Distant Barcodes

Distant Barcodes - Problem

Array Hash Table Greedy Sorting Heap (Priority Queue) Counting Medium

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.

Input & Output

Example 1 — Basic Case

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

› Output: [2,1,2,1,2,3]

💡 Note: Rearrange so no adjacent elements are equal. We place the most frequent element (2) in alternating positions, then fill remaining spots with other elements.

Example 2 — Minimum Size

$ Input: barcodes = [1,2]

› Output: [1,2]

💡 Note: Already arranged correctly - no two adjacent elements are the same.

Example 3 — All Same Frequency

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

› Output: [1,2,1,3,2,3]

💡 Note: When all elements have same frequency, any valid alternating arrangement works.

Constraints

1 ≤ barcodes.length ≤ 10⁴
1 ≤ barcodes[i] ≤ 10⁴

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15 f Facebook 12

The key insight is to place the most frequent elements first in alternating positions to prevent adjacency. The optimal approach uses frequency counting with alternating placement - count frequencies, sort by frequency, then fill even positions first followed by odd positions. Time: O(n log k), Space: O(n).

Common Approaches

✓ Brute Force - Try All Permutations

⏱️ Time: O(n! × n) Space: O(n)

Generate every possible permutation of the barcodes array and check if any permutation satisfies the condition that no two adjacent elements are equal.

Frequency Count with Alternating Placement

⏱️ Time: O(n log k) Space: O(n)

Count the frequency of each barcode, then place the most frequent element in all even positions first, followed by remaining elements in odd positions.

Priority Queue (Max Heap)

⏱️ Time: O(n log k) Space: O(k)

Use a priority queue to always select the most frequent element that is different from the last placed element, ensuring no adjacent duplicates while optimally using frequencies.

Brute Force - Try All Permutations — Algorithm Steps

Step 1: Generate all permutations of the input array
Step 2: For each permutation, check if adjacent elements are different
Step 3: Return the first valid permutation found

Visualization

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Step-by-Step Walkthrough

Generate Permutations

Create all possible arrangements

Check Adjacent Elements

Verify no two adjacent elements are equal

Return Valid Arrangement

First valid permutation is the answer

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int isValid(int* arr, int size) {
    for (int i = 0; i < size - 1; i++) {
        if (arr[i] == arr[i + 1]) return 0;
    }
    return 1;
}

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int permute(int* arr, int start, int size) {
    if (start >= size) {
        return isValid(arr, size);
    }
    
    for (int i = start; i < size; i++) {
        swap(&arr[start], &arr[i]);
        if (permute(arr, start + 1, size)) return 1;
        swap(&arr[start], &arr[i]);
    }
    return 0;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int barcodes[1000];
    int size = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        barcodes[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    permute(barcodes, 0, size);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        printf("%d", barcodes[i]);
        if (i < size - 1) printf(",");
    }
    printf("]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n! × n)

Generate n! permutations, check each one takes O(n) time

⚠ Quadratic Growth

Space Complexity

O(n)

Store one permutation at a time plus recursion stack

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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