Posts published in June 2018

花花酱 LeetCode 845. Longest Mountain in Array

By zxi on June 3, 2018

Problem

题目大意：找出最长的山形子数组。

https://leetcode.com/problems/longest-mountain-in-array/description/

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:

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

Solution: DP

Three passes

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int longestMountain(vector<int>& A) {

vector<int> inc(A.size());

vector<int> dec(A.size());

for (int i = 1; i < A.size(); ++i)

if (A[i] > A[i - 1]) inc[i] = inc[i - 1] + 1;

for (int i = A.size() - 2; i > 0; --i)

if (A[i] > A[i + 1]) dec[i] = dec[i + 1] + 1;

int ans = 0;

for (int i = 0; i < A.size(); ++i)

if (inc[i] && dec[i])

ans = max(ans, inc[i] + dec[i] + 1);

return ans >= 3 ? ans : 0;

}

};

One pass

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int longestMountain(vector<int>& A) {

int inc = 0;

int dec = 0;

int ans = 0;

for (int i = 1; i < A.size(); ++i) {

if (dec && A[i] > A[i - 1] || A[i] == A[i - 1])

dec = inc = 0;

inc += A[i] > A[i - 1];

dec += A[i] < A[i - 1];

if (inc && dec)

ans = max(ans, inc + dec + 1);

}

return ans >= 3 ? ans : 0;

}

};

花花酱 LeetCode 844. Backspace String Compare

By zxi on June 3, 2018

Problem

题目大意：给你2个字符串表示打字顺序，判断它们的结果是否相同，’#’表示退格键。

https://leetcode.com/problems/backspace-string-compare/description/

Given two strings S and T, return if they are equal when both are typed into empty text editors. # means a backspace character.

Example 1:

Input: S = "ab#c", T = "ad#c"
Output: true
Explanation: Both S and T become "ac".

Example 2:

Input: S = "ab##", T = "c#d#"
Output: true
Explanation: Both S and T become "".

Example 3:

Input: S = "a##c", T = "#a#c"
Output: true
Explanation: Both S and T become "c".

Example 4:

Input: S = "a#c", T = "b"
Output: false
Explanation: S becomes "c" while T becomes "b".

Note:

1 <= S.length <= 200
1 <= T.length <= 200
S and T only contain lowercase letters and '#' characters.

Solution: Simulation

Time complexity: O(|S| + |T|)

Space complexity: O(|S| + |T|)

C++

// Author: Huahua

// Running time: 5 ms

class Solution {

public:

bool backspaceCompare(string S, string T) {

return type(S) == type(T);

}

private:

string type(string S) {

string o;

for (char c : S)

if (c == '#') {

if (!o.empty())

o.pop_back();

} else {

o.push_back(c);

}

return o;

}

};

Java

// Author: Huahua, 7 ms

class Solution {

public boolean backspaceCompare(String S, String T) {

return type(S).equals(type(T));

}

private String type(String s) {

StringBuilder sb = new StringBuilder();

for (char c : s.toCharArray()) {

if (c == '#') {

if (sb.length() > 0)

sb.setLength(sb.length() - 1);

} else {

sb.append(c);

}

return sb.toString();

}

Python3

# Author: Huahua, 40 ms

class Solution:

def backspaceCompare(self, S, T):

def sim(S):

ans = ''

for c in S:

if c == '#':

if len(ans) > 0: ans = ans[:-1]

else:

ans += c

return ans

return sim(S) == sim(T)

花花酱 LeetCode 492. Construct the Rectangle

By zxi on June 2, 2018

Problem

For a web developer, it is very important to know how to design a web page’s size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Note:

The given area won’t exceed 10,000,000 and is a positive integer
The web page’s width and length you designed must be positive integers.

Solution

Time complexity: O(sqrt(n))

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 3 ms (<98.48%)

class Solution {

public:

vector<int> constructRectangle(int area) {

for (int i = sqrt(area); i >= 1; --i)

if (area % i == 0) return {area / i, i};

}

};

花花酱 LeetCode 576. Out of Boundary Paths

By zxi on June 2, 2018

Problem

There is an m by n grid with a ball. Given the start coordinate (i,j) of the ball, you can move the ball to adjacent cell or cross the grid boundary in four directions (up, down, left, right). However, you can at most move N times. Find out the number of paths to move the ball out of grid boundary. The answer may be very large, return it after mod 10⁹ + 7.

