Contiguous Array

Updated on 08 May, 2025
Contiguous Array header image

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

  • Java
java
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;
    }
}

The provided Java code addresses the problem of finding the maximum length of a contiguous subarray with an equal number of 0's and 1's. This approach efficiently uses a hashmap to track the first occurrences of balance values, where the "balance" tracks the difference between the count of 1's and 0's.

Here's a breakdown of how the code achieves this:

  • A hashmap indexMap is initialized to store balance values and their corresponding indices. It starts by mapping a balance of 0 to an index of -1, covering the base case before the array begins.
  • A variable maxLen is initialized to keep track of the maximum length of any found subarray with equal numbers of 0's and 1's.
  • Another variable, balance, is used to keep a running difference between the number of 1's and 0's encountered as the array is iterated through. If a 1 is encountered, the balance increases by 1; if 0 is encountered, it decreases by 1.
  • As the loop progresses through the array:
    • If the current balance has occurred before (found in indexMap), it means a subarray with an equal number of 0's and 1's is found. Update maxLen by calculating the length of this subarray.
    • If this is the first occurrence of the current balance, store this balance with its index to reference in potential future matches.

The function finally returns the value of maxLen, which holds the length of the longest contiguous subarray fulfilling the condition.

To use this class and method:

  1. Ensure the class and method are accessible in the Java application or testing environment.
  2. Create an instance of the class Solution.
  3. Call the method maxSubArrayLength passing the binary array (only containing 0's and 1's) as the parameter.

This implementation is efficient, achieving the solution in linear time complexity, O(n), where n is the length of the input array, due to the single pass and constant-time hashmap operations.

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