XOR Operation in an Array - Practice Coding Problems

XOR Operation in an Array - Problem

You are given an integer n and an integer start.

Define an array nums where nums[i] = start + 2 * i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Input & Output

Example 1 — Basic Case

$ Input: n = 5, start = 0

› Output: 8

💡 Note: Array is [0,2,4,6,8]. XOR: 0⊕2⊕4⊕6⊕8 = 8

Example 2 — Different Start

$ Input: n = 4, start = 3

› Output: 8

💡 Note: Array is [3,5,7,9]. XOR: 3⊕5⊕7⊕9 = 8

Example 3 — Single Element

$ Input: n = 1, start = 7

› Output: 7

💡 Note: Array is [7]. XOR of single element is itself: 7

Constraints

1 ≤ n ≤ 1000
0 ≤ start ≤ 1000

Visualization

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Understanding the Visualization

Input

n=5, start=0 - generate 5 elements starting from 0

Generate Sequence

Create [0,2,4,6,8] using formula start + 2*i

XOR All

0⊕2⊕4⊕6⊕8 = 8

Key Takeaway

🎯 Key Insight: Generate arithmetic sequence with formula start + 2*i, then XOR all elements efficiently

Asked in

a Amazon 15 M Microsoft 12 🍎 Apple 8

The key insight is to recognize this as an arithmetic sequence where each element is start + 2*i. The optimal approach computes and XORs elements directly without storing them. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Generate Array and XOR	O(n)	O(n)	Generate the array nums then XOR all elements
Direct XOR without Array	O(n)	O(1)	XOR elements directly without generating array

Generate Array and XOR — Algorithm Steps

Generate array with formula nums[i] = start + 2*i
Initialize result = 0
XOR all elements: result ^= nums[i]

Visualization

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Step-by-Step Walkthrough

Generate Array

Create nums[i] = start + 2*i

XOR Elements

XOR all elements together

Return Result

Final XOR value

Code -

solution.c — C

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

int solution(int n, int start) {
    // Generate the array
    int* nums = (int*)malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        nums[i] = start + 2 * i;
    }
    
    // XOR all elements
    int result = 0;
    for (int i = 0; i < n; i++) {
        result ^= nums[i];
    }
    
    free(nums);
    return result;
}

int main() {
    int n, start;
    scanf("%d", &n);
    scanf("%d", &start);
    
    int result = solution(n, start);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop to generate array plus single loop to XOR elements

✓ Linear Growth

Space Complexity

O(n)

Extra array to store n elements

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Generate Array and XOR — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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