CF 104635B3 - You Can Go Your Own Way - B3

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Solve time: 50s
Verified: yes

Solution

Problem Understanding

We are given a square grid of size n by n. A person named Lydia has already chosen a path from the top-left corner to the bottom-right corner, moving only right or down. This path is encoded as a string of length 2n − 2, where each character represents a single move: either a step to the right or a step downward.

Our task is to construct any other valid path from the same start to the same destination, with the same move constraints, such that our path is not identical to Lydia’s path.

The output is also a string of length 2n − 2 consisting of the same move symbols, but it must differ from the given one in at least one position.

The key constraint is that both paths must be valid grid paths, meaning they contain exactly n − 1 moves in each direction. This ensures we always stay within the grid and reach the bottom-right corner.

Even though the input size is small enough that linear scanning is trivial, the constraint on path validity means we cannot arbitrarily permute moves unless we preserve counts.

A naive approach would try to generate all valid paths and compare them against Lydia’s. This immediately fails because the number of valid paths is combinatorial, on the order of $\binom{2n-2}{n-1}$, which grows exponentially with n.

A more subtle failure case appears when trying to “slightly modify” Lydia’s path greedily while preserving feasibility. For example, changing a move early without checking feasibility can force an impossible suffix.

Example failure scenario:

Input:

3
E S E S

If we try to flip the first character to match the opposite direction without considering counts, we might break validity by producing too many of one move later, making it impossible to reach the destination.

The real requirement is simpler: we only need a valid path that differs somewhere, not a lexicographically constrained or locally optimal one.

Approaches

A brute-force method would enumerate all paths from (0, 0) to (n − 1, n − 1), checking each against Lydia’s path and stopping at the first mismatch. This is correct because every valid path is considered, but the number of such paths grows exponentially with n, making this infeasible even for moderate values. In the worst case, there are roughly $O(2^{2n})$ candidate sequences.

The key observation is that we do not need any structure beyond validity and difference. Any valid path that is not identical is acceptable. This removes all ordering constraints except maintaining the correct number of E and S moves.

Since Lydia’s path contains exactly n − 1 E moves and n − 1 S moves, we can construct another valid path by transforming each move independently while preserving counts globally. The simplest consistent transformation is to swap every move: replace each E with S and each S with E. This guarantees that we still produce exactly n − 1 of each move, because the multiset of moves is preserved under swapping.

This construction also guarantees difference from Lydia’s path, because every position is flipped, so equality is impossible unless the string is symmetric under swapping E and S, which cannot happen for a non-empty valid path.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force Enumeration	O(2^(2n))	O(2n)	Too slow
Swap Each Move	O(n)	O(n)	Accepted

Algorithm Walkthrough

Read the integer n and the path string s of length 2n − 2. This string encodes Lydia’s movements on the grid.
Create an empty result string r. We will build our alternative path character by character.
Iterate through each character in s. For each position, apply a deterministic transformation: if the character is 'E', append 'S' to r, and if it is 'S', append 'E' to r. This ensures that every move is flipped independently.
After processing all characters, output r as the final path.

Why it works

The construction preserves the total number of each move because every E is converted into an S and every S is converted into an E. Since Lydia’s path already contains exactly n − 1 of each direction, the transformed path also contains exactly n − 1 of each, ensuring it remains a valid grid path from start to finish. The two paths differ at every position due to the bijective per-character flip, so equality with the original path is impossible.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input().strip())
    s = input().strip()
    
    res = []
    for c in s:
        if c == 'E':
            res.append('S')
        else:
            res.append('E')
    
    print("".join(res))

if __name__ == "__main__":
    solve()

The implementation reads the input path and constructs the output in a single pass. The transformation logic is local per character, which avoids any need for tracking prefix constraints or grid coordinates. The only subtle point is ensuring that the input string is consumed entirely and no additional parsing is needed beyond reading n and the path.

Worked Examples

Example 1

Input:

3
EESSSS

We process each character and flip it.

Step	Input Char	Output Char	Partial Result
1	E	S	S
2	E	S	SS
3	S	E	SSE
4	S	E	SSEE
5	S	E	SSEEE
6	S	E	SSEEEE

Output:

SSEEEE

This demonstrates that the transformation preserves counts and produces a different path at every position.

Example 2

Input:

4
ESESEESE

Step	Input Char	Output Char	Partial Result
1	E	S	S
2	S	E	SE
3	E	S	SES
4	S	E	SESE
5	E	S	SESES
6	E	S	SESESS
7	S	E	SESESSE
8	E	S	SESESSSE

Output:

SESESSSE

The trace shows consistent flipping, ensuring validity and guaranteed mismatch.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each character is processed once with constant work
Space	O(n)	Output string stores a single transformed path

The algorithm fits comfortably within typical constraints for n up to 10^5 or larger, since it requires only a single linear scan with no auxiliary data structures.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    
    n = int(input().strip())
    s = input().strip()
    
    res = []
    for c in s:
        res.append('S' if c == 'E' else 'E')
    return "".join(res)

# basic sample
assert run("3\nEESSSS\n") == "SSEEEE"

# minimal case
assert run("2\nES\n") == "SE"

# all same direction alternating edge
assert run("3\nESES\n") == "SESE"

# larger balanced case
assert run("4\nEESSEESS\n") == "SSEESSEE"

# symmetric pattern check
assert run("3\nEEE SSS".replace(" ","")) == "SSS EEE".replace(" ","")

Test input	Expected output	What it validates
n=2, ES	SE	minimal grid correctness
n=3, EESSSS	SSEEEE	full flip correctness
n=4 alternating	reversed pattern	alternating structure stability
n=3 all E then S	all S then E	boundary grouping correctness

Edge Cases

One edge case is the smallest grid, where n = 2 and the path length is 2. The input might be something like ES. The algorithm flips each character, producing SE, which remains a valid path because it still contains one E and one S.

Another case is when Lydia’s path alternates strictly, such as ESESESE. The transformation still works because each character is independent of neighbors, so the alternation pattern is preserved structurally but inverted, producing SESESES, which is still valid.

A third case is when the path is grouped, such as all E moves followed by all S moves. Even here, flipping produces all S followed by all E, maintaining exact counts and validity while guaranteeing a completely different route.