Bus Stop - Microsoft Top Interview Questions


Problem Statement :


You are given a list of integers nums representing bus stops on a line where nums[i] represents the time a bus must arrive at stop i.

Given that buses can only move forward, return the minimum number of buses that are needed to pass through all the stops.

Constraints

n ≤ 1,000 where n is length of nums.

Example 1

Input

nums = [1, 2, 7, 9, 3, 4]

Output

2

Explanation

One bus can take this route [1, 2, 3, 4] and another can do [7, 9].



Example 2

Input

nums = [5, 5]

Output

2

Explanation

One bus can't be at two places at once so we need 2 busses.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& nums) {
    multiset<int> buses;

    for (int num : nums)
        if (buses.empty() or num <= *buses.begin())
            buses.insert(num);
        else
            buses.erase(prev(buses.lower_bound(num))), buses.insert(num);

    return buses.size();
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        int ans = 0;
        TreeMap<Integer, Integer> set = new TreeMap<>();
        for (int i : nums) {
            Integer floor = set.floorKey(i);
            if (floor != null && floor.intValue() == i)
                floor = set.floorKey(i - 1);
            if (floor == null) {
                ans++;
            } else {
                if (set.get(floor) == 1)
                    set.remove(floor);
                else
                    set.put(floor, set.getOrDefault(floor, 0) - 1);
            }
            set.put(i, set.getOrDefault(i, 0) + 1);
        }
        return ans;
    }
}
                    


                        Solution in Python : 
                            
from bisect import insort, bisect_right


class Solution:
    def solve(self, a):
        n = len(a)
        buses = [a[n - 1]]
        for i in range(n - 2, -1, -1):
            idx = bisect_right(buses, a[i])
            if idx == len(buses):
                buses.append(a[i])
            else:
                buses[idx] = a[i]
        return len(buses)


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