Asked in: AURIGO
Image of the Question
All Testcases Passed ✔
Solution
def countofFriends(input1, input2,input3,input4):
answer = 0
def helper(first, second):
count = 0
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting how many of John's friends have tickets whose binary representations differ from John's ticket by at most K bits, it’s important to break down the problem into smaller logical components, focusing on bitwise operations and binary representations.
First, understand what is meant by “differ by at most K bits.” This relates to the Hamming distance between two integers. The Hamming distance between two numbers is the number of bit positions at which the corresponding bits are different. For example, if you look at the binary forms of two tickets, each differing bit contributes 1 to the total difference count.
The problem gives you:
- N, the number of friends,
- M, the total types of tickets (which can guide you on the range of ticket values),
- K, the maximum allowed bit difference,
- and the ticket array with N+1 elements, where the last element corresponds to John’s ticket.
Here is how you should think about the approach:
1. **Understand the Binary Representation:**
Each ticket number can be represented as a binary number. The maximum number M helps you gauge the maximum bit length you might encounter. Since the tickets are non-negative integers and limited by M, it is important to consider that the binary representation length will not exceed the bit length of M. For example, if M is 8, you’ll consider at most 4 bits since 8 in binary is 1000.
2. **Identify John’s Ticket:**
The last element of the ticket array is John’s ticket. This is your reference point. The binary representation of John’s ticket is the baseline against which all friends’ tickets are compared.
3. **Calculate Bit Difference (Hamming Distance):**
To find how many bits differ between two integers, you can use the XOR (exclusive OR) operation. XOR between two bits is 1 if the bits differ and 0 if they are the same. So, XOR of two numbers will produce a number whose binary representation has 1s in positions where the original two numbers differ. For example:
- Tickets 5 (0101) and 7 (0111) XOR to 2 (0010), which has one bit set to 1, indicating a difference of one bit.
- Tickets 6 (0110) and 7 (0111) XOR to 1 (0001), also one bit difference.
4. **Counting Set Bits (Bits that Differ):**
Once you get the XOR value, you need to count how many bits are set to 1. This count gives the number of differing bits. Think about efficient ways to count set bits in a binary number. Techniques include repeatedly checking the least significant bit or using bit manipulation tricks such as Brian Kernighan’s algorithm, which efficiently counts set bits by repeatedly clearing the lowest set bit until the number becomes zero.
5. **Compare to K:**
After determining the bit difference for a friend’s ticket and John’s ticket, compare it to K. If it is less than or equal to K, this friend should be included in the count.
6. **Iterate Over All Friends:**
Loop through the first N tickets (excluding John’s last ticket), compute the bit difference for each with John’s ticket, and count how many satisfy the difference condition.
7. **Return the Final Count:**
After processing all friends’ tickets, the accumulated count will represent how many friends’ tickets differ in at most K bits from John’s.
Consider some additional details and nuances:
- **Handling Large Ticket Values:**
Since ticket values could be large, ensure that the bit operations can handle integers of that size efficiently. Usually, standard integer operations in programming languages can handle 32 or 64 bits easily.
- **Performance Considerations:**
Since you only need to compare N friends’ tickets, and each comparison involves XOR and bit count, the time complexity will be roughly O(N * number_of_bits_in_ticket), which is efficient for reasonable constraints.
- **Edge Cases:**
Think about when K is zero, meaning only tickets exactly equal to John’s should be counted. Or when K is large, allowing more friends to qualify. Also consider cases where N is zero, meaning no friends exist to compare.
- **Input Validations:**
Since M is given, it can hint at the range of tickets, but you may not need to use M explicitly beyond understanding bit length limits.
In summary, the approach revolves around using XOR to find differing bits, efficiently counting these bits, comparing the count with K, and iterating over all friends to get the total count. This method leverages fundamental bitwise operations and is both conceptually clear and efficient to implement.
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