LeetCode 3616 - Number of Student Replacements

LeetCode Problem 3616

Difficulty: 🟡 Medium
Topics: Array, Simulation

Solution

Problem Understanding

The problem models a simple selection process among students arriving one by one. Each student has a rank, where a smaller rank value represents a better student.

The first arriving student is automatically selected. As additional students arrive, we compare their rank against the rank of the currently selected student. If the new student has a strictly better rank, meaning a smaller rank value, they replace the current selection. Every such replacement increases our answer by one.

Our goal is to count how many times this replacement event occurs throughout the entire arrival sequence.

Another way to view the problem is that we are tracking the best rank seen so far. Whenever we encounter a new minimum value in the array after the first element, a replacement occurs.

For example, given ranks = [4, 1, 2]:

• The first student with rank 4 is selected.
• Rank 1 is better than 4, so a replacement occurs.
• Rank 2 is not better than 1, so no replacement occurs.

The answer is 1.

The constraints are:

• 1 <= ranks.length <= 100000
• 1 <= ranks[i] <= 100000

These constraints indicate that the input can be quite large, up to one hundred thousand elements. This strongly suggests that we should aim for a linear time solution. Any quadratic approach would be too slow in the worst case.

Important edge cases include:

• An array containing only one student. Since there are no later arrivals, the answer must be 0.
• Multiple students with identical ranks. Since replacements require a strictly better rank, equal ranks do not trigger replacements.
• A strictly decreasing sequence. Every new student is better than all previous students, producing the maximum possible number of replacements.
• A strictly increasing sequence. The first student remains selected forever, producing zero replacements.

Approaches

Brute Force

A straightforward approach is to process each student and determine whether they are better than the current selection by explicitly searching through all previously seen students to identify the best rank so far.

For each position i, we could scan the prefix ranks[0...i-1] to find the minimum rank seen before. If ranks[i] is smaller than that minimum, we count a replacement.

This approach is correct because a replacement occurs exactly when the current rank is smaller than every rank seen earlier. However, finding the previous minimum from scratch for every position requires repeated scans of the array.

With n students, this results in approximately:

• First position scans 0 elements.
• Second position scans 1 element.
• Third position scans 2 elements.
• And so on.

The total work becomes proportional to 1 + 2 + 3 + ... + (n - 1), which is O(n²).

For n = 100000, this is far too slow.

Optimal Approach

The key observation is that we do not need to repeatedly recompute the best rank seen so far.

At any moment, the only information needed is the current selected student's rank, which is simply the minimum rank encountered so far.

As we iterate through the array:

• Maintain best_rank, the smallest rank seen so far.

• For each new student:

• If their rank is smaller than best_rank, a replacement occurs.

• Increment the answer.

• Update best_rank to this new rank.

Since each element is processed exactly once and each operation is constant time, the solution runs in linear time.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n²)	O(1)	Recomputes the minimum rank for every position
Optimal	O(n)	O(1)	Tracks the minimum rank seen so far during a single pass

Algorithm Walkthrough

1. Initialize best_rank with the rank of the first student because the first student is selected by default.
2. Initialize a counter replacements to 0. This variable will store the total number of replacement events.
3. Iterate through the remaining students from index 1 to the end of the array.
4. For each student's rank:

• Compare it with best_rank.
• If the current rank is strictly smaller, the student is better than the currently selected student.

5. When a strictly smaller rank is found:

• Increment replacements by one.
• Update best_rank to the new smaller rank because this student becomes the new selection.

6. If the current rank is equal to or larger than best_rank, do nothing because no replacement occurs.
7. After processing all students, return replacements.

Why it works

The crucial invariant is that best_rank always stores the smallest rank among all students processed so far. Since the currently selected student must always be the best student seen up to that point, a replacement occurs exactly when a newly arriving student has a rank smaller than best_rank. By updating best_rank whenever this happens, the invariant remains true throughout the scan, guaranteeing that every replacement is counted exactly once.

Python Solution

from typing import List

class Solution:
    def totalReplacements(self, ranks: List[int]) -> int:
        best_rank = ranks[0]
        replacements = 0

        for rank in ranks[1:]:
            if rank < best_rank:
                replacements += 1
                best_rank = rank

        return replacements

The implementation begins by storing the first student's rank in best_rank, since that student is selected initially.

The variable replacements tracks how many times a better student arrives.

The loop processes every remaining student exactly once. Whenever the current rank is smaller than best_rank, a replacement occurs. The counter is incremented and best_rank is updated to reflect the new selected student.

