Huahua’s Tech Road

花花酱 LeetCode 1984. Minimum Difference Between Highest and Lowest of K Scores

By zxi on December 31, 2021

You are given a 0-indexed integer array nums, where nums[i] represents the score of the i^th student. You are also given an integer k.

Pick the scores of any k students from the array so that the difference between the highest and the lowest of the k scores is minimized.

Return the minimum possible difference.

Example 1:

Input: nums = [90], k = 1
Output: 0
Explanation: There is one way to pick score(s) of one student:
- [90]. The difference between the highest and lowest score is 90 - 90 = 0.
The minimum possible difference is 0.

Example 2:

Input: nums = [9,4,1,7], k = 2
Output: 2
Explanation: There are six ways to pick score(s) of two students:
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 4 = 5.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 1 = 8.
- [9,4,1,7]. The difference between the highest and lowest score is 9 - 7 = 2.
- [9,4,1,7]. The difference between the highest and lowest score is 4 - 1 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 4 = 3.
- [9,4,1,7]. The difference between the highest and lowest score is 7 - 1 = 6.
The minimum possible difference is 2.

Constraints:

1 <= k <= nums.length <= 1000
0 <= nums[i] <= 10⁵

Solution: Sliding Window

Sort the array, to minimize the difference, k numbers must be consecutive (i.e, from a subarray). We use a sliding window size of k and try all possible subarrays.
Ans = min{(nums[k – 1] – nums[0]), (nums[k] – nums[1]), … (nums[n – 1] – nums[n – k])}

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

C++

// Author: Huahua

class Solution {

public:

int minimumDifference(vector<int>& nums, int k) {

int ans = INT_MAX;

sort(begin(nums), end(nums));

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

ans = min(ans, nums[i] - nums[i - k + 1]);

return ans;

}

};

花花酱 LeetCode 1980. Find Unique Binary String

By zxi on December 31, 2021

Given an array of strings nums containing n unique binary strings each of length n, return a binary string of length n that does not appear in nums. If there are multiple answers, you may return any of them.

Example 1:

Input: nums = ["01","10"]
Output: "11"
Explanation: "11" does not appear in nums. "00" would also be correct.

Example 2:

Input: nums = ["00","01"]
Output: "11"
Explanation: "11" does not appear in nums. "10" would also be correct.

Example 3:

Input: nums = ["111","011","001"]
Output: "101"
Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.

Constraints:

n == nums.length
1 <= n <= 16
nums[i].length == n
nums[i] is either '0' or '1'.
All the strings of nums are unique.

Solution 1: Hashtable

We can use bitset to convert between integer and binary string.

Time complexity: O(n²)
Space complexity: O(n²)

C++

// Author: Huahua

class Solution {

public:

string findDifferentBinaryString(vector<string>& nums) {

const int n = nums.size();

unordered_set<int> seen(1 << n);

for (const string& num : nums)

seen.insert(bitset<16>(num).to_ulong());

for (int i = 0; i < 1 << n; ++i)

if (!seen.count(i)) return bitset<16>(i).to_string().substr(16 - n);

return "";

}

};

Solution 2: One bit a time

Let ans[i] = ‘1’ – nums[i][i], s.t. ans is at least one bit different from any strings.

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

C++

// Author: Huahua

class Solution {

public:

string findDifferentBinaryString(vector<string>& nums) {

const int n = nums.size();

string ans(n, '0');

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

ans[i] = '1' - nums[i][i] + '0';

return ans;

}

};

花花酱 LeetCode 1979. Find Greatest Common Divisor of Array

By zxi on December 31, 2021

Given an integer array nums, return the greatest common divisor of the smallest number and largest number in nums.

The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

Example 1:

Input: nums = [2,5,6,9,10]
Output: 2
Explanation:
The smallest number in nums is 2.
The largest number in nums is 10.
The greatest common divisor of 2 and 10 is 2.

Example 2:

Input: nums = [7,5,6,8,3]
Output: 1
Explanation:
The smallest number in nums is 3.
The largest number in nums is 8.
The greatest common divisor of 3 and 8 is 1.

Example 3:

Input: nums = [3,3]
Output: 3
Explanation:
The smallest number in nums is 3.
The largest number in nums is 3.
The greatest common divisor of 3 and 3 is 3.