Example 1:

Input:m = 2, n = 2, N = 2, i = 0, j = 0
Output: 6
Explanation:

Example 2:

Input:m = 1, n = 3, N = 3, i = 0, j = 1
Output: 12
Explanation:

Note:

Once you move the ball out of boundary, you cannot move it back.
The length and height of the grid is in range [1,50].
N is in range [0,50].

Solution

Time complexity: O(m*n + N * m * n)

Space complexity: O(N*m*n) -> O(m*n)

C++

// Author: Huahua

// Running time: 19 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

constexpr int kMod = 1000000007;

vector<vector<vector<int>>> dp(N + 1, vector<vector<int>>(m, vector<int>(n, 0)));

vector<int> dirs{-1, 0, 1, 0, -1};

for (int s = 1; s <= N; ++s)

for (int y = 0; y < m; ++y)

for (int x = 0; x < n; ++x)

for (int d = 0; d < 4; ++d) {

int tx = x + dirs[d];

int ty = y + dirs[d + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= m)

dp[s][y][x] += 1;

else

dp[s][y][x] = (dp[s][y][x] + dp[s - 1][ty][tx]) % kMod;

}

return dp[N][i][j];

}

};

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

constexpr int kMod = 1000000007;

vector<vector<int>> dp(m, vector<int>(n, 0));

vector<int> dirs{-1, 0, 1, 0, -1};

for (int s = 1; s <= N; ++s) {

vector<vector<int>> tmp(m, vector<int>(n, 0));

for (int y = 0; y < m; ++y)

for (int x = 0; x < n; ++x)

for (int d = 0; d < 4; ++d) {

int tx = x + dirs[d];

int ty = y + dirs[d + 1];

if (tx < 0 || ty < 0 || tx >= n || ty >= m)

tmp[y][x] += 1;

else

tmp[y][x] = (tmp[y][x] + dp[ty][tx]) % kMod;

}

dp.swap(tmp);

}

return dp[i][j];

}

};

Recursion with memorization

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

m_ = m;

n_ = n;

dp_ = vector<vector<vector<int>>>(N + 1, vector<vector<int>>(m, vector<int>(n, INT_MIN)));

return paths(N, j, i);

}

private:

vector<vector<vector<int>>> dp_;

int m_;

int n_;

// number of paths starts from out-of-boundary to (x, y) by moving at most N steps.

int paths(int N, int x, int y) {

static vector<int> dirs{-1, 0, 1, 0, -1};

static const int kMod = 1000000007;

// Out of boundary, one path.

if (x < 0 || x >= n_ || y < 0 || y >= m_) return 1;

// Can not move but still in the grid, no path.

if (N == 0) return 0;

// Already computed.

if (dp_[N][y][x] != INT_MIN) return dp_[N][y][x];

dp_[N][y][x] = 0;

for (int d = 0; d < 4; ++d) {

int tx = x + dirs[d];

int ty = y + dirs[d + 1];

dp_[N][y][x] = (dp_[N][y][x] + paths(N - 1, tx, ty)) % kMod;

}

return dp_[N][y][x];

}

};

no loops

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int findPaths(int m, int n, int N, int i, int j) {

m_ = m;

n_ = n;

dp_ = vector<vector<vector<int>>>(N + 1, vector<vector<int>>(m, vector<int>(n, INT_MIN)));

return paths(N, j, i);

}

private:

vector<vector<vector<int>>> dp_;

int m_;

int n_;

// number of paths start from (x, y) and move at most N steps.

int paths(int N, int x, int y) {

static const int kMod = 1000000007;

if (x < 0 || x >= n_ || y < 0 || y >= m_) return 1;

if (N == 0) return 0;

if (dp_[N][y][x] != INT_MIN) return dp_[N][y][x];

long ans = 0;

ans += paths(N - 1, x + 1, y);

ans += paths(N - 1, x - 1, y);

ans += paths(N - 1, x, y + 1);

ans += paths(N - 1, x, y - 1);

ans %= kMod;

return dp_[N][y][x] = ans;

}

};

Posts published in June 2018

花花酱 LeetCode 845. Longest Mountain in Array

Problem

Solution: DP

花花酱 LeetCode 844. Backspace String Compare

Problem

Solution: Simulation

C++

Java

Python3

花花酱 LeetCode 492. Construct the Rectangle

Problem

Solution

花花酱 LeetCode 576. Out of Boundary Paths

Problem

Solution

Related Problems