Contiguous array leetcode 525 python
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certainly! the "contiguous array" problem is a classic coding problem that can be found on platforms like leetcode. the problem is defined as follows:
problem statement
given a binary array `nums`, find the maximum length of a contiguous subarray with an equal number of 0 and 1.
example
approach
to solve this problem efficiently, we can use the concept of prefix sums combined with a hash map (dictionary in python). the idea is to treat `0` as `-1`, and `1` as `1`. this way, we can find the longest contiguous subarray with a sum of `0`.
1. **transform the array**: convert all `0`s to `-1`s. this means that finding a subarray with an equal number of `0`s and `1`s translates to finding a subarray with a sum of `0`.
2. **use a prefix sum**: keep a running sum of the elements. if the same sum has been seen before, it means that the elements between those two indices sum to `0`.
3. **use a hash map to store the first occurrence of each prefix sum**: this helps in determining the length of the subarray when we encounter the same sum again.
implementation
here’s the python code to implement the above logic:
explanation of the code
we create a dictionary `prefix_sum_index` to store the first index where each prefix sum is encountered.
we initialize `max_length` to store the maximum length of the contiguous subarray found so far.
we iterate through the `nums` array, calculating the `current_sum` by treating `0` as `-1` and `1` as `1`.
if the `current_sum` has been seen before, we compute the length of the subarray from the previous index to the current index. we update `max_length` if this length is greater than the previously recorded maximum.
if `current_sum` has not been seen, we record its index in the dictionary.
finally, we return the `max_length`.
time complexity
the time complexity of this solution is \(o(n)\), where \(n\) is the length of the input array. we only make a single pass through the array.
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