Watch the video solution for Number of Student Replacements, a medium level problem involving Array, Simulation. This walkthrough by AlgorithmicIQ has 9 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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:
ranks[0] = 4 is initially selected.ranks[1] = 1 is better than the current selection, so a replacement occurs.Example 2:
Input: ranks = [2,2,3]
Output: 0
Explanation:
ranks[0] = 2 is initially selected.ranks[1] = 2 or ranks[2] = 3 is better than the current selection.
Constraints:
1 <= ranks.length <= 1051 <= ranks[i] <= 105Problem 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.
| Approach | Time | Space | When 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 Simulation | O(n) | O(1) | Best for production or interview settings where linear performance is expected |