Topics
To move the robot from (0,0) to (x,y), Vasya wants to change the sequence of operations so that the length of the changed subsegment is minimized. The length of the changed subsegment is calculated as maxID−minID+1, where maxID and minID are the maximum and minimum indices of the changed operations. The goal is to find the minimum possible length of the changed subsegment, or determine that it’s impossible.
Idea
The problem can be approached using binary search on the answer (minimum window size) combined with a sliding window check. The key observation is that the answer has a monotonic property: if a window of size k works, any larger window will also work, allowing binary search.
For a given window size k, we use the sliding window framework to check if there exists any subarray of length k that can be modified to reach the target. The modification check involves calculating the required changes in x and y movements within the window. The total changes must satisfy two conditions:
- their sum must be ≤
k(window size) - the parity must match (since each change affects the sum by ±2)
For each window [l ,..., r], we can use prefix sums to find position after going through the sequence [0 ,..., l-1] (i.e. before window). Also, we need to perform the remaining of the sequence [r+1 ,..., n] (found using prefix sums as well). The required number of changes needed to reach target can then be found simply by doing: target - interim where interim is the position reached after we did all the moves before and after the window (skipping the window itself).
Time Complexity:
Space Complexity:
An alternative implementation is to compute interim on the fly during the sliding window’s expansion and contraction processes. The sliding window with fixed window size (snake framework) can be used as an implementation template to follow. This is more space-efficient , as it avoids storing prefix arrays.
Code
// Approach 1 (Binary Search + 2 pointers + prefix sums)
bool canReachDestination(int window_size, const vi &prefix_x,
const vi &prefix_y, int target_x, int target_y) {
int n = prefix_x.size() - 1; // Adjust for 1-indexed prefix arrays
// Slide a window of fixed size 'window_size' through the path
for (int left = 1; left <= n - window_size + 1; ++left) {
int right = left + window_size - 1;
// Position before window
int curr_x = prefix_x[left - 1];
int curr_y = prefix_y[left - 1];
// Remaining moves after window
int remaining_x_moves = prefix_x[n] - prefix_x[right];
int remaining_y_moves = prefix_y[n] - prefix_y[right];
// position reached by doing all moves except the window
int interim_x = (curr_x + remaining_x_moves);
int interim_y = (curr_y + remaining_y_moves);
// Required changes within window to reach target
int needed_x = target_x - interim_x;
int needed_y = target_y - interim_y;
// Check if window can be modified to reach target
int total_changes_needed = abs(needed_x) + abs(needed_y);
if (total_changes_needed <= window_size &&
total_changes_needed % 2 == window_size % 2) {
return true;
}
}
return false;
}
void solve() {
int n;
cin >> n;
string path;
cin >> path;
int target_x, target_y;
cin >> target_x >> target_y;
// Early exit: if Manhattan distance > n, impossible
if (abs(target_x) + abs(target_y) > n ||
(abs(target_x) + abs(target_y)) % 2 != n % 2) {
cout << -1 << endl;
return;
}
// Calculate prefix sums for x and y coordinates
vi prefix_x(n + 1, 0), prefix_y(n + 1, 0);
for (int i = 0; i < n; ++i) {
prefix_x[i + 1] = prefix_x[i];
prefix_y[i + 1] = prefix_y[i];
if (path[i] == 'R')
prefix_x[i + 1]++;
else if (path[i] == 'L')
prefix_x[i + 1]--;
else if (path[i] == 'U')
prefix_y[i + 1]++;
else if (path[i] == 'D')
prefix_y[i + 1]--;
}
// Early check
if (prefix_x[n] == target_x && prefix_y[n] == target_y) {
cout << 0 << endl;
return;
}
// Binary search for minimum window size
int lo = 1, hi = n;
int min_window_size = -1;
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (canReachDestination(mid, prefix_x, prefix_y, target_x, target_y)) {
min_window_size = mid;
hi = mid - 1; // Try to find smaller window
} else {
lo = mid + 1; // Try larger window
}
}
cout << min_window_size << endl;
}Alternative implementation that doesn’t use extra space and aligns with the sliding window with fixed window size (snake framework):
bool canReachDestination(int window_size, const string &path, int target_x,
int target_y) {
int n = path.size();
// Snake framework for fixed window size
int tail = 0, head = -1;
// Track positions before and after the window
int curr_x = 0, curr_y = 0; // Position before window
int remaining_x = 0, remaining_y = 0; // Remaining moves after window
// Calculate initial remaining moves (all moves)
// Note that this can be pre-computed once and passed to this func
for (char c : path) {
if (c == 'R')
remaining_x++;
else if (c == 'L')
remaining_x--;
else if (c == 'U')
remaining_y++;
else if (c == 'D')
remaining_y--;
}
while (tail < n) {
// Expansion phase: Expand until window_size or end of array
while (head + 1 < n && head - tail + 1 < window_size) {
head++;
// Update remaining moves (subtract moves in window)
if (path[head] == 'R')
remaining_x--;
else if (path[head] == 'L')
remaining_x++;
else if (path[head] == 'U')
remaining_y--;
else if (path[head] == 'D')
remaining_y++;
}
// Process window
if (head - tail + 1 == window_size) {
int interim_x = curr_x + remaining_x;
int interim_y = curr_y + remaining_y;
// Required changes within window to reach target
int needed_x = target_x - interim_x;
int needed_y = target_y - interim_y;
// Check if window can be modified to reach target
int total_changes_needed = abs(needed_x) + abs(needed_y);
if (total_changes_needed <= window_size &&
total_changes_needed % 2 == window_size % 2) {
return true;
}
}
// Contraction phase: Remove element at tail
if (tail <= head) {
// Update current position
if (path[tail] == 'R') {
curr_x++;
} else if (path[tail] == 'L') {
curr_x--;
} else if (path[tail] == 'U') {
curr_y++;
} else if (path[tail] == 'D') {
curr_y--;
}
}
// Advance tail
tail++;
head = max(head, tail - 1);
}
return false;
}