Contiguous Array

Problem Statement

In this task, we are presented with a binary array called nums. Our goal is to determine the length of the longest contiguous subarray where the number of 0s and the number 1s are the same. This problem tests our ability to efficiently traverse and analyze subsections of an array to balance specified criteria.

Examples

Example 1

Input:

nums = [0,1]

Output:

2

Explanation:

[0, 1] is the longest contiguous subarray with an equal number of 0 and 1.

Example 2

Input:

nums = [0,1,0]

Output:

2

Explanation:

[0, 1] (or [1, 0]) is a longest contiguous subarray with equal number of 0 and 1.

Constraints

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Approach and Intuition

Given the nature of our problem, solving it requires an understanding of subarray dynamics and the delicate balance between occurrences of elements. Here’s a systematic breakdown of potential approaches and intuitions drawn from the examples:

1. Understanding the Basic Idea:

   • We need a subarray (continuous part of the original array) where the count of 0s equals the count of 1s. This implies that for any length of the subarray, the number of 1s we add is exactly offset by the number of 0s.

2. Illustrative Examples Explanation:

   • Example 1: Here, nums = [0,1]. This array itself is the largest subarray where the count of 0s and 1s are equal, giving us the maximum length of 2.
   • Example 2: In nums = [0,1,0], the optimal subarrays could be [0,1] or [1,0], both providing a balanced subarray with a length of 2.

3. Using a HashMap to Track Balances:

   • A possible approach involves using a hashmap to keep track of the balance between 0s and 1s as we iterate through the array. Here balance can be calculated by considering 0 as -1 and 1 as +1. Hence, the balance changes as per the element we encounter. This balance helps us understand if and when the subarray returns to a previous state.
   • Specifically, the balance at each index i is defined as the number of 1s minus the number of 0s from the start to index i. If this balance repeats at index j, it means the subarray between these indices has an equal number of 0s and 1s.

4. Selecting the Right Data Structure:

   • Use a map to store the first occurrence of each balance level. If the same balance level is reached again at a later index, then we can calculate the length of the subarray that lies between these repetitive balance indices.

5. Edge Cases:

   • Single element arrays or arrays where all elements are 0 or all are 1 can quickly be checked at the start as they will not contribute to a valid required subarray. This can enhance efficiency by reducing unnecessary computation.

By considering these steps and underlying methodologies, the problem can be addressed systematically, ensuring that we efficiently find the required subarray without exhaustive search.

Solutions

public class Solution {
    
    public int maxSubArrayLength(int[] nums) {
        Map<Integer, Integer> indexMap = new HashMap<>();
        indexMap.put(0, -1);
        int maxLen = 0, balance = 0;
        for (int index = 0; index < nums.length; index++) {
            balance += (nums[index] == 1 ? 1 : -1);
            if (indexMap.containsKey(balance)) {
                maxLen = Math.max(maxLen, index - indexMap.get(balance));
            } else {
                indexMap.put(balance, index);
            }
        }
        return maxLen;
    }
}