Find Xor-Beauty of Array

Find Xor-Beauty of Array - Problem

You are given a 0-indexed integer array nums.

The effective value of three indices i, j, and k is defined as ((nums[i] | nums[j]) & nums[k]).

The xor-beauty of the array is the XORing of the effective values of all the possible triplets of indices (i, j, k) where 0 <= i, j, k < n.

Return the xor-beauty of nums.

Note that:

val1 | val2 is bitwise OR of val1 and val2.
val1 & val2 is bitwise AND of val1 and val2.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,4]

› Output: 1

💡 Note: All possible triplets: (0,0,0)→1, (0,0,1)→0, (0,1,0)→1, (0,1,1)→4, (1,0,0)→1, (1,0,1)→4, (1,1,0)→0, (1,1,1)→4. XOR result: 1⊕0⊕1⊕4⊕1⊕4⊕0⊕4 = 1

Example 2 — Single Element

$ Input: nums = [15]

› Output: 15

💡 Note: Only one triplet possible: (0,0,0) gives (15|15)&15 = 15&15 = 15

Example 3 — Three Elements

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

› Output: 2

💡 Note: Calculate all 27 possible triplets and XOR their effective values together

Constraints

1 ≤ nums.length ≤ 300
1 ≤ nums[i] ≤ 10⁸

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is recognizing that XOR operations cause most effective values to cancel out in pairs, leaving only specific diagonal terms. The optimal approach uses mathematical properties to reduce O(n³) brute force to O(n). However, the mathematical derivation is complex and the actual pattern requires careful analysis of all triplet combinations.

Common Approaches

✓ Binary Search Answer

⏱️ Time: N/A Space: N/A

Brute Force - Triple Nested Loop

⏱️ Time: O(n³) Space: O(1)

Generate all possible combinations of three indices and compute ((nums[i] | nums[j]) & nums[k]) for each triplet, then XOR all results together.

Mathematical Optimization

⏱️ Time: O(n) Space: O(1)

Analyze the mathematical structure of XOR operations to discover that when we XOR all possible effective values, most pairs cancel out due to XOR properties, leaving only the case where i=j=k.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int result = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < n; k++) {
                int effective_value = (nums[i] | nums[j]) & nums[k];
                result ^= effective_value;
            }
        }
    }
    
    return result;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    char* ptr = strchr(line, '[') + 1;
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%d\n", solution(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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