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Posts published in “Algorithms”

花花酱 LeetCode 1184. Distance Between Bus Stops

A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.

The bus goes along both directions i.e. clockwise and counterclockwise.

Return the shortest distance between the given start and destination stops.

Example 1:

Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.

Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.

Constraints:

  • 1 <= n <= 10^4
  • distance.length == n
  • 0 <= start, destination < n
  • 0 <= distance[i] <= 10^4

Solution: Summation

  1. compute the total sum
  2. compute the sum from s to d, c
  3. ans = min(c, sum – c)

Time complexity: O(d-s)
Space complexity: O(1)

C++

花花酱 LeetCode 34. Find First and Last Position of Element in Sorted Array

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Solution: Binary Search

Basically this problem asks you to implement lower_bound and upper_bound using binary search.

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

C++

花花酱 LeetCode 1157. Online Majority Element In Subarray

Implementing the class MajorityChecker, which has the following API:

  • MajorityChecker(int[] arr) constructs an instance of MajorityChecker with the given array arr;
  • int query(int left, int right, int threshold) has arguments such that:
    • 0 <= left <= right < arr.length representing a subarray of arr;
    • 2 * threshold > right - left + 1, ie. the threshold is always a strict majority of the length of the subarray

Each query(...) returns the element in arr[left], arr[left+1], ..., arr[right]that occurs at least threshold times, or -1 if no such element exists.

Example:

MajorityChecker majorityChecker = new MajorityChecker([1,1,2,2,1,1]);
majorityChecker.query(0,5,4); // returns 1
majorityChecker.query(0,3,3); // returns -1
majorityChecker.query(2,3,2); // returns 2

Constraints:

  • 1 <= arr.length <= 20000
  • 1 <= arr[i] <= 20000
  • For each query, 0 <= left <= right < len(arr)
  • For each query, 2 * threshold > right - left + 1
  • The number of queries is at most 10000

Solution 0: Brute Force TLE

For each range, find the majority element in O(n) time / O(n) or O(1) space.

Solution 1: Cache the range result

Cache the result for each range: mode and frequency

Time complexity:
init: O(1)
query: O(|right – left|)
total queries: O(sum(right_i – left_i)), right_i, left_i are bounds of unique ranges.
Space complexity:
init: O(1)
query: O(20000)

C++

Solution 2: HashTable + binary search

Preprocessing: use a hashtable to store the indices of each element. And sort by frequency of the element in descending order.
Query: iterator over (total_freq, num) pair, if freq >= threshold do a binary search to find the frequency of the given range for num in O(logn).

Time complexity:
Init: O(nlogn)
Query: worst case: O(nlogn), best case: O(logn)

C++

Solution 2+: Randomization

Randomly pick a num in range [left, right] and check its freq, try k (e.g. 30) times. If a majority element exists, you should find it with error rate 1/2^k.

Time complexity:
Init: O(n)
Query: O(30 * logn)
Space complexity: O(n)

C++

花花酱 1054. Distant Barcodes

In a warehouse, there is a row of barcodes, where the i-th barcode is barcodes[i].

Rearrange the barcodes so that no two adjacent barcodes are equal.  You may return any answer, and it is guaranteed an answer exists.

Example 1:

Input: [1,1,1,2,2,2]
Output: [2,1,2,1,2,1]

Example 2:

Input: [1,1,1,1,2,2,3,3]
Output: [1,3,1,3,2,1,2,1]

Note:

  1. 1 <= barcodes.length <= 10000
  2. 1 <= barcodes[i] <= 10000

Soluton: Sorting

Sort the element by their frequency in descending order. Fill the most frequent element first in even positions, if reach the end of the array, start from position 1 then 3, 5, …

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

C++

Solution 2: Find the most frequent

Actually, we only need to find the most frequent element and put in the even positions, then put the rest of the groups of elements in any order.

e.g. [1, 1, 2, 2, 2, 2, 2, 3, 4]
Can be
5*2 [2 – 2 – 2 – 2 – 2]
4*1 [2 4 2 – 2 – 2 – 2]
3*1 [2 4 2 3 2 – 2 – 2]
1*2 [ 2 3 2 3 2 1 2 1 2]

if we start with any other groups rather than 2, if will become:
[3 2 2 – 2 – 2 – 2 ] which is wrong…

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

C++

花花酱 LeetCode 1051. Height Checker

Students are asked to stand in non-decreasing order of heights for an annual photo.

Return the minimum number of students not standing in the right positions.  (This is the number of students that must move in order for all students to be standing in non-decreasing order of height.)

Example 1:

Input: [1,1,4,2,1,3]
Output: 3
Explanation: 
Students with heights 4, 3 and the last 1 are not standing in the right positions.

Note:

  1. 1 <= heights.length <= 100
  2. 1 <= heights[i] <= 100

Solution: Sorting

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

C++