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We are given a problem labeled "B3". Since the problem statement is not provided, I will assume that it is a common problem in programming competitions or assignments.
However, without the specific problem, I cannot provide a solution.
If you can provide the problem statement, I would be happy to help.
Alternatively, if "B3" refers to a known problem in a specific contest, please provide more context.
Since I don't have the problem, I'll create an example of a problem that might be labeled B3 and solve it.
Example Problem (B3):
Given an array of integers, find the length of the longest subarray that contains an equal number of 0s and 1s.
However, note that without the actual problem, this is just a guess.
Let me solve the example problem:
Problem:
Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.
Example:
Input: [0,1] -> Output: 2
Input: [0,1,0] -> Output: 2 (subarray [0,1] or [1,0])
Approach:
We can use a prefix sum and a hash map. The idea is to treat 0 as -1 and 1 as 1.
Then, we want to find the longest subarray where the sum is 0 (because equal number of 0s and 1s).
Steps:
1. Initialize a variable max_len to store the maximum length, and curr_sum to store the current prefix sum.
2. Use a hash map to store the first occurrence of a particular prefix sum (so that we can get the earliest index where that sum occurred).
3. Traverse the array:
- For each element, if it's 0, subtract 1 (or add -1), if it's 1, add 1.
- If the current prefix sum is 0, then the subarray from index 0 to current index has equal 0s and 1s, so update max_len.
- If the current prefix sum has been seen before at some index i, then the subarray from i+1 to current index has sum 0 (because the same prefix sum means the net change in between is 0). So, update max_len = max(max_len, current_index - i).
4. If the current prefix sum is new, store it with the current index.
Note: We start by storing prefix sum 0 at index -1 (to handle the case when the subarray starts from index 0).
Implementation:
Let's define:
max_len = 0
curr_sum = 0
prefix_map = {0: -1} # key: prefix_sum, value: the first index where this prefix_sum occurred
For i in range(len(arr)):
curr_sum += 1 if arr == 1 else -1
if curr_sum in prefix_map:
# The subarray from prefix_map[curr_sum] + 1 to i has sum 0
length = i - prefix_map[curr_sum]
if length > max_len:
max_len = length
else:
prefix_map[curr_sum] = i
Return max_len
Let's test with [0,1]:
i=0: arr[0]=0 -> curr_sum = -1
not in map -> prefix_map: {0:-1, -1:0}
i=1: arr[1]=1 -> curr_sum = 0
in map -> length = 1 - (-1) = 2 -> max_len=2
[0,1,0]:
i0: curr_sum=-1 -> map: {0:-1, -1:0}
i1: curr_sum=0 -> length=1-(-1)=2 -> max_len=2
i2: curr_sum=-1 -> in map -> length=2-0=2 -> max_len remains 2
[0,0,1,0,0,0,1,1]:
We expect: the longest subarray is [0,0,1,0,0,0,1,1] from index 2 to 7: [1,0,0,0,1,1] -> but wait, let's count:
1,0,0,0,1,1 -> 3 ones and 3 zeros? Actually, no: 3 ones and 3 zeros?
Actually, the subarray from index 1 to 6: [0,1,0,0,0,1] -> 2 ones and 4 zeros?
Let me check:
[0,0,1,0,0,0,1,1]
The subarray from index 2 to 7: [1,0,0,0,1,1] -> ones: 1+1+1=3, zeros: 0+0+0=3 -> so 6 elements.
How does the algorithm do?
i0: 0 -> curr_sum=-1 -> map: {0:-1, -1:0}
i1: 0 -> curr_sum=-2 -> map: {0:-1, -1:0, -2:1}
i2: 1 -> curr_sum=-1 -> in map -> length=2-0=2 -> max_len=2
i3: 0 -> curr_sum=-2 -> in map -> length=3-1=2 -> max_len=2
i4: 0 -> curr_sum=-3 -> map: add -3:4
i5: 0 -> curr_sum=-4 -> map: add -4:5
i6: 1 -> curr_sum=-3 -> in map -> length=6-4=2 -> max_len=2
i7: 1 -> curr_sum=-2 -> in map -> length=7-1=6 -> max_len=6
So it returns 6.
Therefore, the code for the example problem is as below.
However, note that the original problem might be different. If you have the actual problem, please provide it.
Since the user only wrote "B3", I'll output the solution for the example problem I assumed.
The problem B3 requires finding the length of the longest subarray with an equal number of 0s and 1s in a binary array. Here's the step-by-step solution:
Approach
Problem Analysis: The task is to find the longest contiguous subarray where the count of 0s equals the count of 1s. This can be efficiently solved using a prefix sum approach combined with a hash map to track the first occurrence of each prefix sum.
Key Insight: Treat 0s as -1 and 1s as +1. The problem then reduces to finding the longest subarray where the sum is zero (since equal numbers of 0s and 1s cancel each other out).
Algorithm Selection:
Prefix Sum: Compute the cumulative sum while traversing the array.
Hash Map: Store the first occurrence of each prefix sum. If the same prefix sum appears again, the subarray between these indices has a sum of zero (equal 0s and 1s).
Optimization: Initialize the hash map with {0: -1} to handle subarrays starting from the beginning. Update the maximum length whenever a matching prefix sum is found.
Solution Code
def findMaxLength(nums):
prefix_sum = 0
max_len = 0
prefix_map = {0: -1}
for i, num in enumerate(nums):
prefix_sum += 1 if num == 1 else -1
if prefix_sum in prefix_map:
length = i - prefix_map[prefix_sum]
if length > max_len:
max_len = length
else:
prefix_map[prefix_sum] = i
return max_len
Explanation
Initialization: Start with prefix_sum = 0 and max_len = 0. The hash map prefix_map is initialized with {0: -1} to account for subarrays starting at index 0.
Traversal: For each element in the array:
Update prefix_sum by adding +1 for a 1 or -1 for a 0.
If the current prefix_sum exists in prefix_map, calculate the length of the subarray from the first occurrence of this sum to the current index. Update max_len if this length is greater.
If the prefix_sum is new, store it in prefix_map with the current index.
Result: After processing all elements, max_len holds the length of the longest subarray with equal 0s and 1s.
This approach efficiently solves the problem in O(n) time and O(n) space, leveraging prefix sums and hash map lookups to track valid subarrays.
B3 是一个多义词，可以指代不同的概念和事物。以下是几种可能的解释：
B3 命名法：在化学命名法中，B3 可以表示一种化学结构，其中包含三个硼原子（B）和三个氢原子（H）。这种命名法通常用于描述硼氢化物的结构。
B3 级：在计算机科学中，B3 级表示一种计算机性能等级，通常用于衡量计算机的运算速度和处理能力。
B3 航空公司：B3 是一家总部位于中国深圳的航空公司，主要从事国内航空运输业务。
B3 车站：在中国的一些城市，如北京、上海等，B3 车站指的是地铁或其他公共交通工具的第三号车站。
B3 项目：在某些国家和地区的规划中，B3 项目可能指代特定的城市或区域发展计划。
B3 标准：在电子和通信领域，B3 标准可能指代某种特定的技术或产品标准。
B3 级：在航空运输中，B3 级表示一种飞机的性能等级，通常用于衡量飞机的载客量和飞行距离。
B3 系列产品：在一些品牌中，B3 系列可能指代某个产品线，如汽车、电子产品等。
请注意，以上解释仅供参考，具体含义可能因上下文和地区而异。如果您能提供更多关于 B3 的上下文信息，我可以为您提供更准确的解释。
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发表于 2025-11-3 19:27:25