Given an input array of integers, find the length of the longest subarray without repeating integers.
Example
Input: nums = [2, 5, 6, 2, 3, 1, 5, 6]
Output: 5
Explanation: [5, 6, 2, 3, 1] or [6, 2, 3, 1, 5] are both valid and of maximum length 5
For this lesson, your algorithm should run in O(n^3) time and use O(1) extra space.
(There exist faster solutions which we will discuss in future lessons)
The problem requires finding the length of the longest subarray in a given array of integers where no integer repeats. The input is an array of integers, and the output is a single integer representing the length of the longest subarray without repeating integers.
nums
.Input: nums = [2, 5, 6, 2, 3, 1, 5, 6]
Output: 5
Explanation: [5, 6, 2, 3, 1] or [6, 2, 3, 1, 5] are both valid and of maximum length 5
The core challenge is to identify the longest contiguous subarray where all elements are unique. This problem is significant in various applications such as data stream processing, substring problems in strings, and more.
Potential pitfalls include misunderstanding the requirement for contiguous subarrays and not handling edge cases like empty arrays or arrays with all unique elements.
To solve this problem, we can start with a naive approach and then discuss more optimized solutions.
The naive solution involves checking every possible subarray and determining if it contains all unique elements. This approach is not optimal due to its high time complexity.
We can use a sliding window approach to optimize the solution. This involves maintaining a window of unique elements and expanding or contracting the window as needed.
Here is a step-by-step breakdown of the sliding window approach:
start
and end
, both set to the beginning of the array.end
pointer and adding elements to the set until a duplicate is found.start
pointer to the right until the duplicate is removed from the window.// Function to find the length of the longest subarray without repeating integers
function longestSubarrayWithoutRepeating(nums) {
let maxLength = 0; // Variable to store the maximum length of subarray
let start = 0; // Start pointer of the sliding window
// Iterate through the array with the end pointer
for (let end = 0; end < nums.length; end++) {
// Check for duplicates in the current window
for (let i = start; i < end; i++) {
if (nums[i] === nums[end]) {
// Move the start pointer to the right of the duplicate
start = i + 1;
break;
}
}
// Update the maximum length of the subarray
maxLength = Math.max(maxLength, end - start + 1);
}
return maxLength; // Return the maximum length found
}
// Example usage
const nums = [2, 5, 6, 2, 3, 1, 5, 6];
console.log(longestSubarrayWithoutRepeating(nums)); // Output: 5
The time complexity of the naive approach is O(n^3)
due to the nested loops checking for duplicates. The space complexity is O(1)
as we are not using any extra space proportional to the input size.
Consider the following edge cases:
To test the solution comprehensively, consider the following test cases:
nums = []
(Output: 0)nums = [1, 2, 3, 4, 5]
(Output: 5)nums = [1, 1, 1, 1]
(Output: 1)nums = [2, 5, 6, 2, 3, 1, 5, 6]
(Output: 5)When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the length of the longest subarray without repeating integers using a naive approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for coding interviews.
For further reading and practice, consider the following resources:
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