3372. Longest Strictly Increasing Or Strictly Decreasing Subarray¶
Difficulty: Easy
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3372. Longest Strictly Increasing or Strictly Decreasing Subarray
Easy
You are given an array of integers nums. Return the length of the longest subarray of nums which is either strictly increasing or strictly decreasing.
Example 1:
Input: nums = [1,4,3,3,2]
Output: 2
Explanation:
The strictly increasing subarrays of nums are [1], [2], [3], [3], [4], and [1,4].
The strictly decreasing subarrays of nums are [1], [2], [3], [3], [4], [3,2], and [4,3].
Hence, we return 2.
Example 2:
Input: nums = [3,3,3,3]
Output: 1
Explanation:
The strictly increasing subarrays of nums are [3], [3], [3], and [3].
The strictly decreasing subarrays of nums are [3], [3], [3], and [3].
Hence, we return 1.
Example 3:
Input: nums = [3,2,1]
Output: 3
Explanation:
The strictly increasing subarrays of nums are [3], [2], and [1].
The strictly decreasing subarrays of nums are [3], [2], [1], [3,2], [2,1], and [3,2,1].
Hence, we return 3.
Constraints:
1 <= nums.length <= 501 <= nums[i] <= 50
Solution¶
class Solution {
public int longestMonotonicSubarray(int[] nums) {
return Math.max(si(nums) , sd(nums));
}
public static int si(int arr[]) {
int n = arr.length;
int count = 0, maxi = 0;
for (int i = 0; i < n; i++) {
count = 0;
for (int j = i; j < n; j++) {
if (j == i) count++;
else {
if(arr[j] > arr[j - 1]) count++;
else break;
}
maxi = Math.max(maxi, count);
}
}
return maxi;
}
public static int sd(int arr[]) {
int n = arr.length;
int count = 0, maxi = 0;
for (int i = 0; i < n; i++) {
count = 0;
for (int j = i; j < n; j++) {
if (j == i) count++;
else {
if(arr[j] < arr[j - 1]) count++;
else break;
}
maxi = Math.max(maxi, count);
}
}
return maxi;
}
}
Complexity Analysis¶
- Time Complexity:
O(?) - Space Complexity:
O(?)
Approach¶
Detailed explanation of the approach will be added here