2024-09-22
440. K-th Smallest in Lexicographical Order

Topic: Trie
Difficulty: Hard

Problem:
Given two integers n and k, return the k^th lexicographically smallest integer in the range [1, n].

Example 1:

Input: n = 13, k = 2
Output: 10
Explanation: The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.

Example 2:

Input: n = 1, k = 1
Output: 1

Constraints:

• 1 <= k <= n <= 10^9

2024-09-23
2707. Extra Characters in a String

Topic: Array, Hash Table, String, Dynamic Programming, Trie
Difficulty: Medium

Problem:
You are given a 0-indexed string s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings.

Return the minimum number of extra characters left over if you break up s optimally.

Example 1:

Input: s = "leetscode", dictionary = ["leet","code","leetcode"]
Output: 1
Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.

Example 2:

Input: s = "sayhelloworld", dictionary = ["hello","world"]
Output: 3
Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.

Constraints:

• 1 <= s.length <= 50
• 1 <= dictionary.length <= 50
• 1 <= dictionary[i].length <= 50
• dictionary[i] and s consists of only lowercase English letters
• dictionary contains distinct words

2024-09-24
3043. Find the Length of the Longest Common Prefix

Topic: Array, Hash Table, String, Trie
Difficulty: Medium

Problem:
You are given two arrays with positive integers arr1 and arr2.

A prefix of a positive integer is an integer formed by one or more of its digits, starting from its leftmost digit. For example, 123 is a prefix of the integer 12345, while 234 is not.

A common prefix of two integers a and b is an integer c, such that c is a prefix of both a and b. For example, 5655359 and 56554 have a common prefix 565 while 1223 and 43456 do not have a common prefix.

You need to find the length of the longest common prefix between all pairs of integers (x, y) such that x belongs to arr1 and y belongs to arr2.

Return the length of the longest common prefix among all pairs. If no common prefix exists among them, return 0.

Example 1:

Input: arr1 = [1,10,100], arr2 = [1000]
Output: 3
Explanation: There are 3 pairs (arr1[i], arr2[j]):
- The longest common prefix of (1, 1000) is 1.
- The longest common prefix of (10, 1000) is 10.
- The longest common prefix of (100, 1000) is 100.
The longest common prefix is 100 with a length of 3.

Example 2:

Input: arr1 = [1,2,3], arr2 = [4,4,4]
Output: 0
Explanation: There exists no common prefix for any pair (arr1[i], arr2[j]), hence we return 0.
Note that common prefixes between elements of the same array do not count.

Constraints:

• 1 <= arr1.length, arr2.length <= 5 * 10^4
• 1 <= arr1[i], arr2[i] <= 10^8

2024-09-25
2416. Sum of Prefix Scores of Strings

Topic: Array, String, Trie, Counting
Difficulty: Hard

Problem:
You are given an array words of size n consisting of non-empty strings.

We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].

• For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".

Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].

Note that a string is considered as a prefix of itself.

Example 1:

Input: words = ["abc","ab","bc","b"]
Output: [5,4,3,2]
Explanation: The answer for each string is the following:
- "abc" has 3 prefixes: "a", "ab", and "abc".
- There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc".
The total is answer[0] = 2 + 2 + 1 = 5.
- "ab" has 2 prefixes: "a" and "ab".
- There are 2 strings with the prefix "a", and 2 strings with the prefix "ab".
The total is answer[1] = 2 + 2 = 4.
- "bc" has 2 prefixes: "b" and "bc".
- There are 2 strings with the prefix "b", and 1 string with the prefix "bc".
The total is answer[2] = 2 + 1 = 3.
- "b" has 1 prefix: "b".
- There are 2 strings with the prefix "b".
The total is answer[3] = 2.

Example 2:

Input: words = ["abcd"]
Output: [4]
Explanation:
"abcd" has 4 prefixes: "a", "ab", "abc", and "abcd".
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length <= 1000
• words[i] consists of lowercase English letters.

2024-09-26
729. My Calendar I

Topic: Array, Binary Search, Design, Segment Tree, Ordered Set
Difficulty: Medium

Problem:
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a double booking.

A double booking happens when two events have some non-empty intersection (i.e., some moment is common to both events.).

The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end.