Constraints:

2 <= nums.length <= 1000
1 <= nums[i] <= 1000

Solution:

Use std::minmax_element and std::gcd

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

C++

// Author: Huahua

class Solution {

public:

int findGCD(vector<int>& nums) {

auto p = minmax_element(begin(nums), end(nums));

return gcd(*p.first, *p.second);

}

};

花花酱 LeetCode 1975. Maximum Matrix Sum

By zxi on December 31, 2021

You are given an n x n integer matrix. You can do the following operation any number of times:

Choose any two adjacent elements of matrix and multiply each of them by -1.

Two elements are considered adjacent if and only if they share a border.

Your goal is to maximize the summation of the matrix’s elements. Return the maximum sum of the matrix’s elements using the operation mentioned above.

Example 1:

Input: matrix = [[1,-1],[-1,1]]
Output: 4
Explanation: We can follow the following steps to reach sum equals 4:
- Multiply the 2 elements in the first row by -1.
- Multiply the 2 elements in the first column by -1.

Example 2:

Input: matrix = [[1,2,3],[-1,-2,-3],[1,2,3]]
Output: 16
Explanation: We can follow the following step to reach sum equals 16:
- Multiply the 2 last elements in the second row by -1.

Constraints:

n == matrix.length == matrix[i].length
2 <= n <= 250
-10⁵ <= matrix[i][j] <= 10⁵

Solution: Math

Count the number of negative numbers.
1. Even negatives, we can always flip all the negatives to positives. ans = sum(abs(matrix)).
2. Odd negatives, there will be one negative left, we found the smallest abs(element) and let it become negative. ans = sum(abs(matrix))) – 2 * min(abs(matrix))

Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

long long maxMatrixSum(vector<vector<int>>& matrix) {

const int n = matrix.size();

long long ans = 0;

int count = 0;

int lo = INT_MAX;

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

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

ans += abs(matrix[i][j]);

lo = min(lo, abs(matrix[i][j]));

count += matrix[i][j] < 0;

}

return ans - (count & 1) * 2 * lo;

}

};

花花酱 LeetCode 1974. Minimum Time to Type Word Using Special Typewriter

By zxi on December 31, 2021

There is a special typewriter with lowercase English letters 'a' to 'z' arranged in a circle with a pointer. A character can only be typed if the pointer is pointing to that character. The pointer is initially pointing to the character 'a'.

Each second, you may perform one of the following operations:

Move the pointer one character counterclockwise or clockwise.
Type the character the pointer is currently on.

Given a string word, return the minimum number of seconds to type out the characters in word.

Example 1:

Input: word = "abc"
Output: 5
Explanation: 
The characters are printed as follows:
- Type the character 'a' in 1 second since the pointer is initially on 'a'.
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer clockwise to 'c' in 1 second.
- Type the character 'c' in 1 second.

Example 2:

Input: word = "bza"
Output: 7
Explanation:
The characters are printed as follows:
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer counterclockwise to 'z' in 2 seconds.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'a' in 1 second.
- Type the character 'a' in 1 second.

Example 3:

Input: word = "zjpc"
Output: 34
Explanation:
The characters are printed as follows:
- Move the pointer counterclockwise to 'z' in 1 second.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'j' in 10 seconds.
- Type the character 'j' in 1 second.
- Move the pointer clockwise to 'p' in 6 seconds.
- Type the character 'p' in 1 second.
- Move the pointer counterclockwise to 'c' in 13 seconds.
- Type the character 'c' in 1 second.

Constraints:

1 <= word.length <= 100
word consists of lowercase English letters.

Solution: Clockwise or Counter-clockwise?

For each pair of (prev, curr), choose the shortest distance.
One is abs(p – c), another is 26 – abs(p – c).
Don’t forget to add 1 for typing itself.

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

C++

// Author: Huahua

class Solution {

public:

int minTimeToType(string word) {

int ans = 0;

char p = 'a';

for (char c : word) {

ans += 1 + min(abs(c - p), 26 - abs(c - p));

p = c;

}

return ans;

}

};

Huahua's Tech Road

花花酱 LeetCode 1984. Minimum Difference Between Highest and Lowest of K Scores

Solution: Sliding Window

C++

花花酱 LeetCode 1980. Find Unique Binary String

Solution 1: Hashtable

C++

Solution 2: One bit a time

C++

花花酱 LeetCode 1979. Find Greatest Common Divisor of Array

Solution:

C++

花花酱 LeetCode 1975. Maximum Matrix Sum

Solution: Math

C++

花花酱 LeetCode 1974. Minimum Time to Type Word Using Special Typewriter

Solution: Clockwise or Counter-clockwise?

C++