Beautiful Arrangement II

Beautiful Arrangement II - Problem

Array Math Medium

Given two integers n and k, construct a list answer that contains n different positive integers ranging from 1 to n and obeys the following requirement:

Suppose this list is answer = [a₁, a₂, a₃, ..., aₙ], then the list [|a₁ - a₂|, |a₂ - a₃|, |a₃ - a₄|, ..., |aₙ₋₁ - aₙ|] has exactly k distinct integers.

Return the list answer. If there are multiple valid answers, return any of them.

Input & Output

Example 1 — Basic Case

$ Input: n = 4, k = 3

› Output: [1,4,2,3]

💡 Note: Differences are |1-4|=3, |4-2|=2, |2-3|=1, giving exactly 3 distinct values {1,2,3}

Example 2 — Small k

$ Input: n = 5, k = 2

› Output: [1,3,2,4,5]

💡 Note: Differences are |1-3|=2, |3-2|=1, |2-4|=2, |4-5|=1, giving exactly 2 distinct values {1,2}

Example 3 — Minimum Case

$ Input: n = 2, k = 1

› Output: [1,2]

💡 Note: Only one difference |1-2|=1, exactly k=1 distinct value

Constraints

1 ≤ k < n ≤ 10⁴

Visualization

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Asked in

G Google 35 f Facebook 28 a Amazon 22

The key insight is to use an alternating pattern between minimum and maximum values to create exactly k distinct differences efficiently. The mathematical construction approach alternates between low and high values for k+1 positions, then fills remaining positions sequentially. Time: O(n), Space: O(n).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force Permutations

⏱️ Time: O(n! × n) Space: O(n)

Try every possible arrangement of numbers 1 to n, calculate absolute differences between adjacent elements, count distinct values, and return first arrangement with exactly k distinct differences.

Mathematical Construction

⏱️ Time: O(n) Space: O(n)

Use mathematical insight: alternate between smallest and largest remaining numbers to generate differences k, k-1, k-2, ..., 1, then fill remaining positions sequentially to avoid new differences.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

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

int hasDistinctDifferences(int* perm, int n, int k) {
    int differences[1000];
    int diffCount = 0;
    
    // Calculate absolute differences
    for (int i = 0; i < n - 1; i++) {
        int diff = abs(perm[i] - perm[i + 1]);
        differences[diffCount++] = diff;
    }
    
    // Count distinct differences
    int distinctCount = 0;
    for (int i = 0; i < diffCount; i++) {
        int isDistinct = 1;
        for (int j = 0; j < i; j++) {
            if (differences[i] == differences[j]) {
                isDistinct = 0;
                break;
            }
        }
        if (isDistinct) {
            distinctCount++;
        }
    }
    
    return distinctCount == k;
}

int generatePermutations(int* arr, int start, int end, int k, int* result) {
    if (start == end) {
        if (hasDistinctDifferences(arr, end + 1, k)) {
            for (int i = 0; i <= end; i++) {
                result[i] = arr[i];
            }
            return 1; // Found valid permutation
        }
        return 0;
    }
    
    for (int i = start; i <= end; i++) {
        swap(&arr[start], &arr[i]);
        if (generatePermutations(arr, start + 1, end, k, result)) {
            return 1;
        }
        swap(&arr[start], &arr[i]); // backtrack
    }
    return 0;
}

void solution(int n, int k, int* result) {
    int arr[n];
    for (int i = 0; i < n; i++) {
        arr[i] = i + 1;
    }
    
    generatePermutations(arr, 0, n - 1, k, result);
}

int main() {
    int n, k;
    scanf("%d", &n);
    scanf("%d", &k);
    
    int result[n];
    solution(n, k, result);
    
    printf("[");
    for (int i = 0; i < n; i++) {
        printf("%d", result[i]);
        if (i < n - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

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Smart Actions

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