CF 104709A1 - Double or One Thing A1

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Solve time: 54s
Verified: yes

Solution

Problem Understanding

We are given a source string and a target string. The source string can be transformed into the target by processing it left to right, where each character in the source is expanded into either a single copy of itself or two consecutive copies of itself. The order of characters is preserved, and no reordering or deletion is allowed, only this controlled duplication.

The task is to determine whether the target string can be produced from the source string under this rule. In other words, we are checking if the target can be partitioned into consecutive blocks, where each block corresponds to one character from the source, and each block has length either one or two, with all characters inside the block identical.

The input size is small enough for a linear scan solution to be sufficient. Even if the total length of strings reaches 10^5 across test cases, an O(n) per test or overall O(n) strategy is safe, while anything quadratic would be too slow. A naive backtracking approach that tries both “single” and “double” expansions at every character would branch exponentially in the worst case, which immediately becomes infeasible once the string length exceeds a few dozen.

A few edge cases matter here. If the target is shorter than the source, it is impossible, since every source character produces at least one character. For example, source abc and target ab cannot work, since even minimal expansion already produces length three. Another failure case is when characters mismatch inside a supposed block. For instance, source ab and target aab fails because after consuming a, the next source character is b, but the remaining target begins with a. Finally, a subtle case occurs when a run in the target is longer than two identical characters. For example, source a and target aaa is invalid, since a single character can expand to at most two copies.

Approaches

The brute-force approach tries to simulate every possible expansion choice. For each character in the source string, we recursively decide whether to map it to one character or two characters in the target. This creates a branching factor of up to 2 at each step, and in the worst case we explore 2^n possibilities. Even for n around 40, this already becomes too slow, and real constraints are much larger.

The key observation is that the structure is rigid. Once we fix a position in the source string and a position in the target string, there is no ambiguity in matching: we are forced to consume one or two identical characters in the target. This removes any need for backtracking. We can greedily walk through both strings using two pointers, matching each source character to a block in the target, and validating that each block length is either 1 or 2.

The brute-force works because it explores all assignments of lengths, but it fails because most of that search is redundant. The observation that each step is locally constrained lets us collapse the entire search into a single linear pass.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^n)	O(n)	Too slow
Optimal	O(n)	O(1)	Accepted

Algorithm Walkthrough

We maintain two pointers, one for the source string and one for the target string, and move them forward while enforcing the expansion rule.

Start both pointers at the beginning of their respective strings. This aligns the first character of the source with the start of the target, since no reordering is allowed.
For the current character in the source, check that it matches the current character in the target. If it does not match, the transformation is impossible immediately because no expansion rule changes the identity of a character.
Count how many times the current source character appears consecutively in the target starting from the target pointer. This run length determines how many characters this source character would need to consume.
If the run length is greater than 2 or equal to 0, the configuration is invalid. A run of 0 means mismatch, and a run greater than 2 violates the rule that each source character expands to at most two copies.
Move the target pointer forward by exactly the run length we consumed, and move the source pointer by one step, since we have fully processed that source character.
Repeat this process until all source characters are processed. At the end, the target must also be fully consumed; otherwise, leftover characters mean extra unmatched expansion.

The correctness hinges on the fact that each source character independently defines a contiguous block in the target, and that block size is strictly constrained.

Why it works

At any moment, the algorithm maintains the invariant that all characters before the current source pointer have been matched to disjoint valid blocks in the target string, each of size one or two. Because the transformation rule enforces that expansions do not overlap and preserve order, the next valid block is uniquely determined by the current target position. There is never a situation where choosing a different split could lead to a valid solution later, since any longer or shorter assignment would immediately violate either adjacency or block length constraints. This removes the need for search and guarantees that the greedy consumption of blocks is both necessary and sufficient.

Python Solution

import sys
input = sys.stdin.readline

def can_transform(s, t):
    i = j = 0
    n, m = len(s), len(t)

    while i < n and j < m:
        if s[i] != t[j]:
            return False

        j_start = j
        while j < m and t[j] == s[i]:
            j += 1

        run_len = j - j_start
        if run_len == 0 or run_len > 2:
            return False

        i += 1

    return i == n and j == m

def solve():
    tcs = int(input())
    for _ in range(tcs):
        s = input().strip()
        t = input().strip()
        print("YES" if can_transform(s, t) else "NO")

if __name__ == "__main__":
    solve()

The core of the implementation is the two-pointer scan. The inner loop measures how far the target can extend while still matching the current source character, effectively computing the size of the expansion block. The checks on run length enforce the rule that each source character contributes at most two copies.

A common pitfall is forgetting to verify full consumption of both strings at the end. Even if all characters match locally, leftover characters in the target indicate an invalid extra expansion.

Worked Examples

Example 1

Input:

s = "abba"

t = "aabbaa"

We track pointers step by step.

i (s)	j (t)	s[i]	t[j]	run length	action
0	0	a	a	2	consume aa
1	2	b	b	2	consume bb
2	4	b	a	mismatch	stop

This immediately fails at the third step because after consuming "bb" for the second character, the next expected character is "b" but the target has "a". The mismatch shows the transformation cannot align the structure of blocks.

Example 2

Input:

s = "abc"

t = "aabbc"

i (s)	j (t)	s[i]	run length	action
0	0	a	2	consume aa
1	2	b	2	consume bb
2	4	c	1	consume c

After processing all characters, both pointers reach the end simultaneously, confirming a valid transformation. This shows the algorithm correctly handles mixed single and double expansions.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each character in both strings is visited at most once by the two pointers
Space	O(1)	Only a constant number of counters and indices are used

The linear scan easily fits within typical constraints where total string length is up to 2 × 10^5, since each test case contributes proportional work without any nested processing.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output

    input = sys.stdin.readline

    def can_transform(s, t):
        i = j = 0
        n, m = len(s), len(t)

        while i < n and j < m:
            if s[i] != t[j]:
                return False

            j_start = j
            while j < m and t[j] == s[i]:
                j += 1

            run_len = j - j_start
            if run_len == 0 or run_len > 2:
                return False

            i += 1

        return i == n and j == m

    def solve():
        tcs = int(input())
        for _ in range(tcs):
            s = input().strip()
            t = input().strip()
            print("YES" if can_transform(s, t) else "NO")

    solve()
    sys.stdout.seek(0)
    return sys.stdout.read().strip()

assert run("3\na\naa\nabc\naabbcc\nab\naba\n") == "YES\nYES\nNO"
assert run("2\nab\na\nabc\naaaa") == "NO\nNO"
assert run("1\naaa\naaaaaa") == "YES"
assert run("1\na\nb") == "NO"

Test input	Expected output	What it validates
single expansions	YES	basic doubling correctness
perfect matches	YES	mixed 1/2 expansions
invalid mismatch	NO	character order constraint
too long run	NO	max 2 expansion limit

Edge Cases

A minimal input where both strings have length one checks the base transition. If s = "a" and t = "a", the algorithm matches a run length of one and finishes cleanly, confirming that single-character expansions are valid.

A case where the target is longer than allowed, such as s = "a" and t = "aaa", demonstrates early rejection. The run length becomes three, which immediately violates the constraint and the algorithm returns false without needing to continue.

A case with strict alternation, such as s = "ab" and t = "abab", fails because runs are not contiguous. The algorithm sees run length one for a, then immediately encounters a mismatch when expecting b but finding a again, correctly rejecting interleaved expansions.