Problem
Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.
Example 1:
Input: nums = [8,2,4,7], limit = 4
Output: 2
Explanation: All subarrays are:
[8] with maximum absolute diff |8-8| = 0 <= 4.
[8,2] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
[2] with maximum absolute diff |2-2| = 0 <= 4.
[2,4] with maximum absolute diff |2-4| = 2 <= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
[4] with maximum absolute diff |4-4| = 0 <= 4.
[4,7] with maximum absolute diff |4-7| = 3 <= 4.
[7] with maximum absolute diff |7-7| = 0 <= 4.
Therefore, the size of the longest subarray is 2.
Example 2:
Input: nums = [10,1,2,4,7,2], limit = 5
Output: 4
Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
Example 3:
Input: nums = [4,2,2,2,4,4,2,2], limit = 0
Output: 3
Constraints:
1 <= nums.length <= 10^51 <= nums[i] <= 10^90 <= limit <= 10^9
Solution (Java)
class Solution {
public int longestSubarray(int[] nums, int limit) {
ArrayDeque<Integer> maxQ = new ArrayDeque<>();
ArrayDeque<Integer> minQ = new ArrayDeque<>();
int best = 0;
int left = 0;
for (int right = 0; right < nums.length; right++) {
while (!maxQ.isEmpty() && nums[right] > nums[maxQ.peekLast()]) {
maxQ.removeLast();
}
maxQ.offerLast(right);
while (!minQ.isEmpty() && nums[right] < nums[minQ.peekLast()]) {
minQ.removeLast();
}
minQ.offerLast(right);
while (nums[maxQ.peekFirst()] - nums[minQ.peekFirst()] > limit) {
if (maxQ.peekFirst() == left) {
maxQ.removeFirst();
}
if (minQ.peekFirst() == left) {
minQ.removeFirst();
}
left++;
}
best = Math.max(best, right - left + 1);
}
return best;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).