Posts tagged as “easy”

花花酱 LeetCode 1763. Longest Nice Substring

By zxi on February 20, 2021

A string s is nice if, for every letter of the alphabet that s contains, it appears both in uppercase and lowercase. For example, "abABB" is nice because 'A' and 'a' appear, and 'B' and 'b' appear. However, "abA" is not because 'b' appears, but 'B' does not.

Given a string s, return the longest substring of s that is nice. If there are multiple, return the substring of the earliest occurrence. If there are none, return an empty string.

Example 1:

Input: s = "YazaAay"
Output: "aAa"
Explanation: "aAa" is a nice string because 'A/a' is the only letter of the alphabet in s, and both 'A' and 'a' appear.
"aAa" is the longest nice substring.

Example 2:

Input: s = "Bb"
Output: "Bb"
Explanation: "Bb" is a nice string because both 'B' and 'b' appear. The whole string is a substring.

Example 3:

Input: s = "c"
Output: ""
Explanation: There are no nice substrings.

Example 4:

Input: s = "dDzeE"
Output: "dD"
Explanation: Both "dD" and "eE" are the longest nice substrings.
As there are multiple longest nice substrings, return "dD" since it occurs earlier.

Constraints:

1 <= s.length <= 100
s consists of uppercase and lowercase English letters.

Solution: Brute Force

Time complexity: O(n^3)
Space complexity: O(1)

C++

class Solution {

public:

string longestNiceSubstring(string_view s) {

const int n = s.length();

auto isNice = [](string_view s) {

vector<int> u(26), l(26);

for (int c : s)

if (isupper(c)) u[c - 'A'] = 1;

else l[c - 'a'] = 1;

return u == l;

};

string_view ans;

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

for (int j = i + 1; j < n; ++j) {

string_view ss = s.substr(i, j - i + 1);

if (isNice(ss) && ss.length() > ans.length())

ans = ss;

}

return string(ans);

}

};

Optimized 1:

Time complexity: O(n^2*26)
Space complexity: O(1)

C++

class Solution {

public:

string longestNiceSubstring(string_view s) {

const int n = s.length();

string_view ans;

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

vector<int> u(26), l(26);

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

const char c = s[j];

if (isupper(c)) u[c - 'A'] = 1;

else l[c - 'a'] = 1;

if (u == l && j - i + 1 > ans.length())

ans = s.substr(i, j - i + 1);

}

return string(ans);

}

};

花花酱 LeetCode 1758. Minimum Changes To Make Alternating Binary String

By zxi on February 13, 2021

You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa.

The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not.

Return the minimum number of operations needed to make s alternating.

Example 1:

Input: s = "0100"
Output: 1
Explanation: If you change the last character to '1', s will be "0101", which is alternating.

Example 2:

Input: s = "10"
Output: 0
Explanation: s is already alternating.

Example 3:

Input: s = "1111"
Output: 2
Explanation: You need two operations to reach "0101" or "1010".

Constraints:

1 <= s.length <= 10⁴
s[i] is either '0' or '1'.

Solution: Two Counters

The final string is either 010101… or 101010…
We just need two counters to record the number of changes needed to transform the original string to those two final strings.

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

C++

// Author: Huahua

class Solution {

public:

int minOperations(string s) {

int c1 = 0, c2 = 0;

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

c1 += (s[i] - '0' == i % 2);

c2 += (s[i] - '0' != i % 2);

}

return min(c1, c2);

}

};

花花酱 LeetCode 1752. Check if Array Is Sorted and Rotated

By zxi on February 7, 2021

Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false.

There may be duplicates in the original array.

Note: An array A rotated by x positions results in an array B of the same length such that A[i] == B[(i+x) % A.length], where % is the modulo operation.

Example 1:

Input: nums = [3,4,5,1,2]
Output: true
Explanation: [1,2,3,4,5] is the original sorted array.
You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].

Example 2:

Input: nums = [2,1,3,4]
Output: false
Explanation: There is no sorted array once rotated that can make nums.

Example 3:

Input: nums = [1,2,3]
Output: true
Explanation: [1,2,3] is the original sorted array.
You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.

Example 4:

Input: nums = [1,1,1]
Output: true
Explanation: [1,1,1] is the original sorted array.
You can rotate any number of positions to make nums.

Example 5:

Input: nums = [2,1]
Output: true
Explanation: [1,2] is the original sorted array.
You can rotate the array by x = 5 positions to begin on the element of value 2: [2,1].

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Solution: Counting and checking

Count how many turning points (nums[i] < nums[i – 1]) in the array. Return false if there are more than 1.
For the turning point r, (nums[r] < nums[r – 1), return true if both of the following conditions are satisfied:
1. nums[r – 1] is the largest number, e.g. nums[r – 1] >= nums[n – 1]
2. nums[r] is the smallest number, e.g. nums[r] <= nums[0]

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

C++

// Author: Huahua

class Solution {

public:

bool check(vector<int>& nums) {

int count = 0;

int r = -1;

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

if (nums[i] < nums[i - 1]) {

++count;

r = i;

}

if (count == 0) return true;

if (count > 1) return false;

return nums[r] <= nums[0] && nums[r - 1] >= nums.back();

}

};

花花酱 LeetCode 1748. Sum of Unique Elements

By zxi on February 6, 2021

You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.

Return the sum of all the unique elements of nums.

Example 1:

Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.

Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.

Example 3:

Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Solution: Hashtable

Time complexity: O(n)
Space complexity: O(100)

C++

// Author: Huahua

class Solution {

public:

int sumOfUnique(vector<int>& nums) {

vector<int> seen(101);

int ans = 0;

for (int x : nums)

++seen[x];

for (int x : nums)

if (seen[x] == 1) ans += x;

return ans;

}

};

花花酱 LeetCode 1736. Latest Time by Replacing Hidden Digits

By zxi on January 23, 2021

You are given a string time in the form of hh:mm, where some of the digits in the string are hidden (represented by ?).

The valid times are those inclusively between 00:00 and 23:59.

Return the latest valid time you can get from time by replacing the hidden digits.

Example 1:

Input: time = "2?:?0"
Output: "23:50"
Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.

Example 2:

Input: time = "0?:3?"
Output: "09:39"

Example 3:

Input: time = "1?:22"
Output: "19:22"

Constraints:

time is in the format hh:mm.
It is guaranteed that you can produce a valid time from the given string.

Solution 1: Brute Force

Enumerate all possible clock in reverse order and find the first matching one.

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

C++

// Author: Huahua

class Solution {

public:

string maximumTime(string time) {

for (int m = 60 * 24 - 1; m >= 0; --m) {

int hh = m / 60;

int mm = m % 60;

if (time[0] != '?' && time[0] - '0' != hh / 10) continue;

if (time[1] != '?' && time[1] - '0' != hh % 10) continue;

if (time[3] != '?' && time[3] - '0' != mm / 10) continue;

if (time[4] != '?' && time[4] - '0' != mm % 10) continue;

time[0] = (hh / 10) + '0';

time[1] = (hh % 10) + '0';

time[3] = (mm / 10) + '0';

time[4] = (mm % 10) + '0';

break;

}

return time;

}

};

Solution 2: Rules

Using rules, fill from left to right.

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

C++

// Author: Huahua

class Solution {

public:

string maximumTime(string time) {

if (time[0] == '?') time[0] = time[1] >= '4' && time[1] <= '9' ? '1' : '2';

if (time[1] == '?') time[1] = time[0] == '2' ? '3' : '9';

if (time[3] == '?') time[3] = '5';

if (time[4] == '?') time[4] = '9';

return time;

}

};