Huahua’s Tech Road

花花酱 LeetCode 1818. Minimum Absolute Sum Difference

By zxi on April 6, 2021

You are given two positive integer arrays nums1 and nums2, both of length n.

The absolute sum difference of arrays nums1 and nums2 is defined as the sum of |nums1[i] - nums2[i]| for each 0 <= i < n (0-indexed).

You can replace at most one element of nums1 with any other element in nums1 to minimize the absolute sum difference.

Return the minimum absolute sum difference after replacing at most oneelement in the array nums1. Since the answer may be large, return it modulo 10⁹ + 7.

|x| is defined as:

x if x >= 0, or
-x if x < 0.

Example 1:

Input: nums1 = [1,7,5], nums2 = [2,3,5]
Output: 3
Explanation: There are two possible optimal solutions:
- Replace the second element with the first: [1,7,5] => [1,1,5], or
- Replace the second element with the third: [1,7,5] => [1,5,5].
Both will yield an absolute sum difference of |1-2| + (|1-3| or |5-3|) + |5-5| = 3.

Example 2:

Input: nums1 = [2,4,6,8,10], nums2 = [2,4,6,8,10]
Output: 0
Explanation: nums1 is equal to nums2 so no replacement is needed. This will result in an 
absolute sum difference of 0.

Example 3:

Input: nums1 = [1,10,4,4,2,7], nums2 = [9,3,5,1,7,4]
Output: 20
Explanation: Replace the first element with the second: [1,10,4,4,2,7] => [10,10,4,4,2,7].
This yields an absolute sum difference of |10-9| + |10-3| + |4-5| + |4-1| + |2-7| + |7-4| = 20

Constraints:

n == nums1.length
n == nums2.length
1 <= n <= 10⁵
1 <= nums1[i], nums2[i] <= 10⁵

Solution: Binary Search

Greedy won’t work, e.g. finding the max diff pair and replace it. Counter example:
nums1 = [7, 5], nums2 = [1, -2]
pair1 = abs(7 – 1) = 6
pair2 = abs(5 – (-2)) = 7
If we replace 5 with 7, we got pair2′ = abs(7 – (-2)) = 9 > 7.

Every pair of numbers can be the candidate, we just need to find the closest number for each nums2[i].

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

C++

class Solution {

public:

int minAbsoluteSumDiff(vector<int>& nums1, vector<int>& nums2) {

constexpr int kMod = 1e9 + 7;

const int n = nums1.size();

long ans = 0;

int gain = 0;

set<int> s(begin(nums1), end(nums1));

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

const int diff = abs(nums1[i] - nums2[i]);

ans += diff;

if (diff <= gain) continue;

auto it = s.lower_bound(nums2[i]);

if (it != s.end())

gain = max(gain, diff - abs(*it - nums2[i]));

if (it != s.begin())

gain = max(gain, diff - abs(*prev(it) - nums2[i]));

}

return (ans - gain) % kMod;

}

};

花花酱 LeetCode 1817. Finding the Users Active Minutes

By zxi on April 6, 2021

You are given the logs for users’ actions on LeetCode, and an integer k. The logs are represented by a 2D integer array logs where each logs[i] = [ID_i, time_i] indicates that the user with ID_i performed an action at the minute time_i.

Multiple users can perform actions simultaneously, and a single user can perform multiple actions in the same minute.

The user active minutes (UAM) for a given user is defined as the number of unique minutes in which the user performed an action on LeetCode. A minute can only be counted once, even if multiple actions occur during it.

You are to calculate a 1-indexed array answer of size k such that, for each j (1 <= j <= k), answer[j] is the number of users whose UAM equals j.

Return the array answer as described above.

Example 1:

Input: logs = [[0,5],[1,2],[0,2],[0,5],[1,3]], k = 5
Output: [0,2,0,0,0]
Explanation:
The user with ID=0 performed actions at minutes 5, 2, and 5 again. Hence, they have a UAM of 2 (minute 5 is only counted once).
The user with ID=1 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
Since both users have a UAM of 2, answer[2] is 2, and the remaining answer[j] values are 0.

Example 2:

Input: logs = [[1,1],[2,2],[2,3]], k = 4
Output: [1,1,0,0]
Explanation:
The user with ID=1 performed a single action at minute 1. Hence, they have a UAM of 1.
The user with ID=2 performed actions at minutes 2 and 3. Hence, they have a UAM of 2.
There is one user with a UAM of 1 and one with a UAM of 2.
Hence, answer[1] = 1, answer[2] = 1, and the remaining values are 0.

Constraints:

1 <= logs.length <= 10⁴
0 <= ID_i <= 10⁹
1 <= time_i <= 10⁵
k is in the range [The maximum UAM for a user, 10⁵].

Solution: Hashsets in a Hashtable

key: user_id, value: set{time}

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

C++

class Solution {

public:

vector<int> findingUsersActiveMinutes(vector<vector<int>>& logs, int k) {

unordered_map<int, unordered_set<int>> m;

vector<int> ans(k);

for (const auto& log : logs)

m[log[0]].insert(log[1]);

for (const auto& [id, s] : m)

++ans[s.size() - 1];

return ans;

}

};

花花酱 LeetCode 1816. Truncate Sentence

By zxi on April 6, 2021

A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each of the words consists of only uppercase and lowercase English letters (no punctuation).

For example, "Hello World", "HELLO", and "hello world hello world" are all sentences.

You are given a sentence s and an integer k. You want to truncate s such that it contains only the first k words. Return s after truncating it.

Example 1:

Input: s = "Hello how are you Contestant", k = 4
Output: "Hello how are you"
Explanation:
The words in s are ["Hello", "how" "are", "you", "Contestant"].
The first 4 words are ["Hello", "how", "are", "you"].
Hence, you should return "Hello how are you".

Example 2:

Input: s = "What is the solution to this problem", k = 4
Output: "What is the solution"
Explanation:
The words in s are ["What", "is" "the", "solution", "to", "this", "problem"].
The first 4 words are ["What", "is", "the", "solution"].
Hence, you should return "What is the solution".

Example 3:

Input: s = "chopper is not a tanuki", k = 5
Output: "chopper is not a tanuki"

Constraints:

1 <= s.length <= 500
k is in the range [1, the number of words in s].
s consist of only lowercase and uppercase English letters and spaces.
The words in s are separated by a single space.
There are no leading or trailing spaces.

Solution:

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

C++

class Solution {

public:

string truncateSentence(string s, int k) {

string ans;

stringstream ss(s);

for (int i = 0; i < k && ss; ++i) {

string word;

ss >> word;

ans += (ans.empty() ? "" : " ") + word;

}

return ans;

}

};

Python3

class Solution:

def truncateSentence(self, s: str, k: int) -> str:

return " ".join(s.split()[:k])

花花酱 LeetCode 1815. Maximum Number of Groups Getting Fresh Donuts

By zxi on April 5, 2021

There is a donuts shop that bakes donuts in batches of batchSize. They have a rule where they must serve all of the donuts of a batch before serving any donuts of the next batch. You are given an integer batchSize and an integer array groups, where groups[i] denotes that there is a group of groups[i] customers that will visit the shop. Each customer will get exactly one donut.

When a group visits the shop, all customers of the group must be served before serving any of the following groups. A group will be happy if they all get fresh donuts. That is, the first customer of the group does not receive a donut that was left over from the previous group.

You can freely rearrange the ordering of the groups. Return the maximum possible number of happy groups after rearranging the groups.

Example 1:

Input: batchSize = 3, groups = [1,2,3,4,5,6]
Output: 4
Explanation: You can arrange the groups as [6,2,4,5,1,3]. Then the 1^st, 2^nd, 4^th, and 6^th groups will be happy.

Example 2:

Input: batchSize = 4, groups = [1,3,2,5,2,2,1,6]
Output: 4

Constraints:

1 <= batchSize <= 9
1 <= groups.length <= 30
1 <= groups[i] <= 10⁹

Solution 0: Binary Mask DP

Time complexity: O(n*2ⁿ) TLE
Space complexity: O(2ⁿ)

C++

// Author: Huahua, TLE, 42/67 Passed

class Solution {

public:

int maxHappyGroups(int b, vector<int>& groups) {

const int n = groups.size();

vector<int> dp(1 << n);

for (int mask = 0; mask < 1 << n; ++mask) {

int s = 0;

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

if (mask & (1 << i)) s = (s + groups[i]) % b;

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

if (!(mask & (1 << i)))

dp[mask | (1 << i)] = max(dp[mask | (1 << i)],

dp[mask] + (s == 0));

}

return dp[(1 << n) - 1];

}

};

Solution 1: Recursion w/ Memoization

State: count of group size % batchSize

C++

// Author: Huahua, 888 ms, 61.9 MB

class Solution {

public:

int maxHappyGroups(int b, vector<int>& groups) {

int ans = 0;

vector<int> count(b);

for (int g : groups) ++count[g % b];

map<vector<int>, int> cache;

function<int(int)> dp = [&](int s) {

auto it = cache.find(count);

if (it != cache.end()) return it->second;

int ans = 0;

for (int i = 1; i < b; ++i) {

if (!count[i]) continue;

--count[i];

ans = max(ans, (s == 0) + dp((s + i) % b));

++count[i];

}

return cache[count] = ans;

};

return count[0] + dp(0);

}

};

C++/Hashtable

// Author: Huahua 380 ms, 59.2 MB

struct VectorHasher {

size_t operator()(const vector<int>& V) const {

size_t hash = V.size();

for (auto i : V)

hash ^= i + 0x9e3779b9 + (hash << 6) + (hash >> 2);

return hash;

}

};

class Solution {

public:

int maxHappyGroups(int b, vector<int>& groups) {

int ans = 0;

vector<int> count(b);

for (int g : groups) ++count[g % b];

unordered_map<vector<int>, int, VectorHasher> cache;

function<int(int)> dp = [&](int s) {

auto it = cache.find(count);

if (it != cache.end()) return it->second;

int ans = 0;

for (int i = 1; i < b; ++i) {

if (!count[i]) continue;

--count[i];

ans = max(ans, (s == 0) + dp((s + b - i) % b));

++count[i];

}

return cache[count] = ans;

};

return count[0] + dp(0);

}

};

C++/OPT

// Author: Huahua, 0 ms, 8.1 MB

class Solution {

public:

int maxHappyGroups(int b, vector<int>& groups) {

int ans = 0;

vector<int> count(b);

for (int g : groups) {

const int r = g % b;

if (r == 0) { ++ans; continue; }

if (count[b - r]) {

--count[b - r];

++ans;

} else {

++count[r];

}

map<vector<int>, int> cache;

function<int(int, int)> dp = [&](int r, int n) {

if (n == 0) return 0;

auto it = cache.find(count);

if (it != cache.end()) return it->second;

int ans = 0;

for (int i = 1; i < b; ++i) {

if (!count[i]) continue;

--count[i];

ans = max(ans, (r == 0) + dp((r + b - i) % b, n - 1));

++count[i];

}

return cache[count] = ans;

};

const int n = accumulate(begin(count), end(count), 0);

return ans + (n ? dp(0, n) : 0);

}

};

花花酱 LeetCode 1814. Count Nice Pairs in an Array

By zxi on April 3, 2021

You are given an array nums that consists of non-negative integers. Let us define rev(x) as the reverse of the non-negative integer x. For example, rev(123) = 321, and rev(120) = 21. A pair of indices (i, j) is nice if it satisfies all of the following conditions:

0 <= i < j < nums.length
nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])

Return the number of nice pairs of indices. Since that number can be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [42,11,1,97]
Output: 2
Explanation: The two pairs are:
 - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.
 - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.

Example 2:

Input: nums = [13,10,35,24,76]
Output: 4

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Solution: Two Sum

Key = x – rev(x)

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

C++

// Author: Huahua

class Solution {

public:

int countNicePairs(vector<int>& nums) {

constexpr int kMod = 1e9 + 7;

unordered_map<int, int> m;

long ans = 0;

for (int x : nums) {

string s = to_string(x);

reverse(begin(s), end(s));

ans += m[x - stoi(s)]++;

}

return ans % kMod;

}

};

Huahua's Tech Road

花花酱 LeetCode 1818. Minimum Absolute Sum Difference

Solution: Binary Search

C++

花花酱 LeetCode 1817. Finding the Users Active Minutes

Solution: Hashsets in a Hashtable

C++

花花酱 LeetCode 1816. Truncate Sentence

Solution:

C++

Python3

花花酱 LeetCode 1815. Maximum Number of Groups Getting Fresh Donuts

Solution 0: Binary Mask DP

C++

Solution 1: Recursion w/ Memoization

C++

C++/Hashtable

C++/OPT

花花酱 LeetCode 1814. Count Nice Pairs in an Array

Solution: Two Sum

C++