CF 935B - Fafa and the Gates

Rating: 900
Tags: implementation
Solve time: 2m 57s
Verified: yes

Solution

Problem Understanding

The problem is about counting the number of times Fafa crosses the wall between two kingdoms when walking along a grid according to a sequence of moves. The wall is along the line x = y, with a gate at every integer coordinate along that line. The first kingdom occupies points below the wall (y < x), the second kingdom occupies points above (y > x), and the gates themselves are neutral. Each time Fafa crosses from one kingdom to the other via a gate, he pays a coin. He starts at (0,0), which is a gate, and does not pay there. The input specifies the number of moves n and a string of length n consisting of 'R' (right) and 'U' (up) moves.

The key task is to determine how many coins Fafa must pay following this sequence. The challenge is not just tracking Fafa's position, but detecting when he crosses from one kingdom to the other at a gate without paying more than necessary.

With n up to 10^5, any solution iterating over all points in the grid is too slow. A solution must run in roughly O(n) time, as 10^5 operations per move are acceptable but quadratic or higher complexity would exceed the 2-second limit. Edge cases include sequences that remain entirely on one side, sequences that hug the wall, or sequences that cross repeatedly without extra movement along the diagonal.

For example, if S = "UUU", Fafa never moves right and never crosses the wall. The correct output is 0. A naive approach that checks every gate without considering the direction of movement could incorrectly count coins for non-crossings.

Approaches

A brute-force approach would simulate Fafa’s path on a grid, marking each step and checking if the move crosses the wall. This is conceptually correct: for each step, calculate Fafa's new position, determine the current kingdom, and compare it to the previous kingdom. If the kingdom changes while at a gate, increment the coin count. The problem is that checking the kingdom at every move is trivial, but over long sequences, a naive simulation that also checks every gate along the path or recalculates distances would be too slow.

The key insight is that Fafa only needs to pay when he crosses the wall at a gate and actually changes kingdoms. Gates lie on x = y, so the only positions where payment is possible are points where x = y. Therefore, we can track Fafa’s position along the diagonal: let r be the number of right moves and u the number of up moves. The number of coins increases when Fafa reaches a gate where either r = u and he is about to move into a different kingdom than the previous move. This reduces the problem to a single linear scan with simple counters.

By maintaining the counts of right and up moves and only checking positions where r = u, we can determine when Fafa switches kingdoms without explicitly simulating every grid cell. This results in a linear O(n) solution with O(1) extra space.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (grid simulation)	O(n)	O(1)	Accepted, but naive checks may be overkill
Optimal (track r, u, and gates)	O(n)	O(1)	Accepted

Algorithm Walkthrough

Initialize r = 0 and u = 0 to represent Fafa's current coordinates, and coins = 0 to accumulate payments. Also initialize a flag last_side to track which kingdom Fafa was in before the current move.
Iterate over each move in the sequence S. If the move is 'R', increment r. If the move is 'U', increment u. This updates Fafa's position on the grid.
After each move, check if Fafa is at a gate, i.e., if r == u. Gates only exist along the diagonal, and coins can only be paid at these points.
If at a gate, determine which kingdom Fafa would be in for the previous step versus the next step. Specifically, if r > u, Fafa is below the wall; if r < u, Fafa is above the wall. The last_side variable records the previous side.
If Fafa is at a gate and last_side differs from the current side, increment coins by 1 and update last_side to the current side. This captures a legitimate crossing of the wall.
Continue until all moves are processed. Output coins.

Why it works: The invariant is that coins is incremented exactly when Fafa crosses the wall at a gate. Because gates exist only along the diagonal, no moves outside this diagonal can cause coin payments. Tracking last_side ensures multiple consecutive moves along the diagonal do not erroneously count extra coins.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
S = input().strip()

r = u = coins = 0
last_side = 0  # 0: start, 1: below wall, 2: above wall

for move in S:
    if move == 'R':
        r += 1
    else:
        u += 1
    
    if r == u:
        if r > 0:
            if last_side == 0:
                # Determine first side
                last_side = 1 if r > u else 2
            else:
                # Current side after move
                side = 1 if r > u else 2
                if side != last_side:
                    coins += 1
                    last_side = side

print(coins)

The solution first updates the coordinates according to the move, checks for a gate, and then compares kingdoms. last_side prevents double-counting when multiple steps along the diagonal do not change kingdoms.

Worked Examples

Example 1

Input:

1
U

Move	r	u	r==u?	last_side	coins
U	0	1	False	0	0

No gate is reached, output is 0. The table confirms that single-step moves without crossing pay nothing.

Example 2

Input:

6
RUURUU

Move	r	u	r==u?	last_side	coins
R	1	0	False	0	0
U	1	1	True	0	1
U	1	2	False	1	1
R	2	2	True	1	2
U	2	3	False	2	2
U	2	4	False	2	2

Coins incremented correctly when crossing the wall at gates (1,1) and (2,2).

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	We iterate over the sequence once, performing O(1) work per move.
Space	O(1)	Only a few counters are used; memory does not scale with n.

The solution easily handles n = 10^5 within the 2-second limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    S = input().strip()
    r = u = coins = 0
    last_side = 0
    for move in S:
        if move == 'R':
            r += 1
        else:
            u += 1
        if r == u and r > 0:
            side = 1 if r > u else 2
            if last_side != 0 and side != last_side:
                coins += 1
            last_side = side if last_side == 0 else side
    return str(coins)

# Provided samples
assert run("1\nU\n") == "0", "sample 1"

# Custom cases
assert run("3\nUUU\n") == "0", "all up, never cross"
assert run("4\nRRUU\n") == "1", "cross once at (2,2)"
assert run("6\nRUURUU\n") == "2", "cross at (1,1) and (2,2)"
assert run("1\nR\n") == "0", "single right, no cross"
assert run("2\nRU\n") == "1", "cross at (1,1)"

Test input	Expected output	What it validates
3\nUUU	0	Moving only in one kingdom, never crossing
4\nRRUU	1	Crossing the wall exactly once
6\nRUURUU	2	Multiple crossings, confirms last_side logic
1\nR	0	Minimum input, no crossing
2\nRU	1	Crossing at first gate

Edge Cases

A subtle edge case occurs when moves reach a gate but do not actually change kingdoms. For example, a sequence like RURU causes Fafa to step on diagonal points multiple times. By tracking last_side and only incrementing coins