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Formatted question description: https://leetcode.ca/all/845.html

# 845. Longest Mountain in Array (Medium)

Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:

• B.length >= 3
• There exists some 0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]

(Note that B could be any subarray of A, including the entire array A.)

Given an array A of integers, return the length of the longest mountain

Return 0 if there is no mountain.

Example 1:

Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.


Example 2:

Input: [2,2,2]
Output: 0
Explanation: There is no mountain.


Note:

1. 0 <= A.length <= 10000
2. 0 <= A[i] <= 10000

• Can you solve it using only one pass?
• Can you solve it in O(1) space?

Related Topics:
Two Pointers

## Solution 1.

• class Solution {
public int longestMountain(int[] A) {
int length = A.length;
int[] ascending = new int[length];
for (int i = 1; i < length; i++) {
if (A[i] > A[i - 1])
ascending[i] = ascending[i - 1] + 1;
}
int[] descending = new int[length];
for (int i = length - 2; i >= 0; i--) {
if (A[i] > A[i + 1])
descending[i] = descending[i + 1] + 1;
}
int longest = 0;
int start = 1, end = length - 1;
for (int i = start; i < end; i++) {
if (ascending[i] > 0 && descending[i] > 0) {
int mountainLength = ascending[i] + descending[i] + 1;
longest = Math.max(longest, mountainLength);
}
}
return longest;
}
}

############

class Solution {
public int longestMountain(int[] arr) {
int left = 0, right = 0;
int ans = 0;
int status = -1;
while (++right < arr.length) {
if (status == -1 || status == 1) {
if (arr[right] == arr[right - 1]) {
status = -1;
}
if (status == -1) {
if (arr[right] > arr[right - 1]) {
status = 1;
} else {
left = right;
}
}
if (status == 1 && arr[right] < arr[right - 1]) {
status = 2;
}
} else {
if (arr[right] > arr[right - 1]) {
status = 1;
ans = Math.max(right - left, ans);
left = right - 1;
} else if (arr[right] == arr[right - 1]) {
status = -1;
ans = Math.max(right - left, ans);
left = right;
}
}
}
if (status == 2) {
ans = Math.max(ans, right - left);
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/longest-mountain-in-array/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int longestMountain(vector<int>& A) {
int dir = 0, N = A.size(), start = -1, ans = 0;
for (int i = 1; i < N; ++i) {
int d = A[i] == A[i - 1] ? 0 : (A[i] > A[i - 1] ? 1 : -1); // the new direction
if (d == dir) continue; // if the direction is the same, skip
if (dir == -1 && start != -1) ans = max(ans, i - start); // if we have a valid starting point and we are going downwards, we can try to update the answer
dir = d;
if (dir == 1) start = i - 1; // if new direction is upwards, set the starting point
else if (dir == 0) start = -1; // going horizontal will invalidate the starting point.
}
if (dir == -1 && start != -1) ans = max(ans, N - start); // handle the case where the downward section reaches the end.
return ans;
}
};

• class Solution:
def longestMountain(self, arr: List[int]) -> int:
left, right = 0, 1
status = -1
ans = 0
while right < len(arr):
if status == -1 or status == 1:
if arr[right] == arr[right - 1]:
status = -1
if status == -1:
if arr[right] > arr[right - 1]:
status = 1
else:
left = right
if status == 1 and arr[right] < arr[right - 1]:
status = 2
else:
if arr[right] == arr[right - 1]:
status = -1
ans = max(ans, right - left)
left = right
elif arr[right] > arr[right - 1]:
status = 1
ans = max(ans, right - left)
left = right - 1
right += 1
if status == 2:
ans = max(right - left, ans)
return ans

############

class Solution(object):
def longestMountain(self, A):
"""
:type A: List[int]
:rtype: int
"""
N = len(A)
inc = [1] * N
dec = [1] * N
for i in range(1, N):
if A[i] - A[i - 1] > 0:
inc[i] = inc[i - 1] + 1
for i in range(N - 2, -1, -1):
if A[i] - A[i + 1] > 0:
dec[i] = dec[i + 1] + 1
res = 0
for i in range(1, N - 1):
if inc[i] != 1 and dec[i] != 1:
res = max(res, inc[i] + dec[i] - 1)
return res

• func longestMountain(arr []int) (ans int) {
n := len(arr)
for l, r := 0, 0; l+2 < n; l = r {
r = l + 1
if arr[l] < arr[r] {
for r+1 < n && arr[r] < arr[r+1] {
r++
}
if r+1 < n && arr[r] > arr[r+1] {
for r+1 < n && arr[r] > arr[r+1] {
r++
}
ans = max(ans, r-l+1)
} else {
r++
}
}
}
return
}

func max(a, b int) int {
if a > b {
return a
}
return b
}