Asked in: Meesho
Image of the Question
All Testcases Passed ✔
Solution
#include <vector>
using namespace std;
// ... rest of solution available after purchaseExplanation
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To solve this problem, start by understanding the definition of a “beautiful” number. A number `x` is considered beautiful if the bitwise XOR of all numbers from 0 to x is equal to x itself. This immediately suggests a mathematical or bitwise pattern that can be leveraged, rather than performing the XOR computation from 0 to x every time.
So, the key step is to observe the XOR pattern for numbers from 0 up to a certain number. Let’s look at a few values manually and find the XOR of all numbers from 0 to x:
- XOR(0 to 0) = 0
- XOR(0 to 1) = 0 ^ 1 = 1
- XOR(0 to 2) = 0 ^ 1 ^ 2 = 3
- XOR(0 to 3) = 0 ^ 1 ^ 2 ^ 3 = 0
- XOR(0 to 4) = 0 ^ 1 ^ 2 ^ 3 ^ 4 = 4
- XOR(0 to 5) = 1
- XOR(0 to 6) = 7
- XOR(0 to 7) = 0
- XOR(0 to 8) = 8
From this small sample, you can start to notice a pattern: sometimes the XOR from 0 to x equals x, and sometimes it doesn’t. The first step is to identify under what conditions this equality holds. This will let you decide whether a number is beautiful or not without recomputing XOR(0 to x) every time.
Investigate the properties of XOR(0 to x). One way to do this is to find a known mathematical result or pattern for computing XOR from 0 to a given number. Indeed, XOR(0 to x) follows a specific cyclic pattern that depends on x modulo 4. That is:
- If x % 4 == 0 → XOR(0 to x) = x
- If x % 4 == 1 → XOR(0 to x) = 1
- If x % 4 == 2 → XOR(0 to x) = x + 1
- If x % 4 == 3 → XOR(0 to x) = 0
So, using this, you can efficiently determine XOR(0 to x) for any value of x in constant time. Now, using the problem’s definition, a number x is beautiful if XOR(0 to x) == x. Based on the above cyclic pattern, this only happens when x % 4 == 0.
This observation simplifies the problem significantly. Now you do not need to perform any bitwise XOR operations in a loop. You just need to check for each number in the array whether it satisfies x % 4 == 0. If it does, it is beautiful.
Now shift your thinking to the structure of the input and the requirement to process multiple test cases. For each test case, you are given a value N and an array A of size N. Your goal is to count how many elements in A are beautiful using the condition you discovered.
This shifts the focus of the problem from algorithmic complexity to efficient data processing. You will loop through each number in the array A and check whether the number modulo 4 equals 0. If yes, you increment a counter for that test case. Repeat this for each test case and output the result.
Although this is computationally simple, some additional considerations help make your implementation clean and effective:
- Pay attention to reading input, especially when handling multiple test cases.
- Consider edge cases such as when the array is empty or when it contains only one number.
- Be mindful of negative numbers if they are allowed, though in many XOR problems, the domain is limited to non-negative integers.
In summary, the main insight lies in reducing the computational complexity of checking the condition using a mathematical pattern derived from bitwise operations. Recognizing that XOR(0 to x) == x only when x % 4 == 0 allows for a very efficient solution. From that point, the problem becomes a simple filtering operation over an array for each test case. This combination of mathematical insight and clean data handling is key to solving the problem efficiently.
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