comments | difficulty | edit_url | rating | source | tags | |
---|---|---|---|---|---|---|
true |
Easy |
1217 |
Weekly Contest 392 Q1 |
|
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 <= 50
1 <= nums[i] <= 50
We first perform a pass to find the length of the longest strictly increasing subarray, and update the answer. Then we perform another pass to find the length of the longest strictly decreasing subarray, and update the answer again.
The time complexity is
class Solution:
def longestMonotonicSubarray(self, nums: List[int]) -> int:
ans = t = 1
for i, x in enumerate(nums[1:]):
if nums[i] < x:
t += 1
ans = max(ans, t)
else:
t = 1
t = 1
for i, x in enumerate(nums[1:]):
if nums[i] > x:
t += 1
ans = max(ans, t)
else:
t = 1
return ans
class Solution {
public int longestMonotonicSubarray(int[] nums) {
int ans = 1;
for (int i = 1, t = 1; i < nums.length; ++i) {
if (nums[i - 1] < nums[i]) {
ans = Math.max(ans, ++t);
} else {
t = 1;
}
}
for (int i = 1, t = 1; i < nums.length; ++i) {
if (nums[i - 1] > nums[i]) {
ans = Math.max(ans, ++t);
} else {
t = 1;
}
}
return ans;
}
}
class Solution {
public:
int longestMonotonicSubarray(vector<int>& nums) {
int ans = 1;
for (int i = 1, t = 1; i < nums.size(); ++i) {
if (nums[i - 1] < nums[i]) {
ans = max(ans, ++t);
} else {
t = 1;
}
}
for (int i = 1, t = 1; i < nums.size(); ++i) {
if (nums[i - 1] > nums[i]) {
ans = max(ans, ++t);
} else {
t = 1;
}
}
return ans;
}
};
func longestMonotonicSubarray(nums []int) int {
ans := 1
t := 1
for i, x := range nums[1:] {
if nums[i] < x {
t++
ans = max(ans, t)
} else {
t = 1
}
}
t = 1
for i, x := range nums[1:] {
if nums[i] > x {
t++
ans = max(ans, t)
} else {
t = 1
}
}
return ans
}
function longestMonotonicSubarray(nums: number[]): number {
let ans = 1;
for (let i = 1, t = 1; i < nums.length; ++i) {
if (nums[i - 1] < nums[i]) {
ans = Math.max(ans, ++t);
} else {
t = 1;
}
}
for (let i = 1, t = 1; i < nums.length; ++i) {
if (nums[i - 1] > nums[i]) {
ans = Math.max(ans, ++t);
} else {
t = 1;
}
}
return ans;
}