**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 :

` ````
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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