Implement the MyCalendar class:

• MyCalendar() Initializes the calendar object.
• boolean book(int start, int end) Returns true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input
["MyCalendar", "book", "book", "book"]
[[], [10, 20], [15, 25], [20, 30]]
Output
[null, true, false, true]

Explanation
MyCalendar myCalendar = new MyCalendar();
myCalendar.book(10, 20); // return True
myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event.
myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.

Constraints:

• 0 <= start < end <= 10^9
• At most 1000 calls will be made to book.

2024-09-27
731. My Calendar II

Topic: Array, Binary Search, Design, Segment Tree, Prefix Sum, Ordered Set
Difficulty: Medium

Problem:
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.

A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).

The event can be represented as a pair of integers start and end that represents a booking on the half-open interval [start, end), the range of real numbers x such that start <= x < end.

Implement the MyCalendarTwo class:

• MyCalendarTwo() Initializes the calendar object.
• boolean book(int start, int end) Returns true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input
["MyCalendarTwo", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, true, true, true, false, true, true]

Explanation
MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
myCalendarTwo.book(10, 20); // return True, The event can be booked. 
myCalendarTwo.book(50, 60); // return True, The event can be booked. 
myCalendarTwo.book(10, 40); // return True, The event can be double booked. 
myCalendarTwo.book(5, 15);  // return False, The event cannot be booked, because it would result in a triple booking.
myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.
myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event, the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Constraints:

• 0 <= start < end <= 10^9
• At most 1000 calls will be made to book.

2024-09-28
641. Design Circular Deque

Topic: Array, Linked List, Design, Queue
Difficulty: Medium

Problem:
Design your implementation of the circular double-ended queue (deque).

Implement the MyCircularDeque class:

• MyCircularDeque(int k) Initializes the deque with a maximum size of k.
• boolean insertFront() Adds an item at the front of Deque. Returns true if the operation is successful, or false otherwise.
• boolean insertLast() Adds an item at the rear of Deque. Returns true if the operation is successful, or false otherwise.
• boolean deleteFront() Deletes an item from the front of Deque. Returns true if the operation is successful, or false otherwise.
• boolean deleteLast() Deletes an item from the rear of Deque. Returns true if the operation is successful, or false otherwise.
• int getFront() Returns the front item from the Deque. Returns -1 if the deque is empty.
• int getRear() Returns the last item from Deque. Returns -1 if the deque is empty.
• boolean isEmpty() Returns true if the deque is empty, or false otherwise.
• boolean isFull() Returns true if the deque is full, or false otherwise.

Example 1:

Input
["MyCircularDeque", "insertLast", "insertLast", "insertFront", "insertFront", "getRear", "isFull", "deleteLast", "insertFront", "getFront"]
[[3], [1], [2], [3], [4], [], [], [], [4], []]
Output
[null, true, true, true, false, 2, true, true, true, 4]

Explanation
MyCircularDeque myCircularDeque = new MyCircularDeque(3);
myCircularDeque.insertLast(1);  // return True
myCircularDeque.insertLast(2);  // return True
myCircularDeque.insertFront(3); // return True
myCircularDeque.insertFront(4); // return False, the queue is full.
myCircularDeque.getRear();      // return 2
myCircularDeque.isFull();       // return True
myCircularDeque.deleteLast();   // return True
myCircularDeque.insertFront(4); // return True
myCircularDeque.getFront();     // return 4

Constraints:

• 1 <= k <= 1000
• 0 <= value <= 1000
• At most 2000 calls will be made to insertFront, insertLast, deleteFront, deleteLast, getFront, getRear, isEmpty, isFull.

2024-09-29
432. All O`one Data Structure

Topic: Hash Table, Linked List, Design, Doubly-Linked List
Difficulty: Hard

Problem:
Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.

Implement the AllOne class:

• AllOne() Initializes the object of the data structure.
• inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1.
• dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement.
• getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "".
• getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "".

Note that each function must run in O(1) average time complexity.

Example 1:

Input
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
Output
[null, null, null, "hello", "hello", null, "hello", "leet"]

Explanation
AllOne allOne = new AllOne();
allOne.inc("hello");
allOne.inc("hello");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "hello"
allOne.inc("leet");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "leet"

Constraints:

• 1 <= key.length <= 10
• key consists of lowercase English letters.
• It is guaranteed that for each call to dec, key is existing in the data structure.
• At most 5 * 10^4 calls will be made to inc, dec, getMaxKey, and getMinKey.

