CF 104635B1 - You Can Go Your Own Way - B1

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Solve time: 50s
Verified: yes

Solution

Problem Understanding

We are given a square grid of size $N \times N$. Someone else, Lydia, has already chosen a valid path from the top-left corner to the bottom-right corner, moving only right or down at each step. Her path is provided as a string consisting of two characters, where each character represents one move in the grid.

Our task is to construct a different valid path from the same start to the same destination, also using only right and down moves, but with the constraint that our path must not be identical to Lydia’s path. In other words, at least one step in our route must differ from hers in position, so the overall sequence of moves is not exactly the same.

The input encodes a single test case: the size of the grid and Lydia’s path string. The output is any valid alternative path of the same length that still reaches the destination.

The constraints are small enough that we only need linear processing of the string. Since the path length is $2N - 2$, any $O(N)$ construction is sufficient. Anything involving backtracking or enumeration of paths would be unnecessary and would introduce avoidable complexity.

A naive mistake in this type of problem is trying to construct a path that deviates early without guaranteeing it still ends at the bottom-right corner. For example, if we try to “swap” a move without tracking remaining balance of right and down steps, we might produce an invalid prefix.

Another subtle issue appears if one tries to change only a single character arbitrarily. If Lydia’s path is mostly monotonic in one direction early, a naive flip can still accidentally reproduce the same full sequence if not handled carefully in context.

Approaches

The brute-force idea would be to try all possible valid paths from the top-left to the bottom-right using only right and down moves and compare each against Lydia’s path. There are $\binom{2N-2}{N-1}$ such paths, which grows exponentially with $N$. Even for moderate $N$, this becomes completely infeasible since the number of paths is already astronomical.

The key observation is that we do not need a “special” different path, only any path that is not identical. This removes almost all complexity from the problem. Since a valid path is fully determined by the multiset of moves, any rearrangement that preserves the count of right and down moves still leads to a valid solution.

Lydia’s path consists of exactly $N-1$ moves to the right and $N-1$ moves downward. If we construct a path that swaps every move type, turning every right into a down and every down into a right, we automatically preserve validity: the counts remain identical, so the endpoint is still correct, and at each prefix we never exceed grid boundaries because we are simply traversing a symmetric path in the opposite direction pattern.

The important structural insight is that validity depends only on the total counts of each move, not their order, as long as we stay within prefix feasibility, which is preserved under this full inversion.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (enumerate all paths)	$O(\binom{2N}{N})$	$O(N)$	Too slow
Invert moves	$O(N)$	$O(N)$	Accepted

Algorithm Walkthrough

Read the integer $N$ and the string $S$ describing Lydia’s path. The string length is $2N - 2$, so it fully determines the sequence of moves.
Initialize an empty result list that will store our alternative path character by character.
Traverse each character in $S$ from left to right.
For each character, apply a direct transformation: if the character is “E”, append “S” to the result; if it is “S”, append “E”. This guarantees that every step differs from Lydia’s corresponding step while preserving the total number of each move type.
After processing the full string, join the result list into a single string and output it.

Why it works

The constructed path contains exactly the same number of “E” and “S” moves as Lydia’s path, since every occurrence is swapped one-to-one. This ensures the endpoint remains the same, because reaching the bottom-right corner depends only on having the correct counts of horizontal and vertical moves. At the same time, every position in the sequence differs from Lydia’s corresponding position, so the paths cannot be identical.

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 mirrors the construction directly. The loop performs a single pass over the input string, ensuring linear time behavior. The decision to use a list for building the output avoids repeated string concatenation, which would otherwise introduce quadratic overhead in Python.

A subtle point is that no additional validation is needed for grid bounds. Since the transformation preserves the exact number of each move, we implicitly maintain a valid lattice path from start to finish.

Worked Examples

Example 1

Input:

3
EES

We process each character step by step.

Step	Input char	Output built so far
1	E	S
2	E	SS
3	S	SSE

Final output:

SSE

This demonstrates that every move is flipped, guaranteeing a different path while still using exactly the same number of moves.

Example 2

Input:

4
SESEEE

Step	Input char	Output built so far
1	S	E
2	E	ES
3	S	ESE
4	E	ESES
5	E	ESESS
6	E	ESESSS

Final output:

ESESSS

This trace shows that even with alternating patterns, the transformation consistently preserves balance while guaranteeing difference at every index.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(N)$	Each character is processed once in a single pass
Space	$O(N)$	Output string stores one character per input character

The input size is linear in $N$, so a single traversal comfortably fits within typical constraints. Memory usage is also linear and dominated by the output string itself.

Test Cases

import sys, io

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

# custom cases
assert run("3\nEES\n") == "SSE", "basic flip"
assert run("2\nES\n") == "SE", "minimum grid"
assert run("5\nEEEE\n") == "SSSS", "all same direction"
assert run("4\nSESE\n") == "ESES", "alternating pattern"

Test input	Expected output	What it validates
3 / EES	SSE	basic correctness of flip logic
2 / ES	SE	minimum non-trivial grid
5 / EEEE	SSSS	uniform direction handling
4 / SESE	ESES	alternating structure preservation

Edge Cases

A corner case occurs when Lydia’s path consists entirely of one direction early on, such as all “E” moves first. A naive approach that tries to change only a single position might still produce a path that coincides in structure for long prefixes, but the full inversion guarantees every position differs immediately.

Another case is the smallest grid where $N = 2$. There is only one move of each type, so any correct solution must swap that single pair. The algorithm naturally handles this because it processes each character independently.

For highly alternating paths, the transformation still preserves validity because it does not rely on patterns or structure, only on per-character inversion.