If the rank is equal to or larger than best_rank, the current selection remains unchanged, so the algorithm simply continues.

Finally, the total number of replacements is returned.

Go Solution

func totalReplacements(ranks []int) int {
	bestRank := ranks[0]
	replacements := 0

	for i := 1; i < len(ranks); i++ {
		if ranks[i] < bestRank {
			replacements++
			bestRank = ranks[i]
		}
	}

	return replacements
}

The Go implementation follows exactly the same logic as the Python version.

Because the constraints guarantee that the array contains at least one element, accessing ranks[0] is always safe. No special handling for empty slices is required.

The values involved are at most 100000, and the number of replacements is at most n - 1, so standard Go int values are more than sufficient. No overflow concerns exist.

Worked Examples

Example 1

Input:

ranks = [4, 1, 2]

Initial state:

Index	Rank	best_rank	replacements
0	4	4	0

Process remaining students:

Index	Rank	Condition	Action	best_rank	replacements
1	1	1 < 4	Replacement	1	1
2	2	2 < 1 is false	No change	1	1

Final answer:

1

Example 2

Input:

ranks = [2, 2, 3]

Initial state:

Index	Rank	best_rank	replacements
0	2	2	0

Process remaining students:

Index	Rank	Condition	Action	best_rank	replacements
1	2	2 < 2 is false	No change	2	0
2	3	3 < 2 is false	No change	2	0

Final answer:

0

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each student is examined exactly once
Space	O(1)	Only a few variables are maintained regardless of input size

The algorithm performs a single linear scan through the array. Every iteration consists of one comparison and possibly one update, both constant time operations. No auxiliary data structures are used, so memory consumption remains constant regardless of input size.

Test Cases

sol = Solution()

assert sol.totalReplacements([4, 1, 2]) == 1      # provided example
assert sol.totalReplacements([2, 2, 3]) == 0      # provided example

assert sol.totalReplacements([5]) == 0            # single student
assert sol.totalReplacements([1, 2, 3, 4, 5]) == 0  # strictly increasing
assert sol.totalReplacements([5, 4, 3, 2, 1]) == 4  # strictly decreasing

assert sol.totalReplacements([3, 3, 3, 3]) == 0  # all equal ranks
assert sol.totalReplacements([10, 5, 5, 4]) == 2 # equal values do not replace

assert sol.totalReplacements([8, 6, 7, 5, 4]) == 3  # multiple new minima
assert sol.totalReplacements([100000, 1]) == 1       # constraint boundary values
assert sol.totalReplacements([1, 100000]) == 0       # best rank first

assert sol.totalReplacements([7, 2, 6, 1, 5]) == 2  # non-consecutive replacements

Test	Why
[4, 1, 2]	Validates the first example
[2, 2, 3]	Validates that equal ranks do not replace
[5]	Smallest valid input
[1, 2, 3, 4, 5]	No replacements occur
[5, 4, 3, 2, 1]	Every new student replaces the previous one
[3, 3, 3, 3]	All values equal
[10, 5, 5, 4]	Tests strict inequality
[8, 6, 7, 5, 4]	Multiple replacement events
[100000, 1]	Boundary rank values
[1, 100000]	Best student arrives first
[7, 2, 6, 1, 5]	Replacements separated by non-replacements

Edge Cases

Single Student

When the array contains only one student, there are no later arrivals that could trigger a replacement. A common bug is to assume there is always at least one comparison to perform. The implementation handles this naturally because the loop over ranks[1:] executes zero times, leaving the answer as 0.

Equal Ranks

The problem specifies that a replacement occurs only when a student's rank is strictly better. Equal ranks should not trigger replacements. An incorrect implementation might use <= instead of <, causing extra replacements to be counted. The solution correctly uses a strict comparison.

Strictly Decreasing Sequence

For an input such as [5, 4, 3, 2, 1], every arriving student is better than the current selection. This produces the maximum possible number of replacements, which is n - 1. The implementation correctly counts each new minimum as a replacement and updates best_rank after every step.

Strictly Increasing Sequence

For an input such as [1, 2, 3, 4, 5], the first student remains the best throughout the process. No replacements should occur. Since every subsequent rank is larger than best_rank, the replacement counter never increases and the algorithm correctly returns 0.

Duplicate Values Mixed With Improvements

Consider [10, 5, 5, 4]. The second student replaces the first, the third student has the same rank as the current selection and should not replace, and the fourth student replaces because 4 is strictly better than 5. The implementation handles this correctly by updating only when the rank is strictly smaller than the current minimum.