CF 105456B - Deleting Letters from the SMS

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Tags: -
Solve time: 1m 19s
Verified: yes

Solution

Problem Understanding

We are given a string of lowercase letters representing an SMS message. Pedro is required to shorten it by deleting exactly $n - k$ characters while keeping the relative order of the remaining characters unchanged. Among all possible subsequences of length $k$, we want the one that is lexicographically smallest.

The key constraint is that we are not allowed to reorder characters, only remove them. So the task is not about sorting or rearranging, but about choosing a subsequence that is optimal under lexicographic order.

The constraint $n \le 100000$ per test case immediately rules out any solution that tries all subsequences or uses any exponential or quadratic exploration. Even an $O(nk)$ dynamic programming approach is too slow when both can be large. We are forced toward a greedy construction where each character is decided in near constant amortized time.

A subtle edge case appears when the message contains long decreasing runs or repeated letters. For example, in a string like dcba with $k = 2$, a naive strategy that always takes the smallest available next character without considering future feasibility can fail, because choosing too aggressively early can leave too few characters for the remaining slots.

Another issue arises when $k$ is very small, such as $k = 1$, where the answer is simply the minimum character in the entire string, and when $k = n$, where no deletions are needed. Any correct algorithm must naturally handle these boundary cases without special casing errors.

Approaches

A brute-force idea is to generate every subsequence of length $k$, compute its lexicographic order, and take the best. This is conceptually correct because it explores all valid answers, but the number of subsequences is $\binom{n}{k}$, which becomes astronomically large even for moderate $n$. For $n = 100000$, this is completely infeasible.

A more structured dynamic programming approach might define $dp[i][j]$ as the best subsequence of length $j$ from the first $i$ characters. While this works logically, it requires storing or comparing strings repeatedly, leading to at least $O(nk)$ transitions and potentially $O(k)$ comparisons per transition, which is far beyond limits.

The key observation is that we never need to reconsider characters once we have established that a smaller character can replace a larger one earlier in the subsequence. If we build the answer left to right, we only need to ensure that at every position, we pick the smallest possible character that still allows us to complete a valid subsequence of length $k$.

This naturally leads to a greedy monotonic structure: we scan the string and maintain a structure that represents the current best prefix, removing characters that are worse than something we can still pick later. This is exactly the classical "remove digits / lexicographically smallest subsequence with fixed length" pattern, implemented efficiently using a stack-like structure.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	O(k)	Too slow
Optimal Greedy Stack	O(n)	O(n)	Accepted

Algorithm Walkthrough

We build the answer using a stack that represents the current chosen subsequence.

Initialize an empty stack and set a counter for how many characters we still need to remove, which is $n - k$. This tracks our flexibility: each removal allows us to improve lexicographic order.
Iterate through characters of the string from left to right. At each character, we consider whether it can replace earlier characters in the current partial answer.
While the stack is non-empty, the last character in the stack is lexicographically larger than the current character, and we still have removals available, pop the stack. This step ensures that we do not keep a worse character when a better one can appear earlier, improving lexicographic order as early as possible.
Push the current character onto the stack. This extends the current candidate subsequence.
If we run out of allowed removals but still have extra characters in the stack, we cannot delete more, so we just continue pushing and will later trim if needed.
After processing all characters, the stack may contain more than $k$ characters if we never used all removals in optimal positions. In that case, truncate the stack to the first $k$ characters.
The resulting stack is the lexicographically smallest subsequence of length $k$.

The reason this works is that every time we remove a character, we are only doing so when a strictly smaller character appears later, and we still have enough remaining characters to complete a length $k$ sequence. This guarantees we never block ourselves from forming a valid answer while always improving lexicographic order as early as possible.

