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 thatB[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:
0 <= A.length <= 10000
0 <= A[i] <= 10000
Follow up:
- Can you solve it using only one pass?
- Can you solve it in
O(1)
space?
Related Topics:
Two Pointers
Solution 1.
// 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;
}
};
Java
-
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; } }
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// 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; } };
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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