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花花酱 LeetCode 2588. Count the Number of Beautiful Subarrays

By zxi on March 11, 2023

You are given a 0-indexed integer array nums. In one operation, you can:

Choose two different indices i and j such that 0 <= i, j < nums.length.
Choose a non-negative integer k such that the k^th bit (0-indexed) in the binary representation of nums[i] and nums[j] is 1.
Subtract 2^k from nums[i] and nums[j].

A subarray is beautiful if it is possible to make all of its elements equal to 0 after applying the above operation any number of times.

Return the number of beautiful subarrays in the array nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [4,3,1,2,4]
Output: 2
Explanation: There are 2 beautiful subarrays in nums: [4,3,1,2,4] and [4,3,1,2,4].
- We can make all elements in the subarray [3,1,2] equal to 0 in the following way:
  - Choose [3, 1, 2] and k = 1. Subtract 2¹ from both numbers. The subarray becomes [1, 1, 0].
  - Choose [1, 1, 0] and k = 0. Subtract 2⁰ from both numbers. The subarray becomes [0, 0, 0].
- We can make all elements in the subarray [4,3,1,2,4] equal to 0 in the following way:
  - Choose [4, 3, 1, 2, 4] and k = 2. Subtract 2² from both numbers. The subarray becomes [0, 3, 1, 2, 0].
  - Choose [0, 3, 1, 2, 0] and k = 0. Subtract 2⁰ from both numbers. The subarray becomes [0, 2, 0, 2, 0].
  - Choose [0, 2, 0, 2, 0] and k = 1. Subtract 2¹ from both numbers. The subarray becomes [0, 0, 0, 0, 0].

Example 2:

Input: nums = [1,10,4]
Output: 0
Explanation: There are no beautiful subarrays in nums.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶

Solution: Hashtable + Prefix XOR

The problem is asking to find # of subarrays whose element wise xor is 0. We can use a hashtable to store the frequency of each prefix xor value, which reduces this problem to # of Subarray sum equal to k.

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

C++

// Author: Huahua

class Solution {

public:

long long beautifulSubarrays(vector<int>& nums) {

long long ans = 0;

unordered_map<int, int> m{{0, 1}};

int x = 0;

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

x ^= nums[i];

ans += m[x];

m[x] += 1;

}

return ans;

}

};

花花酱 LeetCode 2587. Rearrange Array to Maximize Prefix Score

By zxi on March 11, 2023

You are given a 0-indexed integer array nums. You can rearrange the elements of nums to any order (including the given order).

Let prefix be the array containing the prefix sums of nums after rearranging it. In other words, prefix[i] is the sum of the elements from 0 to i in nums after rearranging it. The score of nums is the number of positive integers in the array prefix.

Return the maximum score you can achieve.

Example 1:

Input: nums = [2,-1,0,1,-3,3,-3]
Output: 6
Explanation: We can rearrange the array into nums = [2,3,1,-1,-3,0,-3].
prefix = [2,5,6,5,2,2,-1], so the score is 6.
It can be shown that 6 is the maximum score we can obtain.

Example 2:

Input: nums = [-2,-3,0]
Output: 0
Explanation: Any rearrangement of the array will result in a score of 0.

Constraints:

1 <= nums.length <= 10⁵
-10⁶ <= nums[i] <= 10⁶

Solution: Greedy

Sort the numbers in descending order.

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

C++

// Author: Huahua

class Solution {

public:

int maxScore(vector<int>& nums) {

sort(rbegin(nums), rend(nums));

int ans = 0;

for (long i = 0, s = 0; i < nums.size(); ++i) {

s += nums[i];

if (s <= 0) break;

++ans;

}

return ans;

}

};

花花酱 LeetCode 2586. Count the Number of Vowel Strings in Range

By zxi on March 11, 2023

You are given a 0-indexed array of string words and two integers left and right.

A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'.

Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Example 1:

Input: words = ["are","amy","u"], left = 0, right = 2
Output: 2
Explanation: 
- "are" is a vowel string because it starts with 'a' and ends with 'e'.
- "amy" is not a vowel string because it does not end with a vowel.
- "u" is a vowel string because it starts with 'u' and ends with 'u'.
The number of vowel strings in the mentioned range is 2.

