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. A common pitfall is to overlook the need for contiguous subarrays and mistakenly consider non-contiguous subarrays.
To solve this problem, we can use a brute-force approach that checks all possible subarrays and determines if they contain unique elements. Although this approach is not optimal, it meets the problem's constraints of O(n^3) time complexity.
The naive solution involves generating all possible subarrays and checking each one for uniqueness. This approach is straightforward but inefficient for large arrays.
While the naive solution is acceptable for this problem, we can discuss a more optimized approach for educational purposes. A sliding window technique can be used to achieve a more efficient solution, but it exceeds the given constraints.
Here is a step-by-step breakdown of the naive algorithm:
maxLength
to store the maximum length of subarrays found.maxLength
if the current subarray length is greater than maxLength
.maxLength
.public class LongestSubarrayWithoutRepeating {
public static int lengthOfLongestSubarray(int[] nums) {
int maxLength = 0;
// Iterate over all possible subarrays
for (int i = 0; i < nums.length; i++) {
for (int j = i; j < nums.length; j++) {
if (allUnique(nums, i, j)) {
maxLength = Math.max(maxLength, j - i + 1);
}
}
}
return maxLength;
}
// Helper function to check if all elements in the subarray are unique
private static boolean allUnique(int[] nums, int start, int end) {
for (int i = start; i <= end; i++) {
for (int j = i + 1; j <= end; j++) {
if (nums[i] == nums[j]) {
return false;
}
}
}
return true;
}
public static void main(String[] args) {
int[] nums = {2, 5, 6, 2, 3, 1, 5, 6};
System.out.println("Length of the longest subarray without repeating integers: " + lengthOfLongestSubarray(nums));
}
}
The time complexity of the naive approach is O(n^3) due to the three nested loops. The space complexity is O(1) as we are not using any additional data structures that grow with input size.
Potential edge cases include:
Input: nums =[]
Output: 0 Input: nums =[1, 2, 3, 4, 5]
Output: 5 Input: nums =[1, 1, 1, 1]
Output: 1
To test the solution comprehensively, consider a variety of test cases:
When approaching such problems, consider the following tips:
In this blog post, we discussed the problem of finding the longest subarray without repeating integers. We explored a naive solution that meets the given constraints and provided a detailed explanation of the algorithm and its implementation in Java. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
For further reading and practice, consider the following resources: