Huahua’s Tech Road

LeetCode 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points

By zxi on November 3, 2021

A critical point in a linked list is defined as either a local maxima or a local minima.

A node is a local maxima if the current node has a value strictly greater than the previous node and the next node.

A node is a local minima if the current node has a value strictly smaller than the previous node and the next node.

Note that a node can only be a local maxima/minima if there exists both a previous node and a next node.

Given a linked list head, return an array of length 2 containing [minDistance, maxDistance] where minDistance is the minimum distance between any two distinct critical points and maxDistance is the maximum distance between any two distinct critical points. If there are fewer than two critical points, return [-1, -1].

Example 1:

Input: head = [3,1]
Output: [-1,-1]
Explanation: There are no critical points in [3,1].

Example 2:

Input: head = [5,3,1,2,5,1,2]
Output: [1,3]
Explanation: There are three critical points:
- [5,3,1,2,5,1,2]: The third node is a local minima because 1 is less than 3 and 2.
- [5,3,1,2,5,1,2]: The fifth node is a local maxima because 5 is greater than 2 and 1.
- [5,3,1,2,5,1,2]: The sixth node is a local minima because 1 is less than 5 and 2.
The minimum distance is between the fifth and the sixth node. minDistance = 6 - 5 = 1.
The maximum distance is between the third and the sixth node. maxDistance = 6 - 3 = 3.

Example 3:

Input: head = [1,3,2,2,3,2,2,2,7]
Output: [3,3]
Explanation: There are two critical points:
- [1,3,2,2,3,2,2,2,7]: The second node is a local maxima because 3 is greater than 1 and 2.
- [1,3,2,2,3,2,2,2,7]: The fifth node is a local maxima because 3 is greater than 2 and 2.
Both the minimum and maximum distances are between the second and the fifth node.
Thus, minDistance and maxDistance is 5 - 2 = 3.
Note that the last node is not considered a local maxima because it does not have a next node.

Example 4:

Input: head = [2,3,3,2]
Output: [-1,-1]
Explanation: There are no critical points in [2,3,3,2].

Constraints:

The number of nodes in the list is in the range [2, 10⁵].
1 <= Node.val <= 10⁵

Solution: One Pass

Track the first and last critical points.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> nodesBetweenCriticalPoints(ListNode* head) {

ListNode* p = nullptr;

int min_dist = INT_MAX;

int max_dist = INT_MIN;

for (int i = 0, l = -1, r = -1; head; ++i, p = head, head = head->next) {

if (p && head->next) {

if ((head->val > p->val && head->val > head->next->val)

|| (head->val < p->val && head->val < head->next->val)) {

if (r != -1) {

min_dist = min(min_dist, i - r);

max_dist = max(max_dist, i - l);

} else {

l = i;

}

r = i;

}

return { min_dist == INT_MAX ? -1 : min_dist,

max_dist == INT_MIN ? -1 : max_dist };

}

};

花花酱 LeetCode 2057. Smallest Index With Equal Value

By zxi on November 3, 2021

Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist.

x mod y denotes the remainder when x is divided by y.

Example 1:

Input: nums = [0,1,2]
Output: 0
Explanation: 
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
All indices have i mod 10 == nums[i], so we return the smallest index 0.

Example 2:

Input: nums = [4,3,2,1]
Output: 2
Explanation: 
i=0: 0 mod 10 = 0 != nums[0].
i=1: 1 mod 10 = 1 != nums[1].
i=2: 2 mod 10 = 2 == nums[2].
i=3: 3 mod 10 = 3 != nums[3].
2 is the only index which has i mod 10 == nums[i].

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9,0]
Output: -1
Explanation: No index satisfies i mod 10 == nums[i].

Example 4:

Input: nums = [2,1,3,5,2]
Output: 1
Explanation: 1 is the only index with i mod 10 == nums[i].

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 9

Solution: Brute Force

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

C++

// Author: Huahua

class Solution {

public:

int smallestEqual(vector<int>& nums) {

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

if (i % 10 == nums[i]) return i;

return -1;

}

};

花花酱 LeetCode 2055. Plates Between Candles

By zxi on November 3, 2021

There is a long table with a line of plates and candles arranged on top of it. You are given a 0-indexed string s consisting of characters '*' and '|' only, where a '*' represents a plate and a '|' represents a candle.

You are also given a 0-indexed 2D integer array queries where queries[i] = [left_i, right_i] denotes the substring s[left_i...right_i] (inclusive). For each query, you need to find the number of plates between candles that are in the substring. A plate is considered between candles if there is at least one candle to its left and at least one candle to its right in the substring.

For example, s = "||**||**|*", and a query [3, 8] denotes the substring "*||**|". The number of plates between candles in this substring is 2, as each of the two plates has at least one candle in the substring to its left and right.

Return an integer array answer where answer[i] is the answer to the i^th query.

Example 1:

Input: s = "**|**|***|", queries = [[2,5],[5,9]]
Output: [2,3]
Explanation:
- queries[0] has two plates between candles.
- queries[1] has three plates between candles.

Example 2:

Input: s = "***|**|*****|**||**|*", queries = [[1,17],[4,5],[14,17],[5,11],[15,16]]
Output: [9,0,0,0,0]
Explanation:
- queries[0] has nine plates between candles.
- The other queries have zero plates between candles.

Constraints:

3 <= s.length <= 10⁵
s consists of '*' and '|' characters.
1 <= queries.length <= 10⁵
queries[i].length == 2
0 <= left_i <= right_i < s.length

Solution: Binary Search

Store the indices of all candles into an array idx.

For each query q:
1. Find the left most candle whose index is greater or equal to left as l.
2. Find the left most candle whose index is greater than right, choose the previous candle as r.

[idx[l], idx[r]] is the range that are elements between two candles , there are (idx[r] – idx[l] + 1) elements in total and there are (r – l + 1) candles in the range. So the number of plates is (idx[r] – idx[l] + 1) – (r – l + 1) or (idx[r] – idx[l]) – (r – l)

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

C++

// Author: huahua

class Solution {

public:

vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) {

vector<int> idx;

for (size_t i = 0; i < s.size(); ++i)

if (s[i] == '|') idx.push_back(i);

vector<int> ans;

for (const auto& q : queries) {

const int l = lower_bound(begin(idx), end(idx), q[0]) - begin(idx);

const int r = upper_bound(begin(idx), end(idx), q[1]) - begin(idx) - 1;

ans.push_back(l >= r ? 0 : (idx[r] - idx[l]) - (r - l));

}

return ans;

}

};

花花酱 LeetCode 2054. Two Best Non-Overlapping Events

By zxi on November 1, 2021

You are given a 0-indexed 2D integer array of events where events[i] = [startTime_i, endTime_i, value_i]. The i^th event starts at startTime_iand ends at endTime_i, and if you attend this event, you will receive a value of value_i. You can choose at most two non-overlapping events to attend such that the sum of their values is maximized.

Return this maximum sum.

Note that the start time and end time is inclusive: that is, you cannot attend two events where one of them starts and the other ends at the same time. More specifically, if you attend an event with end time t, the next event must start at or after t + 1.

Example 1:

Input: events = [[1,3,2],[4,5,2],[2,4,3]]
Output: 4
Explanation: Choose the green events, 0 and 1 for a sum of 2 + 2 = 4.

Example 2:

Input: events = [[1,3,2],[4,5,2],[1,5,5]]
Output: 5
Explanation: Choose event 2 for a sum of 5.

Example 3:

Input: events = [[1,5,3],[1,5,1],[6,6,5]]
Output: 8
Explanation: Choose events 0 and 2 for a sum of 3 + 5 = 8.

Constraints:

2 <= events.length <= 10⁵
events[i].length == 3
1 <= startTime_i <= endTime_i <= 10⁹
1 <= value_i <= 10⁶

Solution: Sort + Heap

Sort events by start time, process them from left to right.

Use a min heap to store the events processed so far, a variable cur to track the max value of a non-overlapping event.

For a given event, pop all non-overlapping events whose end time is smaller than its start time and update cur.

ans = max(val + cur)

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

C++

// Author: Huahua

class Solution {

public:

int maxTwoEvents(vector<vector<int>>& events) {

using E = pair<int,int>;

sort(begin(events), end(events));

priority_queue<E, vector<E>, greater<E>> q; // (end_time, val)

int ans = 0;

int cur = 0;

for (const auto& e : events) {

while (!q.empty() && q.top().first < e[0]) {

cur = max(cur, q.top().second);

q.pop();

}

ans = max(ans, cur + e[2]);

q.emplace(e[1], e[2]);

}

return ans;

}

};

花花酱 LeetCode 2053. Kth Distinct String in an Array

By zxi on November 1, 2021

A distinct string is a string that is present only once in an array.

Given an array of strings arr, and an integer k, return the k^th distinct string present in arr. If there are fewer than k distinct strings, return an empty string "".

Note that the strings are considered in the order in which they appear in the array.

Example 1:

Input: arr = ["d","b","c","b","c","a"], k = 2
Output: "a"
Explanation:
The only distinct strings in arr are "d" and "a".
"d" appears 1^st, so it is the 1^st distinct string.
"a" appears 2^nd, so it is the 2^nd distinct string.
Since k == 2, "a" is returned.

Example 2:

Input: arr = ["aaa","aa","a"], k = 1
Output: "aaa"
Explanation:
All strings in arr are distinct, so the 1^st string "aaa" is returned.

Example 3:

Input: arr = ["a","b","a"], k = 3
Output: ""
Explanation:
The only distinct string is "b". Since there are fewer than 3 distinct strings, we return an empty string "".

Constraints:

1 <= k <= arr.length <= 1000
1 <= arr[i].length <= 5
arr[i] consists of lowercase English letters.

Solution: Hashtable

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

C++

// Author: Huahua

class Solution {

public:

string kthDistinct(vector<string>& arr, int k) {

unordered_map<string, int> m;

for (const string& s : arr)

++m[s];

for (const string& s : arr)

if (m[s] == 1 && --k == 0) return s;

return "";

}

};

Huahua's Tech Road

LeetCode 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points

Solution: One Pass

C++

花花酱 LeetCode 2057. Smallest Index With Equal Value

Solution: Brute Force

C++

花花酱 LeetCode 2055. Plates Between Candles

Solution: Binary Search

C++

花花酱 LeetCode 2054. Two Best Non-Overlapping Events

Solution: Sort + Heap

C++

花花酱 LeetCode 2053. Kth Distinct String in an Array

Solution: Hashtable

C++