2024-09-30
1381. Design a Stack With Increment Operation

Topic: Array, Stack, Design
Difficulty: Medium

Problem:
Design a stack that supports increment operations on its elements.

Implement the CustomStack class:

• CustomStack(int maxSize) Initializes the object with maxSize which is the maximum number of elements in the stack.
• void push(int x) Adds x to the top of the stack if the stack has not reached the maxSize.
• int pop() Pops and returns the top of the stack or -1 if the stack is empty.
• void inc(int k, int val) Increments the bottom k elements of the stack by val. If there are less than k elements in the stack, increment all the elements in the stack.

Example 1:

Input
["CustomStack","push","push","pop","push","push","push","increment","increment","pop","pop","pop","pop"]
[[3],[1],[2],[],[2],[3],[4],[5,100],[2,100],[],[],[],[]]
Output
[null,null,null,2,null,null,null,null,null,103,202,201,-1]
Explanation
CustomStack stk = new CustomStack(3); // Stack is Empty []
stk.push(1);                          // stack becomes [1]
stk.push(2);                          // stack becomes [1, 2]
stk.pop();                            // return 2 --> Return top of the stack 2, stack becomes [1]
stk.push(2);                          // stack becomes [1, 2]
stk.push(3);                          // stack becomes [1, 2, 3]
stk.push(4);                          // stack still [1, 2, 3], Do not add another elements as size is 4
stk.increment(5, 100);                // stack becomes [101, 102, 103]
stk.increment(2, 100);                // stack becomes [201, 202, 103]
stk.pop();                            // return 103 --> Return top of the stack 103, stack becomes [201, 202]
stk.pop();                            // return 202 --> Return top of the stack 202, stack becomes [201]
stk.pop();                            // return 201 --> Return top of the stack 201, stack becomes []
stk.pop();                            // return -1 --> Stack is empty return -1.

Constraints:

• 1 <= maxSize, x, k <= 1000
• 0 <= val <= 100
• At most 1000 calls will be made to each method of increment, push and pop each separately.

2024-10-01
1497. Check If Array Pairs Are Divisible by k

Topic: Array, Hash Table, Counting
Difficulty: Medium

Problem:
Given an array of integers arr of even length n and an integer k.

We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.

Return true If you can find a way to do that or false otherwise.

Example 1:

Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
Output: true
Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).

Example 2:

Input: arr = [1,2,3,4,5,6], k = 7
Output: true
Explanation: Pairs are (1,6),(2,5) and(3,4).

Example 3:

Input: arr = [1,2,3,4,5,6], k = 10
Output: false
Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.

Constraints:

• arr.length == n
• 1 <= n <= 10^5
• n is even.
• -10^9 <= arr[i] <= 10^9
• 1 <= k <= 10^5

2024-10-02
1331. Rank Transform of an Array

Topic: Array, Hash Table, Sorting
Difficulty: Easy

Problem:
Given an array of integers arr, replace each element with its rank.

The rank represents how large the element is. The rank has the following rules:

• Rank is an integer starting from 1.
• The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
• Rank should be as small as possible.

Example 1:

Input: arr = [40,10,20,30]
Output: [4,1,2,3]
Explanation: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.

Example 2:

Input: arr = [100,100,100]
Output: [1,1,1]
Explanation: Same elements share the same rank.

Example 3:

Input: arr = [37,12,28,9,100,56,80,5,12]
Output: [5,3,4,2,8,6,7,1,3]

Constraints:

• 0 <= arr.length <= 10^5
• -10^9 <= arr[i] <= 10^9

2024-10-03
1590. Make Sum Divisible by P

Topic: Array, Hash Table, Prefix Sum
Difficulty: Medium

Problem:
Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.

Return the length of the smallest subarray that you need to remove, or -1 if it's impossible.

A subarray is defined as a contiguous block of elements in the array.

Example 1:

Input: nums = [3,1,4,2], p = 6
Output: 1
Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

Example 2:

Input: nums = [6,3,5,2], p = 9
Output: 2
Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

Example 3:

Input: nums = [1,2,3], p = 3
Output: 0
Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

Constraints:

• 1 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9
• 1 <= p <= 10^9

2024-10-04
2491. Divide Players Into Teams of Equal Skill

Topic: Array, Hash Table, Two Pointers, Sorting
Difficulty: Medium

Problem:
You are given a positive integer array skill of even length n where skill[i] denotes the skill of the i^th player. Divide the players into n / 2 teams of size 2 such that the total skill of each team is equal.

