Posts tagged as “medium”

花花酱 LeetCode 223. Rectangle Area

By zxi on September 19, 2018

Problem

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Rectangle Area

Example:

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

Solution:

area1 + area2 – overlapped area

Time complexity: O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {

int area1 = (C - A) * (D - B);

int area2 = (G - E) * (H - F);

int x1 = max(A, E);

int x2 = max(x1, min(C, G));

int y1 = max(B, F);

int y2 = max(y1, min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1);

}

};

Java

// Author: Huahua

class Solution {

public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {

int area1 = (C - A) * (D - B);

int area2 = (G - E) * (H - F);

int x1 = Math.max(A, E);

int x2 = Math.max(x1, Math.min(C, G));

int y1 = Math.max(B, F);

int y2 = Math.max(y1, Math.min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1);

}

Python3

# Author: Huahua

class Solution:

def computeArea(self, A, B, C, D, E, F, G, H):

area1 = (C - A) * (D - B);

area2 = (G - E) * (H - F);

x1 = max(A, E);

x2 = max(x1, min(C, G));

y1 = max(B, F);

y2 = max(y1, min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1)

花花酱 LeetCode 236. Lowest Common Ancestor of a Binary Tree

By zxi on September 14, 2018

Problem

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes’ values will be unique.
p and q are different and both values will exist in the binary tree.

Solution 1: Recursion

Time complexity: O(n)

Space complexity: O(h)

For a given root, recursively call LCA(root.left, p, q) and LCA(root.right, p, q)

if both returns a valid node which means p, q are in different subtrees, then root will be their LCA.

if only one valid node returns, which means p, q are in the same subtree, return that valid node as their LCA.

C++

// Author: Huahua

class Solution {

public:

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

if (!root || root == p || root == q) return root;

TreeNode* l = lowestCommonAncestor(root->left, p, q);

TreeNode* r = lowestCommonAncestor(root->right, p, q);

if (l && r) return root;

return l ? l : r;

}

};

Java

// Author: Huahua

class Solution {

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

if (root == null || root == p || root == q) return root;

TreeNode l = lowestCommonAncestor(root.left, p, q);

TreeNode r = lowestCommonAncestor(root.right, p, q);

if (l == null || r == null) return l == null ? r : l;

return root;

}

Python3

# Author: Huahua

class Solution:

def lowestCommonAncestor(self, root, p, q):

if any((not root, root == p, root == q)): return root

l = self.lowestCommonAncestor(root.left, p, q)

r = self.lowestCommonAncestor(root.right, p, q)

if not l or not r: return l if l else r

return root

Problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Solution: HashTable + Simulation

Map integer 1,4,5,9,10,40,50,90, …, 1000 to Romain

Start from the largest number y,

if x >= y:
ans += Roman[y]
x -= y

Time complexity: O(x)

Space complexity: O(x)

C++

// Author: Huahua

class Solution {

public:

string intToRoman(int num) {

vector<pair<int,string>> v{{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"},

{ 100, "C"}, { 90, "XC"}, { 50, "L"}, { 40, "XL"},

{ 10, "X"}, { 9, "IX"}, { 5, "V"}, { 4, "IV"},

{ 1, "I"}};

string ans;

auto it = cbegin(v);

while (num) {

if (num >= it->first) {

num -= it->first;

ans += it->second;

} else {

++it;

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def intToRoman(self, num):

m = [[1000, "M"], [900, "CM"], [500, "D"], [400, "CD"],

[ 100, "C"], [ 90, "XC"], [ 50, "L"], [ 40, "XL"],

[ 10, "X"], [ 9, "IX"], [ 5, "V"], [ 4, "IV"],

[ 1, "I"]]

index = 0

ans = ''

while num > 0:

if num >= m[index][0]:

ans += m[index][1]

num -= m[index][0]

else:

index += 1

return ans

花花酱 LeetCode 8. String to Integer (atoi)

By zxi on September 12, 2018

Problem

Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. If the numerical value is out of the range of representable values, INT_MAX (2³¹− 1) or INT_MIN (−2³¹) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−2³¹) is returned.

Solution: Simulation

You need to handle all corner cases in order to pass…

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int myAtoi(const string& s) {

long long ans = 0;

int neg = 0;

int pos = 0;

bool p = false; // has '.'

bool a = false; // has letters

bool n = false; // has number

for (int i = 0; i < s.length(); ++i) {

if (s[i] == ' ') { if (n || neg || pos) break; }

else if (s[i] == '-') { if (n) break; else neg++; }

else if (s[i] == '+') { if (n) break; else pos++; }

else if (s[i] == '.') p = true;

else if (s[i] >= '0' && s[i] <= '9' && !p && !a) {

ans = ans * 10 + (s[i] - '0');

n = true;

} else {

a = true;

}

if (ans / 10 > INT_MAX) break; // ans can be > 64 bit

}

if (neg) ans = -ans;

if (ans > INT_MAX) ans = INT_MAX;

if (ans < INT_MIN) ans = INT_MIN;

if (pos + neg > 1) ans = 0;

return ans;

}

};

花花酱 LeetCode 6. ZigZag Conversion

By zxi on September 12, 2018

Problem

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:
<preclass=”crayon:false”>Input: s = “PAYPALISHIRING”, numRows = 3 Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: Simulation

Store the zigzag results for each row and then output row by row.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

string convert(string s, int nRows) {

if (nRows == 1) return s;

vector<string> ss(nRows);

int l = s.length();

int x = 0, y = 0, i = 0;

bool down = true;

while (i < l) {

ss[y] += s[i];

if (down) {

++y;

if (y == nRows) {

down = false;

y -=2;

}

} else {

--y;

if (y < 0) {

down = true;

y = 1;

}

i++;

}

string ans;

for(const string& r : ss)

ans += r;

return ans;

}

};

Posts tagged as “medium”

花花酱 LeetCode 223. Rectangle Area

Problem

Solution:

C++

Java

Python3

花花酱 LeetCode 236. Lowest Common Ancestor of a Binary Tree

Problem

Solution 1: Recursion

C++

Java

Python3

Related Problems:

花花酱 LeetCode 12. Integer to Roman

Problem

Solution: HashTable + Simulation

C++

Python3

花花酱 LeetCode 8. String to Integer (atoi)

Problem

Solution: Simulation

C++

花花酱 LeetCode 6. ZigZag Conversion

Problem

Solution: Simulation

C++