Beautiful Array

Beautiful Array - Problem

An array nums of length n is beautiful if:

nums is a permutation of the integers in the range [1, n]
For every 0 <= i < j < n, there is no index k with i < k < j where 2 * nums[k] == nums[i] + nums[j]

Given the integer n, return any beautiful array nums of length n. There will be at least one valid answer for the given n.

Input & Output

Example 1 — Small Case

$ Input: n = 4

› Output: [1,3,2,4]

💡 Note: This is a beautiful array because for any i < j, there's no k with i < k < j where 2*nums[k] = nums[i] + nums[j]. For instance, checking all triplets: no middle element equals the average of two others.

Example 2 — Base Case

$ Input: n = 1

› Output: [1]

💡 Note: Single element array is always beautiful since there are no triplets to violate the condition.

Example 3 — Larger Array

$ Input: n = 5

› Output: [1,5,3,2,4]

💡 Note: Built using divide and conquer: odds from n=3 become [1,5,3], evens from n=2 become [2,4], combined to form beautiful array of length 5.

Constraints

1 ≤ n ≤ 1000
It's guaranteed that for the given n, at least one valid answer exists.

Visualization

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Asked in

G Google 15 f Facebook 8

The key insight is that if an array A is beautiful, then both 2*A-1 (odds) and 2*A (evens) are also beautiful. The optimal approach uses divide and conquer to recursively build beautiful arrays. Time: O(n log n), Space: O(n log n)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force Permutation

⏱️ Time: O(n! × n³) Space: O(n)

Generate all possible permutations of numbers 1 to n, then check each permutation to see if it satisfies the beautiful array condition by examining all triplets.

Divide and Conquer (Optimal)

⏱️ Time: O(n log n) Space: O(n log n)

Use the mathematical property that if A is beautiful, then both 2*A-1 (odds) and 2*A (evens) are also beautiful. Recursively build larger beautiful arrays from smaller ones.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

static int tempArray[2000];
static int resultArray[1000];

int beautifulArrayHelper(int n, int* result) {
    if (n == 1) {
        result[0] = 1;
        return 1;
    }
    
    // Get beautiful array for roughly half size
    int oddsSize = beautifulArrayHelper((n + 1) / 2, tempArray);
    int evensSize = beautifulArrayHelper(n / 2, tempArray + 500);
    
    int idx = 0;
    // Transform: odds become 2*x-1
    for (int i = 0; i < oddsSize; i++) {
        result[idx++] = 2 * tempArray[i] - 1;
    }
    // Transform: evens become 2*x
    for (int i = 0; i < evensSize; i++) {
        result[idx++] = 2 * tempArray[500 + i];
    }
    
    return idx;
}

int main() {
    int n;
    scanf("%d", &n);
    
    int size = beautifulArrayHelper(n, resultArray);
    
    printf("[");
    for (int i = 0; i < size; i++) {
        if (i > 0) printf(",");
        printf("%d", resultArray[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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