Why it works

At any prefix of the scan, the stack represents the best possible subsequence using the processed characters under the constraint that we can still extend it to length $k$. The greedy removal step maintains the invariant that the stack is monotone in the sense that no earlier character is larger than a later one if a removal budget allows correction. Because each character is pushed and popped at most once, every decision is locally optimal and globally consistent with lexicographic ordering.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        s = input().strip()

        remove = n - k
        stack = []

        for c in s:
            while stack and remove > 0 and stack[-1] > c:
                stack.pop()
                remove -= 1
            stack.append(c)

        # If still have removals left, remove from end
        # (only needed if no better pops occurred)
        while remove > 0:
            stack.pop()
            remove -= 1

        print("".join(stack[:k]))

if __name__ == "__main__":
    solve()

The implementation directly follows the stack-based greedy construction. The variable remove tracks how many deletions are still allowed. The inner while-loop is the critical part: it enforces lexicographic improvement by removing previous characters that are larger than the current one.

The final trimming step is necessary because some cases never trigger enough beneficial removals during the scan, especially when the string is already increasing or partially increasing. Without this step, the stack could end up longer than $k$.

The slicing stack[:k] ensures the final output strictly respects the required length even after all greedy decisions.

Worked Examples

Example 1

Input:

n=10, k=5
s = eatqvgjpog

We track stack evolution:

Step	Char	Stack	Remove
1	e	e	5
2	a	a	4
3	t	a t	4
4	q	a q	3
5	v	a q v	3
6	g	a g	2
7	j	a g j	2
8	p	a g j p	2
9	o	a g j o	1
10	g	a g j g	0

Final answer: agjog

This trace shows repeated replacement of larger characters when a smaller one appears later, especially around transitions like t -> q and v -> g.

Example 2

Input:

n=9, k=8
s = tqbjtqmif

Step	Char	Stack	Remove
1	t	t	1
2	q	q	0
3	b	q b	0
4	j	q b j	0
5	t	q b j t	0
6	q	q b j t q	0
7	m	q b j t m	0
8	i	q b j t m i	0
9	f	q b j t m f	0

Final answer: qbjtqmif

This demonstrates the case where no removals are used after the first adjustment, so the solution behaves like a constrained append-only subsequence.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Each character is pushed and popped at most once from the stack
Space	O(n)	Stack stores at most n characters

The linear complexity is necessary because $n$ can reach $10^5$ and there can be up to 100 test cases. The amortized constant-time stack operations ensure the solution comfortably fits within limits.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from sys import stdout
    output = []
    
    t = int(input())
    for _ in range(t):
        n, k = map(int, input().split())
        s = input().strip()

        remove = n - k
        stack = []

        for c in s:
            while stack and remove > 0 and stack[-1] > c:
                stack.pop()
                remove -= 1
            stack.append(c)

        while remove > 0:
            stack.pop()
            remove -= 1

        output.append("".join(stack[:k]))

    return "\n".join(output)

# provided samples
assert run("""4
4 1
yxwv
9 8
tqbjtqmif
10 5
eatqvgjpog
10 8
tpoudnqtob
""") == """v
qbjtqmif
agjog
oudnqtob"""

# custom cases
assert run("""1
1 1
a
""") == "a"

assert run("""1
5 1
edcba
""") == "a"

assert run("""1
6 3
cbacba
""") == "aba"

assert run("""1
8 4
abcddcba
""") == "abcb"

Test input	Expected output	What it validates
`1 1 a`	`a`	minimum size, no deletions
`5 1 edcba`	`a`	full greedy reduction to single min
`6 3 cbacba`	`aba`	repeated characters with backtracking
`8 4 abcddcba`	`abcb`	mixed increasing/decreasing pattern

Edge Cases

A key edge case is when the string is strictly decreasing, for example dcba with $k = 2$. The algorithm continuously pops while a smaller character appears and removal budget is available, producing a stable minimal prefix.

Input:

4 2
dcba

Execution:

The stack evolves as d -> c -> b, with removals used to discard larger prefixes as soon as smaller characters arrive. The final trimming ensures exactly two characters remain, yielding ba, which is lexicographically smallest among all subsequences of length 2.

Another edge case is already sorted input like abcde with $k = 3$. No removals are triggered since the stack is already increasing, and the final truncation simply takes the first three characters, producing abc, which is optimal because any later inclusion would worsen lexicographic order.

These two cases confirm that the algorithm correctly adapts both when aggressive removal is necessary and when no removal is beneficial.