CF 1203D2 - Remove the Substring (hard version)

Rating: 1700
Tags: binary search, greedy, implementation, two pointers
Solve time: 5m 33s
Verified: yes

Solution

Problem Understanding

We are given two strings, s and t. The second string is guaranteed to already appear inside the first one as a subsequence, meaning we can pick characters from s in order and obtain t.

The task is to remove one contiguous segment from s so that after deletion, t is still obtainable as a subsequence of the remaining string. Among all valid deletions, we want the maximum possible length of the removed segment.

So conceptually, we are searching for a “hole” in s: we cut out a continuous block, and we need the remaining left and right parts to still contain all characters of t in order.

The constraints allow both strings to be up to 200,000 characters. Any solution that tries all substrings would require checking O(n²) candidates, which is far too slow because each check would also require subsequence verification. This pushes us toward an O(n log n) or O(n) strategy.

A key subtle edge case appears when t is very small or even a single character. In that case, removing almost the entire string is often possible, but only if at least one occurrence of that character remains outside the removed segment. For example, if s = "aaaaa" and t = "a", we can remove any contiguous block except one position, so the answer becomes 4.

Another corner case is when occurrences of t are tightly packed inside s. Then any removal that overlaps all valid embeddings of t fails, even if it seems large locally.

Approaches

A brute-force approach would try every pair of indices l and r, remove s[l:r], and then check if t is still a subsequence of the remaining string. There are O(n²) choices for the segment, and subsequence checking costs O(n), so this becomes O(n³), which is infeasible even for moderate input sizes.

We can improve by observing that we do not need to fully recheck subsequences for each removal. Instead, we want to know how far we can “push” a split between left and right parts while still being able to match t.

The key idea is to fix how much of t is matched from the prefix of s, and how much from the suffix. If we know that the first i characters of t can be matched in the prefix of s, and the remaining suffix of t can be matched in the suffix of s, then everything between those two matched regions can be removed.

So we precompute two structures: the earliest positions in s where prefixes of t end, and the latest positions in s where suffixes of t start. Then we try all splits of t into prefix and suffix.

This reduces the problem to linear scanning with two pointers.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n³)	O(1)	Too slow
Prefix-Suffix Matching	O(n)	O(n)	Accepted

Algorithm Walkthrough

Build an array pref[i] meaning the earliest index in s where we can match the prefix t[0:i]. We scan s from left to right, advancing a pointer in t whenever characters match. This guarantees we always match as early as possible, leaving maximum room on the right.
Build an array suf[i] meaning the latest index in s where we can match the suffix t[i:]. We scan s from right to left, similarly greedily matching characters of t from the end. This ensures suffixes are packed as far right as possible.
Initialize answer as 0. We interpret removing a segment between two positions in s.
For every split point i in t, interpret it as taking prefix t[0:i] from the left side of s, and suffix t[i:] from the right side of s.
If prefix t[0:i] can be matched ending at position pref[i-1], and suffix t[i:] can be matched starting at position suf[i], then we can remove everything strictly between these two positions.
The removable length is suf[i] - pref[i-1] - 1. Track the maximum over all valid splits.
Also consider the case where we take empty prefix or empty suffix of t, which corresponds to removing a prefix or suffix of s.

Why it works

The construction forces every character of t to be assigned either to the left or right side of the removed segment. The greedy prefix and suffix matches ensure that if a split is feasible anywhere, it is feasible at these extremal placements. Any valid deletion corresponds to some partition of t into two subsequences, and the algorithm checks all such partitions while minimizing overlap in s, which maximizes the removable gap.

Python Solution

PythonRun

The pref array is built greedily so that each prefix of t is matched as early as possible in s. This is crucial because any later matching would only reduce the removable gap, never increase it.

The suf array is symmetric but from the right side, ensuring suffix matches are as far right as possible.

The loop over split points tries every way to divide t into two parts. The computed gap between the last matched prefix position and first matched suffix position is exactly the removable substring.

Worked Examples

Example 1

Input:

split i	prefix matched end	suffix matched start	removable
0	-1	0	0
1	0	1	0
2	1	5	3

The best split takes one b from the left and one from the right. The middle segment "aba" is removed.

This shows how splitting t allows us to isolate a large middle block while preserving order.

Example 2

Input:

split i	prefix end	suffix start	removable
0	-1	0	0
1	0	5	4

Only one character is needed for t, so removing everything except one occurrence is optimal.

This demonstrates the boundary case where prefix or suffix can be empty.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n + m)	One forward scan and one backward scan over `s`, plus a linear scan over `t` splits
Space	O(m)	Arrays storing prefix and suffix match positions

The constraints allow up to 200,000 characters, so linear scanning is necessary. Any solution worse than O(n log n) risks timing out due to large constant factors.

Test Cases

PythonRun

Test input	Expected output	What it validates
`bbaba / bb`	3	sample correctness
`aaaaa / a`	4	single character extreme
`abcde / ace`	1	interleaved subsequence matching
`abcd / d`	3	prefix removal extreme

Edge Cases

When t has only one character, the solution must correctly allow removing everything except one occurrence. A naive approach might incorrectly assume we need to preserve structure around multiple matches, but the split logic naturally handles empty prefix or suffix.

When s contains many repeated characters, greedy matching must always prefer earliest and latest placements. If matching is not extremal, the computed removable segment shrinks artificially. The prefix-suffix construction guarantees maximal gap.

When t is identical to s, every valid split collapses to zero removable length because prefix and suffix positions coincide. The formula right_pos - left_pos - 1 naturally produces 0 in all cases.