1054. Distant Barcodes LeetCode Solution

In this guide, you will get 1054. Distant Barcodes LeetCode Solution with the best time and space complexity. The solution to Distant Barcodes problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Distant Barcodes solution in C++
4. Distant Barcodes solution in Java
5. Distant Barcodes solution in Python
6. Additional Resources

Problem Statement of Distant Barcodes

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.

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

1054. Distant Barcodes LeetCode Solution in C++

class Solution {
 public:
  vector<int> rearrangeBarcodes(vector<int>& barcodes) {
    vector<int> ans(barcodes.size());
    vector<int> count(10001);
    int i = 0;  // ans' index

    for (const int b : barcodes)
      ++count[b];

    const auto maxIt = ranges::max_element(count);
    const int maxNum = maxIt - count.begin();

    auto fillAns = [&](int num) {
      while (count[num]-- > 0) {
        ans[i] = num;
        i = i + 2 < barcodes.size() ? i + 2 : 1;
      }
    };

    fillAns(maxNum);
    for (int num = 1; num < 10001; ++num)
      fillAns(num);

    return ans;
  }
};
/* code provided by PROGIEZ */

1054. Distant Barcodes LeetCode Solution in Java

class Solution {
  public int[] rearrangeBarcodes(int[] barcodes) {
    int[] ans = new int[barcodes.length];
    int[] count = new int[10001];
    int maxCount = 0;
    int maxNum = 0;

    for (final int b : barcodes)
      ++count[b];

    for (int i = 1; i < 10001; ++i)
      if (count[i] > maxCount) {
        maxCount = count[i];
        maxNum = i;
      }

    fillAns(ans, count, maxNum, barcodes.length);
    for (int num = 1; num < 10001; ++num)
      fillAns(ans, count, num, barcodes.length);

    return ans;
  }

  private int i = 0; // ans' index

  private void fillAns(int[] ans, int[] count, int num, int n) {
    while (count[num]-- > 0) {
      ans[i] = num;
      i = i + 2 < n ? i + 2 : 1;
    }
  }
}
// code provided by PROGIEZ

1054. Distant Barcodes LeetCode Solution in Python

class Solution:
  def rearrangeBarcodes(self, barcodes: list[int]) -> list[int]:
    ans = [0] * len(barcodes)
    count = collections.Counter(barcodes)
    i = 0  # ans' index
    maxNum = max(count, key=count.get)

    def fillAns(num: int) -> None:
      nonlocal i
      while count[num]:
        ans[i] = num
        i = i + 2 if i + 2 < len(barcodes) else 1
        count[num] -= 1

    fillAns(maxNum)
    for num in count.keys():
      fillAns(num)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

Happy Coding! Keep following PROGIEZ for more updates and solutions.