The chemistry of a team is equal to the product of the skills of the players on that team.

Return the sum of the chemistry of all the teams, or return -1 if there is no way to divide the players into teams such that the total skill of each team is equal.

Example 1:

Input: skill = [3,2,5,1,3,4]
Output: 22
Explanation: 
Divide the players into the following teams: (1, 5), (2, 4), (3, 3), where each team has a total skill of 6.
The sum of the chemistry of all the teams is: 1 * 5 + 2 * 4 + 3 * 3 = 5 + 8 + 9 = 22.

Example 2:

Input: skill = [3,4]
Output: 12
Explanation: 
The two players form a team with a total skill of 7.
The chemistry of the team is 3 * 4 = 12.

Example 3:

Input: skill = [1,1,2,3]
Output: -1
Explanation: 
There is no way to divide the players into teams such that the total skill of each team is equal.

Constraints:

• 2 <= skill.length <= 10^5
• skill.length is even.
• 1 <= skill[i] <= 1000

2024-10-05
567. Permutation in String

Topic: Hash Table, Two Pointers, String, Sliding Window
Difficulty: Medium

Problem:
Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1's permutations is the substring of s2.

Example 1:

Input: s1 = "ab", s2 = "eidbaooo"
Output: true
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:

Input: s1 = "ab", s2 = "eidboaoo"
Output: false

Constraints:

• 1 <= s1.length, s2.length <= 10^4
• s1 and s2 consist of lowercase English letters.

2024-10-06
1813. Sentence Similarity III

Topic: Array, Two Pointers, String
Difficulty: Medium

Problem:
You are given two strings sentence1 and sentence2, each representing a sentence composed of words. A sentence is a list of words that are separated by a single space with no leading or trailing spaces. Each word consists of only uppercase and lowercase English characters.

Two sentences s1 and s2 are considered similar if it is possible to insert an arbitrary sentence (possibly empty) inside one of these sentences such that the two sentences become equal. Note that the inserted sentence must be separated from existing words by spaces.

For example,

• s1 = "Hello Jane" and s2 = "Hello my name is Jane" can be made equal by inserting "my name is" between "Hello" and "Jane" in s1.
• s1 = "Frog cool" and s2 = "Frogs are cool" are not similar, since although there is a sentence "s are" inserted into s1, it is not separated from "Frog" by a space.

Given two sentences sentence1 and sentence2, return true if sentence1 and sentence2 are similar. Otherwise, return false.

Example 1:

Input: sentence1 = "My name is Haley", sentence2 = "My Haley"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "name is" between "My" and "Haley".

Example 2:

Input: sentence1 = "of", sentence2 = "A lot of words"

Output: false

Explanation:

No single sentence can be inserted inside one of the sentences to make it equal to the other.

Example 3:

Input: sentence1 = "Eating right now", sentence2 = "Eating"

Output: true

Explanation:

sentence2 can be turned to sentence1 by inserting "right now" at the end of the sentence.

Constraints:

• 1 <= sentence1.length, sentence2.length <= 100
• sentence1 and sentence2 consist of lowercase and uppercase English letters and spaces.
• The words in sentence1 and sentence2 are separated by a single space.

2024-10-07
2696. Minimum String Length After Removing Substrings

Topic: String, Stack, Simulation
Difficulty: Easy

Problem:
You are given a string s consisting only of uppercase English letters.

You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s.

Return the minimum possible length of the resulting string that you can obtain.

Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings.

Example 1:

Input: s = "ABFCACDB"
Output: 2
Explanation: We can do the following operations:
- Remove the substring "ABFCACDB", so s = "FCACDB".
- Remove the substring "FCACDB", so s = "FCAB".
- Remove the substring "FCAB", so s = "FC".
So the resulting length of the string is 2.
It can be shown that it is the minimum length that we can obtain.

Example 2:

Input: s = "ACBBD"
Output: 5
Explanation: We cannot do any operations on the string so the length remains the same.

Constraints:

• 1 <= s.length <= 100
• s consists only of uppercase English letters.

2024-10-08
1963. Minimum Number of Swaps to Make the String Balanced

Topic: Two Pointers, String, Stack, Greedy
Difficulty: Medium

Problem:
You are given a 0-indexed string s of even length n. The string consists of exactly n / 2 opening brackets '[' and n / 2 closing brackets ']'.

