algoadvance

Given a binary array nums, return the maximum length of a contiguous subarray with an equal number of 0 and 1.

Clarifying Questions

1. What is the length range of the nums array?
   • The array can have a length ranging from 1 to 10^5.
2. Can the array contain any values other than 0 and 1?
   • No, the array only contains 0s and 1s.
3. What should we return if there is no subarray with an equal number of 0s and 1s?
   • If there is no such subarray, we should return 0.

Strategy

To solve this problem efficiently, we can leverage a prefix sum approach but with a twist.

1. Transform the array: Let’s transform the array such that all 0s become -1. This way, the problem transforms into finding the maximum length subarray with a sum of 0.
2. Prefix Sum with HashMap: We will use a hashmap to store the first occurrence of each prefix sum. We traverse through the array, calculate the prefix sum at each step, and check if this prefix sum has been seen before:
   • If it has been seen before, that means the subarray between the previous occurrence and the current index has a sum of 0.
   • If it has not been seen before, we store the current index in the hashmap.
3. Calculate the Maximum Length: We keep track of the maximum length of subarrays where the prefix sum is zero.

Code

def findMaxLength(nums):
    # Transform the 0s in the array to -1s
    nums = [-1 if num == 0 else 1 for num in nums]

    # Dictionary to store (prefix_sum:index) pairs
    prefix_sum_map = {}
    prefix_sum = 0
    max_length = 0
    
    # Initialize the map with the base case
    prefix_sum_map[0] = -1

    for i in range(len(nums)):
        prefix_sum += nums[i]
        
        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

Time Complexity

• Time Complexity: (O(n)), where (n) is the length of the array. We traverse the array once and perform constant-time operations for each element.
• Space Complexity: (O(n)), since in the worst case, we may store all the prefix sums in the hashmap.

This approach ensures that we solve the problem optimally, given the constraints.

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