Posts tagged as “easy”

花花酱 LeetCode 2129. Capitalize the Title

By zxi on January 8, 2022

You are given a string title consisting of one or more words separated by a single space, where each word consists of English letters. Capitalize the string by changing the capitalization of each word such that:

If the length of the word is 1 or 2 letters, change all letters to lowercase.
Otherwise, change the first letter to uppercase and the remaining letters to lowercase.

Return the capitalized title.

Example 1:

Input: title = "capiTalIze tHe titLe"
Output: "Capitalize The Title"
Explanation:
Since all the words have a length of at least 3, the first letter of each word is uppercase, and the remaining letters are lowercase.

Example 2:

Input: title = "First leTTeR of EACH Word"
Output: "First Letter of Each Word"
Explanation:
The word "of" has length 2, so it is all lowercase.
The remaining words have a length of at least 3, so the first letter of each remaining word is uppercase, and the remaining letters are lowercase.

Example 3:

Input: title = "i lOve leetcode"
Output: "i Love Leetcode"
Explanation:
The word "i" has length 1, so it is lowercase.
The remaining words have a length of at least 3, so the first letter of each remaining word is uppercase, and the remaining letters are lowercase.

Constraints:

1 <= title.length <= 100
title consists of words separated by a single space without any leading or trailing spaces.
Each word consists of uppercase and lowercase English letters and is non-empty.

Solution: Straight forward

Without splitting the sentence into words, we need to take care the word of length one and two.

Tips: use std::tolower, std::toupper to transform letters.

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

C++

// Author: Huahua

class Solution {

public:

string capitalizeTitle(string title) {

const int n = title.size();

for (int i = 0; i < title.size(); ++i) {

if ((i == 0 || title[i - 1] == ' ')

&& i + 2 < n && title[i + 1] != ' ' && title[i + 2] != ' ')

title[i] = toupper(title[i]);

else

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

}

return title;

}

};

Python3

# Author: Huahua

class Solution:

def capitalizeTitle(self, title: str) -> str:

return " ".join(map(lambda w: w.lower() if len(w) <= 2 else w[0].upper() + w[1:].lower(), title.split()))

花花酱 LeetCode 2124. Check if All A’s Appears Before All B’s

By zxi on January 2, 2022

Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears before every 'b' in the string. Otherwise, return false.

Example 1:

Input: s = "aaabbb"
Output: true
Explanation:
The 'a's are at indices 0, 1, and 2, while the 'b's are at indices 3, 4, and 5.
Hence, every 'a' appears before every 'b' and we return true.

Example 2:

Input: s = "abab"
Output: false
Explanation:
There is an 'a' at index 2 and a 'b' at index 1.
Hence, not every 'a' appears before every 'b' and we return false.

Example 3:

Input: s = "bbb"
Output: true
Explanation:
There are no 'a's, hence, every 'a' appears before every 'b' and we return true.

Constraints:

`1 <= s.length <= 100`

`s[i]` is either `'a'` or `'b'`.

Solution: Count bs

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

C++

// Author: Huahua

class Solution {

public:

bool checkString(string s) {

int b = 0;

for (char ch : s) {

b += ch == 'b';

if (ch == 'a' && b) return false;

}

return true;

}

};

花花酱 LeetCode 1995. Count Special Quadruplets

By zxi on December 31, 2021

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

Constraints:

4 <= nums.length <= 50
1 <= nums[i] <= 100

Solution 1: Brute force (224ms)

Enumerate a, b, c, d.

Time complexity: O(C(n, 4)) = O(n⁴/24)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

const int n = nums.size();

int ans = 0;

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c)

for (int d = c + 1; d < n; ++d)

ans += nums[a] + nums[b] + nums[c] == nums[d];

return ans;

}

};

Solution 2: Static frequency table + binary search (39ms)

For each element, we store its indices (sorted).

Given a, b, c, target t = nums[a] + nums[b] + nums[c], we check the hashtable and use binary search to find how many times it occurred after index c.

Time complexity: O(n³/6*logn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<vector<int>> m(kMax + 1);

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

m[nums[i]].push_back(i);

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += end(m[t]) - lower_bound(begin(m[t]), end(m[t]), c + 1);

}

return ans;

}

};

Solution 3: Dynamic frequency table (29ms)

Similar to 花花酱 LeetCode 1. Two Sum, we dynamically add elements (from right to left) into the hashtable.

Time complexity: O(n³/6)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<int> m(kMax + 1);

// for every c we had seen, we can use it as target (nums[d]).

for (int c = n - 1; c >= 0; ++m[nums[c--]])

for (int a = 0; a < c; ++a)

for (int b = a + 1; b < c; ++b) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += m[t];

}

return ans;

}

};

花花酱 LeetCode 1991. Find the Middle Index in Array

By zxi on December 31, 2021

Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones).

A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].

If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0.

Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.

Example 1:

Input: nums = [2,3,-1,8,4]
Output: 3
Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4
The sum of the numbers after index 3 is: 4 = 4

Example 2:

Input: nums = [1,-1,4]
Output: 2
Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0
The sum of the numbers after index 2 is: 0

Example 3:

Input: nums = [2,5]
Output: -1
Explanation: There is no valid middleIndex.

Constraints:

1 <= nums.length <= 100
-1000 <= nums[i] <= 1000

Solution: Pre-compute + prefix sum

Pre-compute the sum of entire array. We scan the array and accumulate prefix sum and we can compute the sum of the rest of array.

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

C++

// Author: Huahua

class Solution {

public:

int findMiddleIndex(vector<int>& nums) {

int sum = accumulate(begin(nums), end(nums), 0);

for (int i = 0, cur = 0; i < nums.size(); ++i) {

if (cur == sum - cur - nums[i]) return i;

cur += nums[i];

}

return -1;

}

};

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;

}

};

Posts tagged as “easy”

花花酱 LeetCode 2129. Capitalize the Title

Solution: Straight forward

C++

Python3

花花酱 LeetCode 2124. Check if All A’s Appears Before All B’s

1 <= s.length <= 100

s[i] is either 'a' or 'b'.Solution: Count bs

C++

花花酱 LeetCode 1995. Count Special Quadruplets

Solution 1: Brute force (224ms)

C++

Solution 2: Static frequency table + binary search (39ms)

C++

Solution 3: Dynamic frequency table (29ms)

C++

花花酱 LeetCode 1991. Find the Middle Index in Array

Solution: Pre-compute + prefix sum

C++

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

Solution: Sliding Window

C++

`1 <= s.length <= 100`

`s[i]` is either `'a'` or `'b'`.

Solution: Count bs