a warehouse, there is a row of barcodes, where the i^th barcode is barcodes[i].

Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.

Example 1:

Input: barcodes = [1,1,1,2,2,2]
Output: [2,1,2,1,2,1]

Example 2:

Input: barcodes = [1,1,1,1,2,2,3,3]
Output: [1,3,1,3,1,2,1,2]

Constraints:

1 <= barcodes.length <= 10000
1 <= barcodes[i] <= 10000

Step 01

Brute Force Baseline

Problem summary: In a warehouse, there is a row of barcodes, where the ith barcode is barcodes[i]. Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.

Baseline thinking

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

Pattern signal: Array · Hash Map · Greedy

Example 1

[1,1,1,2,2,2]

Example 2

[1,1,1,1,2,2,3,3]

Step 02

Core Insight

What unlocks the optimal approach

We want to always choose the most common or second most common element to write next. What data structure allows us to query this effectively?

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

Upper-end input sizes

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 #1054: Distant Barcodes
class Solution {
    public int[] rearrangeBarcodes(int[] barcodes) {
        int n = barcodes.length;
        Integer[] t = new Integer[n];
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            t[i] = barcodes[i];
            mx = Math.max(mx, barcodes[i]);
        }
        int[] cnt = new int[mx + 1];
        for (int x : barcodes) {
            ++cnt[x];
        }
        Arrays.sort(t, (a, b) -> cnt[a] == cnt[b] ? a - b : cnt[b] - cnt[a]);
        int[] ans = new int[n];
        for (int k = 0, j = 0; k < 2; ++k) {
            for (int i = k; i < n; i += 2) {
                ans[i] = t[j++];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1054: Distant Barcodes
func rearrangeBarcodes(barcodes []int) []int {
	mx := slices.Max(barcodes)
	cnt := make([]int, mx+1)
	for _, x := range barcodes {
		cnt[x]++
	}
	sort.Slice(barcodes, func(i, j int) bool {
		a, b := barcodes[i], barcodes[j]
		if cnt[a] == cnt[b] {
			return a < b
		}
		return cnt[a] > cnt[b]
	})
	n := len(barcodes)
	ans := make([]int, n)
	for k, j := 0, 0; k < 2; k++ {
		for i := k; i < n; i, j = i+2, j+1 {
			ans[i] = barcodes[j]
		}
	}
	return ans
}

# Accepted solution for LeetCode #1054: Distant Barcodes
class Solution:
    def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]:
        cnt = Counter(barcodes)
        barcodes.sort(key=lambda x: (-cnt[x], x))
        n = len(barcodes)
        ans = [0] * len(barcodes)
        ans[::2] = barcodes[: (n + 1) // 2]
        ans[1::2] = barcodes[(n + 1) // 2 :]
        return ans

// Accepted solution for LeetCode #1054: Distant Barcodes
struct Solution;
use std::collections::HashMap;

impl Solution {
    fn rearrange_barcodes(barcodes: Vec<i32>) -> Vec<i32> {
        let n = barcodes.len();
        if n == 1 {
            return barcodes;
        }
        let mut hm: HashMap<i32, usize> = HashMap::new();
        let mut max: (usize, i32) = (0, 0);
        for barcode in barcodes {
            let count = hm.entry(barcode).or_default();
            *count += 1;
            if *count > max.0 {
                max = (*count, barcode);
            }
        }
        let mut stack = vec![];
        for (k, v) in hm {
            if k != max.1 {
                for _ in 0..v {
                    stack.push(k);
                }
            }
        }
        for _ in 0..max.0 {
            stack.push(max.1);
        }
        let mut res = vec![0; n];
        let m = if n % 2 == 0 { n / 2 } else { (n + 1) / 2 };
        for i in 0..m {
            res[i * 2] = stack.pop().unwrap();
        }
        let mut i = 1;
        while let Some(top) = stack.pop() {
            res[i] = top;
            i += 2;
        }
        res
    }
}

#[test]
fn test() {
    let barcodes = vec![1, 1, 1, 2, 2, 2];
    let res = vec![1, 2, 1, 2, 1, 2];
    assert_eq!(Solution::rearrange_barcodes(barcodes), res);
    let barcodes = vec![1, 1, 2];
    let res = vec![1, 2, 1];
    assert_eq!(Solution::rearrange_barcodes(barcodes), res);
}

// Accepted solution for LeetCode #1054: Distant Barcodes
function rearrangeBarcodes(barcodes: number[]): number[] {
    const mx = Math.max(...barcodes);
    const cnt = Array(mx + 1).fill(0);
    for (const x of barcodes) {
        ++cnt[x];
    }
    barcodes.sort((a, b) => (cnt[a] === cnt[b] ? a - b : cnt[b] - cnt[a]));
    const n = barcodes.length;
    const ans = Array(n);
    for (let k = 0, j = 0; k < 2; ++k) {
        for (let i = k; i < n; i += 2, ++j) {
            ans[i] = barcodes[j];
        }
    }
    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(M)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.