Example 2:

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
Output: 3
Explanation: 
- "aeo" is a vowel string because it starts with 'a' and ends with 'o'.
- "mu" is not a vowel string because it does not start with a vowel.
- "ooo" is a vowel string because it starts with 'o' and ends with 'o'.
- "artro" is a vowel string because it starts with 'a' and ends with 'o'.
The number of vowel strings in the mentioned range is 3.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 10
words[i] consists of only lowercase English letters.
0 <= left <= right < words.length

Solution:

Iterator overs words, from left to right. Check the first and last element of the string.

Time complexity: O(|right – left + 1|)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int vowelStrings(vector<string>& words, int left, int right) {

auto isVowel = [](char c) {

return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';

};

return count_if(begin(words) + left, begin(words) + right + 1, [&](const string& w) {

return isVowel(w.front()) && isVowel(w.back());

});

}

};

花花酱 LeetCode 2583. Kth Largest Sum in a Binary Tree

By zxi on March 5, 2023

You are given the root of a binary tree and a positive integer k.

The level sum in the tree is the sum of the values of the nodes that are on the same level.

Return the k^th largest level sum in the tree (not necessarily distinct). If there are fewer than k levels in the tree, return -1.

Note that two nodes are on the same level if they have the same distance from the root.

Example 1:

Input: root = [5,8,9,2,1,3,7,4,6], k = 2
Output: 13
Explanation: The level sums are the following:
- Level 1: 5.
- Level 2: 8 + 9 = 17.
- Level 3: 2 + 1 + 3 + 7 = 13.
- Level 4: 4 + 6 = 10.
The 2^nd largest level sum is 13.

Example 2:

Input: root = [1,2,null,3], k = 1
Output: 3
Explanation: The largest level sum is 3.

Constraints:

The number of nodes in the tree is n.
2 <= n <= 10⁵
1 <= Node.val <= 10⁶
1 <= k <= n

Solution: DFS + Sorting

Use DFS to traverse the tree and accumulate level sum. Once done, sort the level sums and find the k-th largest one.

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

Note: sum can be very large, use long long.

C++

// Author: Huahua

class Solution {

public:

long long kthLargestLevelSum(TreeNode* root, int k) {

vector<long long> sums;

function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int d) {

if (!root) return;

while (d >= sums.size()) sums.push_back(0);

sums[d] += root->val;

dfs(root->left, d + 1);

dfs(root->right, d + 1);

};

dfs(root, 0);

if (sums.size() < k) return -1;

sort(rbegin(sums), rend(sums));

return sums[k - 1];

}

};

花花酱 LeetCode 2582. Pass the Pillow

By zxi on March 5, 2023

There are n people standing in a line labeled from 1 to n. The first person in the line is holding a pillow initially. Every second, the person holding the pillow passes it to the next person standing in the line. Once the pillow reaches the end of the line, the direction changes, and people continue passing the pillow in the opposite direction.

For example, once the pillow reaches the n^th person they pass it to the n - 1^th person, then to the n - 2^th person and so on.

Given the two positive integers n and time, return the index of the person holding the pillow after time seconds.

Example 1:

Input: n = 4, time = 5
Output: 2
Explanation: People pass the pillow in the following way: 1 -> 2 -> 3 -> 4 -> 3 -> 2.
Afer five seconds, the pillow is given to the 2^nd person.

Example 2:

Input: n = 3, time = 2
Output: 3
Explanation: People pass the pillow in the following way: 1 -> 2 -> 3.
Afer two seconds, the pillow is given to the 3^r^d person.

Constraints:

2 <= n <= 1000
1 <= time <= 1000

Solution: Math

It takes n – 1 seconds from 1 to n and takes another n – 1 seconds back from n to 1.
So one around takes 2 * (n – 1) seconds. We can mod time with 2 * (n – 1).

After that if time < n – 1 answer is time + 1, otherwise answer is n – (time – (n – 1))

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

C++

// Author: Huahua

class Solution {

public:

int passThePillow(int n, int time) {

time %= (2 * (n - 1));

return time > n - 1 ? n - (time - (n - 1)) : time + 1;

}

};