Is Number of Student Replacements easy or hard?

The problem is typically considered Medium difficulty. The logic itself is straightforward, but recognizing that the process can be simulated in a single pass instead of repeated scans is the key insight.

Number of Student Replacements Python/Java solution

The solution can be implemented in Python, Java, C++, Go, or TypeScript using the same logic. Iterate through the array, apply the replacement rule, increment a counter when a replacement occurs, and update the current state accordingly.

How to solve Number of Student Replacements in O(n)?

Traverse the array once and simulate the replacement rule using a running state. Whenever the condition for a replacement is met, increment the replacement counter and update the state immediately. Because no element needs to be revisited, the entire process completes in linear time.

What is the best approach for Number of Student Replacements?

The best approach is a single-pass simulation over the array. By maintaining the current state while iterating and applying the replacement rule immediately, you can count all replacements in O(n) time with O(1) extra space. This avoids repeated rescans required by naive simulation.

Is Number of Student Replacements asked at Google/Amazon/Meta?

Problems involving array simulation and process-based state updates are common in technical interviews at companies like Google, Amazon, and Meta. While the exact problem may vary, the pattern of scanning an array and applying rule-based updates appears frequently.

What data structure is used in Number of Student Replacements?

The primary data structure is an array. The algorithm processes the array sequentially and maintains a few variables to track the simulation state and the number of replacements.

What is the time complexity of Number of Student Replacements?

The optimal solution runs in O(n) time because each student in the array is processed once during the simulation. Space complexity is O(1) since the algorithm only maintains a few counters or state variables.

Description Hints 2 Solution Video 8:20

3616. Number of Student Replacements

Medium85.1% acceptancePremium elsewhere, free here

You are given an integer array ranks where ranks[i] represents the rank of the i^th student arriving in order. A lower number indicates a better rank.

Initially, the first student is selected by default.

A replacement occurs when a student with a strictly better rank arrives and replaces the current selection.

Return the total number of replacements made.

Example 1:

Input: ranks = [4,1,2]

Output: 1

Explanation:

The first student with ranks[0] = 4 is initially selected.
The second student with ranks[1] = 1 is better than the current selection, so a replacement occurs.
The third student has a worse rank, so no replacement occurs.
Thus, the number of replacements is 1.

Example 2:

Input: ranks = [2,2,3]

Output: 0

Explanation:

The first student with ranks[0] = 2 is initially selected.
Neither of ranks[1] = 2 or ranks[2] = 3 is better than the current selection.
Thus, the number of replacements is 0.

Constraints:

1 <= ranks.length <= 10⁵
1 <= ranks[i] <= 10⁵

Topics

Array Simulation

Hints open one at a time

Hint 1

Iterate through the array from left to right, keeping track of the current best (minimum) rank.

Hint 2

Whenever you see an element strictly lower than your current best, update the best to that element and increment the replacement count.

8:20

Still stuck?Read the written solution, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[4,1,2]

Input

[2,2,3]

Your code is saved on this device.Solution