Skip to main content
Talentd

Number of Student Replacements - Solution & Explanation

MediumPremiumFree on FleetCodeArraySimulation5 min read
Practice this problem

Problem Statement

You are given an integer array ranks where ranks[i] represents the rank of the ith 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 <= 105ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹
  • 1 <= ranks[i] <= 105

Approach Overview

Problem Overview: You are given an array that represents the current state of students. Based on the rules defined in the problem, certain students get replaced during the process. The task is to simulate these changes and return the total number of replacements that occur.

Approach 1: Naive Simulation with Repeated Passes (O(n^2) time, O(1) space)

A straightforward solution directly simulates the process exactly as described. Iterate through the array and apply the replacement rule whenever the condition is satisfied. After each change, restart or continue scanning because a replacement may affect nearby positions. This approach mirrors the real process but can repeatedly revisit the same elements, leading to O(n^2) time in the worst case. Space usage remains O(1) since all updates happen directly inside the array.

Approach 2: Single-Pass Simulation (O(n) time, O(1) space)

The optimized method observes that replacements follow a predictable progression as you traverse the array. Instead of repeatedly rescanning, iterate once from left to right while maintaining the state required to determine whether a replacement should occur. When the rule condition is met, increment a counter and update the state to reflect the replacement. Because each element is processed only once, the algorithm runs in O(n) time and uses constant extra space.

This approach works well because the process does not require backtracking once the local state is updated. By maintaining the current simulation state while scanning the array, you effectively reproduce the same final result as the naive process but avoid redundant passes.

The implementation relies mainly on sequential traversal of an array and careful state updates typical of simulation problems. Many interview problems with process-based rules can be solved using a similar pattern: translate the rules into simple conditional checks during a single iteration.

Recommended for interviews: Start by explaining the naive simulation to show you understand the rules of the process. Then optimize to the single-pass simulation. Interviewers expect the O(n) traversal because it eliminates redundant work while keeping the implementation simple and readable.

Solution

We use a variable cur to record the rank of the currently selected student. We iterate through the array ranks, and if we encounter a student with a better rank (i.e., ranks[i] < cur), we update cur and increment the answer by one.

After the iteration, we return the answer.

The time complexity is O(n), where n is the number of students. The space complexity is O(1).

Code

PythonJavaC++GoTypeScript

Python

Java

C++

Go

TypeScript

Try this approach in the editor ā†’

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Naive Simulation (Repeated Scans)O(n^2)O(1)Useful for understanding the process rules or when constraints are very small
Single-Pass SimulationO(n)O(1)Best for production or interview settings where linear performance is expected

Video Solution

Leetcode 3616 Number of Student Replacements ā€¢ AlgorithmicIQ ā€¢ 10 views views

Frequently Asked Questions

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.

Ready to solve this problem?

Practice Number of Student Replacements with our built-in code editor and test cases.

Practice on FleetCode
Previous

Two Sum

Next

Median of Two Sorted Arrays