花花酱 LeetCode 504. Base 7

By zxi on August 17, 2018

Problem

Given an integer, return its base 7 string representation.

Example 1:

Input: 100
Output: "202"

Example 2:

Input: -7
Output: "-10"

Note: The input will be in range of [-1e7, 1e7].

Solution: Simulation

Time complexity: O(logn)

Space complexity: O(logn)

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

string convertToBase7(int num) {

bool is_nagetive = num<0;

num = abs(num);

string ans;

do {

ans += ('0' + (num % 7));

num /= 7;

} while (num > 0);

if (is_nagetive)

ans += "-";

reverse(begin(ans), end(ans));

return ans;

}

};

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 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 888. Uncommon Words from Two Sentences

By zxi on August 12, 2018

Problem

We are given two sentences A and B. (A sentence is a string of space separated words. Each word consists only of lowercase letters.)

A word is uncommon if it appears exactly once in one of the sentences, and does not appear in the other sentence.

Return a list of all uncommon words.

You may return the list in any order.

Example 1:

Input: A = "this apple is sweet", B = "this apple is sour"
Output: ["sweet","sour"]

Example 2:

Input: A = "apple apple", B = "banana"
Output: ["banana"]

Note:

0 <= A.length <= 200
0 <= B.length <= 200
A and B both contain only spaces and lowercase letters.

Solution: HashTable

Time complexity: O(m+n)

Space complexity: O(m+n)

C++

class Solution {

public:

vector<string> uncommonFromSentences(string A, string B) {

istringstream ss(A + " " + B);

unordered_map<string, int> counts;

string s;

while (ss >> s) ++counts[s];

vector<string> ans;

for (const auto& p : counts)

if (p.second == 1)

ans.push_back(p.first);

return ans;

}

};

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 “string”

花花酱 LeetCode 504. Base 7

Problem

Solution: Simulation

花花酱 LeetCode 551. Student Attendance Record I

Problem

Solution1: Simulation

Solution 2: Regex

花花酱 LeetCode 655. Print Binary Tree

Problem

Solution: Recursion

C++

Python3

花花酱 LeetCode 888. Uncommon Words from Two Sentences

Problem

Solution: HashTable

花花酱 LeetCode 97. Interleaving String

Problem

Solution: DP