花花酱 LeetCode 551. Student Attendance Record I

By zxi on August 17, 2018

Problem

You are given a string representing an attendance record for a student. The record only contains the following three characters:

‘A’ : Absent.
‘L’ : Late.
‘P’ : Present.

A student could be rewarded if his attendance record doesn’t contain more than one ‘A’ (absent) or more than two continuous ‘L’ (late).

You need to return whether the student could be rewarded according to his attendance record.

Example 1:

Input: "PPALLP"
Output: True

Example 2:

Input: "PPALLL"
Output: False

Solution1: Simulation

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

bool checkRecord(string s) {

int a{0};

int l{0};

for(char c: s) {

if (c == 'A') ++a;

if (c == 'L') ++l;

else l=0;

if (a > 1 || l > 2) return false;

}

return true;

}

};

Solution 2: Regex

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

bool checkRecord(string s) {

return !std::regex_search(s, std::regex("A.*A|LLL"));

}

};

花花酱 LeetCode 303. Range Sum Query – Immutable

By zxi on August 17, 2018

Problem

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

You may assume that the array does not change.
There are many calls to sumRange function.

Solution: Prefix sum

sums[i] = nums[0] + nums[1] + … + nums[i]

sumRange(i, j) = sums[j] – sums[i – 1]

Time complexity: pre-compute: O(n), query: O(1)

Space complexity: O(n)

// Author: Huahua

// Running time: 112 ms

class NumArray {

public:

NumArray(vector<int> nums): sums_(nums) {

if (nums.empty()) return;

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

sums_[i] += sums_[i - 1];

}

int sumRange(int i, int j) {

if (i == 0) return sums_[j];

return sums_[j] - sums_[i-1];

}

private:

vector<int> sums_;

};

花花酱 LeetCode 655. Print Binary Tree

By zxi on August 17, 2018

Problem

Print a binary tree in an m*n 2D string array following these rules:

The row number m should be equal to the height of the given binary tree.
The column number n should always be an odd number.
The root node’s value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part). You should print the left subtree in the left-bottom part and print the right subtree in the right-bottom part. The left-bottom part and the right-bottom part should have the same size. Even if one subtree is none while the other is not, you don’t need to print anything for the none subtree but still need to leave the space as large as that for the other subtree. However, if two subtrees are none, then you don’t need to leave space for both of them.
Each unused space should contain an empty string "".
Print the subtrees following the same rules.

Example 1:

Input:
     1
    /
   2
Output:
[["", "1", ""],
 ["2", "", ""]]

Example 2:

Input:
     1
    / \
   2   3
    \
     4
Output:
[["", "", "", "1", "", "", ""],
 ["", "2", "", "", "", "3", ""],
 ["", "", "4", "", "", "", ""]]

Example 3:

Input:
      1
     / \
    2   5
   / 
  3 
 / 
4 
Output:

[["",  "",  "", "",  "", "", "", "1", "",  "",  "",  "",  "", "", ""]
 ["",  "",  "", "2", "", "", "", "",  "",  "",  "",  "5", "", "", ""]
 ["",  "3", "", "",  "", "", "", "",  "",  "",  "",  "",  "", "", ""]
 ["4", "",  "", "",  "", "", "", "",  "",  "",  "",  "",  "", "", ""]]

Note: The height of binary tree is in the range of [1, 10].

Solution: Recursion

Compute the layers h of the tree.

shape of the output matrix: h * w (w = 2^h – 1)

pre-order to fill the output matrix

first layer’s root: y1 = 0, x1 = (l1 + r1) / 2 = (0 + w – 1) / 2 (center)

first layer’s left child (2nd layer): y2 = 1, x2 = (l2 + r2) = (l1 + (x1 – 1)) / 2 (center of the left half)

first layer’s right child(2nd layer): y1 = 2, x2 = (l2′ + r2′) = ((x1+1) + r1) / 2 (center of the right half)

…

Time complexity: O(m*n)

Space complexity: O(m*n)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

vector<vector<string>> printTree(TreeNode* root) {

int h = getHeight(root);

int w = (1 << h) - 1; // 2 ^ h - 1

vector<vector<string>> ans(h, vector<string>(w, ""));

fill(root, ans, 0, 0, w - 1);

return ans;

}

private:

int getHeight(TreeNode* root) {

if (!root) return 0;

return max(getHeight(root->left), getHeight(root->right)) + 1;

}

void fill(TreeNode* root, vector<vector<string>>& ans, int h, int l, int r){

if (!root) return;

int mid = (l + r) / 2;

ans[h][mid] = std::to_string(root->val);

fill(root->left, ans, h + 1, l, mid - 1);

fill(root->right, ans, h + 1, mid + 1, r);

}

};

Python3

"""

Author: Huahua

Running time: 44 ms

"""

class Solution:

def printTree(self, root):

def height(root):

if not root: return 0

return max(height(root.left), height(root.right)) + 1

def fill(root, h, l, r):

if not root: return

mid = (l + r) // 2

m[h][mid] = str(root.val)

if root.left: fill(root.left, h + 1, l, mid - 1)

if root.right: fill(root.right, h + 1, mid + 1, r)

h = height(root)

w = 2**h - 1

m = [[""] * w for _ in range(h)]

fill(root, 0, 0, w - 1)

return m

花花酱 LeetCode 561. Array Partition I

By zxi on August 16, 2018

Problem

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a₁, b₁), (a₂, b₂), …, (a_n, b_n) which makes sum of min(a_i, b_i) for all i from 1 to n as large as possible.

Example 1:

Input: [1,4,3,2]

Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

n is a positive integer, which is in the range of [1, 10000].
All the integers in the array will be in the range of [-10000, 10000].

Solution 1: Sorting

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 52 ms

class Solution {

public:

int arrayPairSum(vector<int>& nums) {

std::sort(nums.begin(), nums.end());

int ans = 0;

for(int i = 0; i < nums.size(); i += 2)

ans += nums[i];

return ans;

}

};

Solution 2: HashTable

Time complexity: O(n + max(nums) – min(nums))

Space complexity: O(max(nums) – min(nums))

// Author: Huahua

// Running time: 39 ms

class Solution {

public:

int arrayPairSum(vector<int>& nums) {

const int max_val = *max_element(begin(nums), end(nums));

const int min_val = *min_element(begin(nums), end(nums));

const int offset = -min_val;

vector<int> c(max_val - min_val + 1);

for (int num : nums)

++c[num + offset];

int ans = 0;

int index = 0;

int n = min_val;

while (n <= max_val) {

if (c[n + offset] == 0) {

++n;

continue;

}

ans += (++index & 1) ? n : 0;

--c[n + offset];

}

return ans;

}

};

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;

}

};

Huahua's Tech Road

花花酱 LeetCode 551. Student Attendance Record I

Problem

Solution1: Simulation

Solution 2: Regex

花花酱 LeetCode 303. Range Sum Query – Immutable

Problem

Solution: Prefix sum

花花酱 LeetCode 655. Print Binary Tree

Problem

Solution: Recursion

C++

Python3

花花酱 LeetCode 561. Array Partition I

Problem

Solution 1: Sorting

Solution 2: HashTable

花花酱 LeetCode 566. Reshape the Matrix

Problem

Solution1: Brute Force