Contiguous Array
1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:
- We are given an array of 0s and 1s.
- We need to find the maximum length of a contiguous subarray with an equal number of 0s and 1s.
2. Analyze the problem: To solve this problem, we can use a prefix sum technique. We iterate through the array and calculate the prefix sum by adding 1 for each 1 encountered and subtracting 1 for each 0 encountered. We store the prefix sum along with its corresponding index in a dictionary. If we encounter the same prefix sum again, it means that the subarray between the two indices has an equal number of 0s and 1s. We can calculate the length of this subarray and update the maximum length if necessary.
3. Design an algorithm: Here is the algorithm to solve the problem:
- Initialize a variable
max_length
to 0. - Initialize a variable
prefix_sum
to 0. - Create an empty dictionary
prefix_sum_map
. - Iterate through the array:
- For each element in the array:
- If the element is 1, add 1 to
prefix_sum
. - If the element is 0, subtract 1 from
prefix_sum
. - If
prefix_sum
is 0, updatemax_length
to the current index + 1. - If
prefix_sum
is already inprefix_sum_map
, updatemax_length
to the maximum ofmax_length
and the difference between the current index and the index stored inprefix_sum_map[prefix_sum]
. - If
prefix_sum
is not inprefix_sum_map
, addprefix_sum
as a key toprefix_sum_map
with the current index as its value.
- If the element is 1, add 1 to
- For each element in the array:
- Return
max_length
.
4. Explain your approach: The approach involves using a prefix sum technique and a dictionary to keep track of the prefix sums encountered so far. We iterate through the array and calculate the prefix sum at each index. If we encounter the same prefix sum again, it means that the subarray between the two indices has an equal number of 0s and 1s. We update the maximum length accordingly.
5. Write clean and readable code:
python
def findMaxLength(nums):
max_length = 0
prefix_sum = 0
prefix_sum_map = {0: -1}
for i in range(len(nums)):
if nums[i] == 1:
prefix_sum += 1
else:
prefix_sum -= 1
if prefix_sum in prefix_sum_map:
max_length = max(max_length, i - prefix_sum_map[prefix_sum])
else:
prefix_sum_map[prefix_sum] = i
return max_length
6. Test your code: Let's test the code with some test cases:
Test case 1:
- nums = [0, 1]
- The expected output is 2 because the subarray [0, 1] has an equal number of 0s and 1s.
Test case 2:
- nums = [0, 1, 0]
- The expected output is 2 because the subarray [0, 1] or [1, 0] has an equal number of 0s and 1s.
python
# Test case 1
nums1 = [0, 1]
print(findMaxLength(nums1)) # Expected output: 2
# Test case 2
nums2 = [0, 1, 0]
print(findMaxLength(nums2)) # Expected output: 2
7. Optimize if necessary:
The current solution has a time complexity of O(n), where n is the length of the input array nums
. We iterate through the array once, and the dictionary operations take constant time on average. The space complexity is O(n) as we store the prefix sums in the dictionary. The solution is already optimal.
8. Handle error cases:
The code assumes that the input nums
is a valid array of 0s and 1s. If the input is None or an empty array, the code will return 0.
9. Discuss complexity analysis:
The time complexity of the solution is O(n), where n is the length of the input array nums
. We iterate through the array once, and the dictionary operations take constant time on average. The space complexity is O(n) as we store the prefix sums in the dictionary. The solution has linear time and space complexity.