A string is called balanced if and only if:

• It is the empty string, or
• It can be written as AB, where both A and B are balanced strings, or
• It can be written as [C], where C is a balanced string.

You may swap the brackets at any two indices any number of times.

Return the minimum number of swaps to make s balanced.

Example 1:

Input: s = "][]["
Output: 1
Explanation: You can make the string balanced by swapping index 0 with index 3.
The resulting string is "[[]]".

Example 2:

Input: s = "]]][[["
Output: 2
Explanation: You can do the following to make the string balanced:
- Swap index 0 with index 4. s = "[]][][".
- Swap index 1 with index 5. s = "[[][]]".
The resulting string is "[[][]]".

Example 3:

Input: s = "[]"
Output: 0
Explanation: The string is already balanced.

Constraints:

• n == s.length
• 2 <= n <= 10^6
• n is even.
• s[i] is either '[' or ']'.
• The number of opening brackets '[' equals n / 2, and the number of closing brackets ']' equals n / 2.

2024-10-09
921. Minimum Add to Make Parentheses Valid

Topic: String, Stack, Greedy
Difficulty: Medium

Problem:
A parentheses string is valid if and only if:

• It is the empty string,
• It can be written as AB (A concatenated with B), where A and B are valid strings, or
• It can be written as (A), where A is a valid string.

You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string.

• For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "())))".

Return the minimum number of moves required to make s valid.

Example 1:

Input: s = "())"
Output: 1

Example 2:

Input: s = "((("
Output: 3

Constraints:

• 1 <= s.length <= 1000
• s[i] is either '(' or ')'.

2024-10-10
962. Maximum Width Ramp

Topic: Array, Stack, Monotonic Stack
Difficulty: Medium

Problem:
A ramp in an integer array nums is a pair (i, j) for which i < j and nums[i] <= nums[j]. The width of such a ramp is j - i.

Given an integer array nums, return the maximum width of a ramp in nums. If there is no ramp in nums, return 0.

Example 1:

Input: nums = [6,0,8,2,1,5]
Output: 4
Explanation: The maximum width ramp is achieved at (i, j) = (1, 5): nums[1] = 0 and nums[5] = 5.

Example 2:

Input: nums = [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: The maximum width ramp is achieved at (i, j) = (2, 9): nums[2] = 1 and nums[9] = 1.

Constraints:

• 2 <= nums.length <= 5 * 10^4
• 0 <= nums[i] <= 5 * 10^4

2024-10-11
1942. The Number of the Smallest Unoccupied Chair

Topic: Array, Hash Table, Heap (Priority Queue)
Difficulty: Medium

Problem:
There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity. When a friend arrives at the party, they sit on the unoccupied chair with the smallest number.

• For example, if chairs 0, 1, and 5 are occupied when a friend comes, they will sit on chair number 2.

When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair.

You are given a 0-indexed 2D integer array times where times[i] = [arrival_i, leaving_i], indicating the arrival and leaving times of the i^th friend respectively, and an integer targetFriend. All arrival times are distinct.

Return the chair number that the friend numbered targetFriend will sit on.

Example 1:

Input: times = [[1,4],[2,3],[4,6]], targetFriend = 1
Output: 1
Explanation: 
- Friend 0 arrives at time 1 and sits on chair 0.
- Friend 1 arrives at time 2 and sits on chair 1.
- Friend 1 leaves at time 3 and chair 1 becomes empty.
- Friend 0 leaves at time 4 and chair 0 becomes empty.
- Friend 2 arrives at time 4 and sits on chair 0.
Since friend 1 sat on chair 1, we return 1.

Example 2:

Input: times = [[3,10],[1,5],[2,6]], targetFriend = 0
Output: 2
Explanation: 
- Friend 1 arrives at time 1 and sits on chair 0.
- Friend 2 arrives at time 2 and sits on chair 1.
- Friend 0 arrives at time 3 and sits on chair 2.
- Friend 1 leaves at time 5 and chair 0 becomes empty.
- Friend 2 leaves at time 6 and chair 1 becomes empty.
- Friend 0 leaves at time 10 and chair 2 becomes empty.
Since friend 0 sat on chair 2, we return 2.

Constraints:

• n == times.length
• 2 <= n <= 10^4
• times[i].length == 2
• 1 <= arrival_i < leaving_i <= 10^5
• 0 <= targetFriend <= n - 1
• Each arrival_i time is distinct.