花花酱 LeetCode 566. Reshape the Matrix

By zxi on August 16, 2018

Problem

In MATLAB, there is a very useful function called ‘reshape’, which can reshape a matrix into a new one with different size but keep its original data.

You’re given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.

The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.

If the ‘reshape’ operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

Example 1:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 1, c = 4
Output: 
[[1,2,3,4]]
Explanation:
The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

Example 2:

Input: 
nums = 
[[1,2],
 [3,4]]
r = 2, c = 4
Output: 
[[1,2],
 [3,4]]
Explanation:
There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

Note:

The height and width of the given matrix is in range [1, 100].
The given r and c are all positive.

Solution1: Brute Force

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 32 ms

class Solution {

public:

vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {

if (nums.size() == 0) return nums;

int m = nums.size();

int n = nums[0].size();

if (m * n != r * c) return nums;

// new matrix r*c

vector<vector<int>> ans(r, vector<int>(c));

for(int i = 0; i < m * n;++i) {

int src_y = i / n;

int src_x = i % n;

int dst_y = i / c;

int dst_x = i % c;

ans[dst_y][dst_x] = nums[src_y][src_x];

}

return ans;

}

};

花花酱 LeetCode 174. Dungeon Game

By zxi on August 16, 2018

Problem

The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0’s) or contain magic orbs that increase the knight’s health (positive integers).

In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

Write a function to determine the knight’s minimum initial health so that he is able to rescue the princess.

For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

-2 (K)	-3	3
-5	-10	1
10	30	-5 (P)

Note:

The knight’s health has no upper bound.
Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

Solution: DP

Time complexity: O(mn)

Space complexity: O(mn) -> O(n)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int calculateMinimumHP(vector<vector<int>>& dungeon) {

const int m = dungeon.size();

const int n = dungeon[0].size();

// hp[y][x]: min hp required to reach bottom right (P).

vector<vector<int>> hp(m + 1, vector<int>(n + 1, INT_MAX));

hp[m][n - 1] = hp[m - 1][n] = 1;

for (int y = m - 1; y >= 0; --y)

for (int x = n - 1; x >= 0; --x)

hp[y][x] = max(1, min(hp[y + 1][x], hp[y][x + 1]) - dungeon[y][x]);

return hp[0][0];

}

};

O(n) space

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int calculateMinimumHP(vector<vector<int>>& dungeon) {

const int m = dungeon.size();

const int n = dungeon[0].size();

// hp[y][x]: min hp required to reach bottom right (P).

vector<int> hp(n + 1, INT_MAX);

hp[n - 1] = 1;

for (int y = m - 1; y >= 0; --y)

for (int x = n - 1; x >= 0; --x)

hp[x] = max(1, min(hp[x], hp[x + 1]) - dungeon[y][x]);

return hp[0];

}

};

花花酱 LeetCode 97. Interleaving String

By zxi on August 5, 2018

Problem

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Solution: DP

Subproblems : whether s3[0:i+j] can be formed by interleaving s1[0:i] and s2[0:j].

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huhaua

// Running time: 0 ms

class Solution {

public:

bool isInterleave(string s1, string s2, string s3) {

int l1 = s1.length();

int l2 = s2.length();

int l3 = s3.length();

if (l1 + l2 != l3) return false;

// dp[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]

vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1));

dp[0][0] = true;

for (int i = 0; i <= l1; ++i)

for (int j = 0; j <= l2; ++j) {

if (i > 0) dp[i][j] |= dp[i - 1][j] && s1[i - 1] == s3[i + j - 1];

if (j > 0) dp[i][j] |= dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];

}

return dp[l1][l2];

}

};

Recursion + Memorization

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

bool isInterleave(string s1, string s2, string s3) {

m_ = vector<vector<int>>(s1.length() + 1, vector<int>(s2.length() + 1, INT_MIN));

return dp(s1, s1.length(), s2, s2.length(), s3, s3.length());

}

private:

// m_[i][j]: whehter s3[0:i+j] is a interleva of s1[0:i] and s2[0:j]

vector<vector<int>> m_;

bool dp(const string& s1, int l1, const string& s2, int l2, const string& s3, int l3) {

if (l1 + l2 != l3) return false;

if (l1 == 0 && l2 == 0) return true;

if (m_[l1][l2] != INT_MIN) return m_[l1][l2];

if (l1 > 0 & s3[l3 - 1] == s1[l1 - 1] && dp(s1, l1 - 1, s2, l2, s3, l3 - 1)

||l2 > 0 & s3[l3 - 1] == s2[l2 - 1] && dp(s1, l1, s2, l2 - 1, s3, l3 - 1))

m_[l1][l2] = 1;

else

m_[l1][l2] = 0;

return m_[l1][l2];

}

};

Posts tagged as “O(mn)”

花花酱 LeetCode 566. Reshape the Matrix

Problem

Solution1: Brute Force

花花酱 LeetCode 174. Dungeon Game

Problem

Solution: DP

花花酱 LeetCode 97. Interleaving String

Problem

Solution: DP