Huahua’s Tech Road

花花酱 LeetCode 2085. Count Common Words With One Occurrence

By zxi on November 27, 2021

Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.

Example 1:

Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"]
Output: 2
Explanation:
- "leetcode" appears exactly once in each of the two arrays. We count this string.
- "amazing" appears exactly once in each of the two arrays. We count this string.
- "is" appears in each of the two arrays, but there are 2 occurrences of it in words1. We do not count this string.
- "as" appears once in words1, but does not appear in words2. We do not count this string.
Thus, there are 2 strings that appear exactly once in each of the two arrays.

Example 2:

Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"]
Output: 0
Explanation: There are no strings that appear in each of the two arrays.

Example 3:

Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"]
Output: 1
Explanation: The only string that appears exactly once in each of the two arrays is "ab".

Constraints:

1 <= words1.length, words2.length <= 1000
1 <= words1[i].length, words2[j].length <= 30
words1[i] and words2[j] consists only of lowercase English letters.

Solution: Hashtable

Time complexity: O(n + m)
Space complexity: O(n + m)

C++

// Author: Huahua

class Solution {

public:

int countWords(vector<string>& words1, vector<string>& words2) {

unordered_map<string, int> c1;

unordered_map<string, int> c2;

for (const string& w : words1) ++c1[w];

for (const string& w : words2) ++c2[w];

int ans = 0;

for (const auto& [w, c] : c1)

if (c == 1 && c2[w] == 1) ++ans;

return ans;

}

};

Solution 1: Binary Search

Find the smallest l such that l * l > x, sqrt(x) = l – 1.

Time complexity: O(logx)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int mySqrt(int x) {

unsigned l = 1;

unsigned r = 1u + x;

while (l < r) {

unsigned m = l + (r - l) / 2;

if (m > x / m) {

r = m;

} else {

l = m + 1;

}

return l - 1;

}

};

花花酱 LeetCode 65. Valid Number

By zxi on November 26, 2021

A valid number can be split up into these components (in order):

A decimal number or an integer.
(Optional) An 'e' or 'E', followed by an integer.

A decimal number can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
One of the following formats:
1. One or more digits, followed by a dot '.'.
2. One or more digits, followed by a dot '.', followed by one or more digits.
3. A dot '.', followed by one or more digits.

An integer can be split up into these components (in order):

(Optional) A sign character (either '+' or '-').
One or more digits.

For example, all the following are valid numbers: ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"], while the following are not valid numbers: ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].

Given a string s, return true if s is a valid number.

Example 1:

Input: s = "0"
Output: true

Example 2:

Input: s = "e"
Output: false

Example 3:

Input: s = "."
Output: false

Example 4:

Input: s = ".1"
Output: true

Constraints:

1 <= s.length <= 20
s consists of only English letters (both uppercase and lowercase), digits (0-9), plus '+', minus '-', or dot '.'.

Solution: Rule checking

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool isNumber(string s) {

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

if (s[i] != ' ') {

s = s.substr(i);

break;

}

while (!s.empty() && s.back() == ' ') s.pop_back();

bool dot = false;

bool e = false;

bool digit = false;

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

s[i] = tolower(s[i]);

if (isdigit(s[i])) {

digit = true;

} else if (s[i] == '.') {

if (e || dot) return false;

dot = true;

} else if (s[i] == 'e') {

if (e || !digit) return false;

e = true;

digit = false;

} else if (s[i] == '-' || s[i] == '+') {

if (i != 0 && s[i - 1] != 'e')

return false;

} else {

return false;

}

return digit;

}

};

花花酱 LeetCode 41. First Missing Positive

By zxi on November 26, 2021

Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses constant extra space.

Example 1:

Input: nums = [1,2,0]
Output: 3

Example 2:

Input: nums = [3,4,-1,1]
Output: 2

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1

Constraints:

1 <= nums.length <= 5 * 10⁵
-2³¹ <= nums[i] <= 2³¹ - 1

Solution: Marking

First pass, marking nums[i] to INT_MAX if nums[i] <= 0
Second pass, use a negative number to mark the presence of a number x at nums[x – 1]
Third pass, the first positive number is the missing index i, return i +1
If not found return n + 1.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int firstMissingPositive(vector<int>& nums) {

const int n = nums.size();

for (int& x : nums)

if (x <= 0) x = INT_MAX;

for (int x : nums) {

x = abs(x);

if (x >= 1 && x <= n && nums[x - 1] > 0)

nums[x - 1] *= -1;

}

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

if (nums[i] > 0)

return i + 1;

return n + 1;

}

};

花花酱 LeetCode 34. Find First and Last Position of Element in Sorted Array

By zxi on November 26, 2021

Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

Constraints:

0 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹
nums is a non-decreasing array.
-10⁹ <= target <= 10⁹

Solution: Binary Search

Use lower_bound to find first
Use upper_bound to find last + 1

Time complexity: O(logn)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

vector<int> searchRange(vector<int>& nums, int target) {

auto it = lower_bound(begin(nums), end(nums), target);

int l = (it == end(nums) || *it != target) ? -1 : it - begin(nums);

auto it2 = upper_bound(begin(nums), end(nums), target);

int r = (it2 == begin(nums) || *prev(it2) != target) ? -1 : prev(it2) - begin(nums);

return {l, r};

}

};

Huahua's Tech Road

花花酱 LeetCode 2085. Count Common Words With One Occurrence

Solution: Hashtable

C++

花花酱 LeetCode 69. Sqrt(x)

Solution 1: Binary Search

C++

花花酱 LeetCode 65. Valid Number

Solution: Rule checking

C++

花花酱 LeetCode 41. First Missing Positive

Solution: Marking

C++

花花酱 LeetCode 34. Find First and Last Position of Element in Sorted Array

C++