Execution of All Suffix Instructions Staying in a Grid - Problem

String Simulation Medium

There is an n × n grid, with the top-left cell at (0, 0) and the bottom-right cell at (n - 1, n - 1). You are given the integer n and an integer array startPos where startPos = [startrow, startcol] indicates that a robot is initially at cell (startrow, startcol).

You are also given a 0-indexed string s of length m where s[i] is the ith instruction for the robot: 'L' (move left), 'R' (move right), 'U' (move up), and 'D' (move down).

The robot can begin executing from any ith instruction in s. It executes the instructions one by one towards the end of s but it stops if either of these conditions is met:

The next instruction will move the robot off the grid.
There are no more instructions left to execute.

Return an array answer of length m where answer[i] is the number of instructions the robot can execute if the robot begins executing from the ith instruction in s.

Input & Output

Example 1 — Basic Grid Navigation

$ Input: n = 3, startPos = [0,1], s = "RRDDLU"

› Output: [1,5,4,3,1,0]

💡 Note: Starting from index 0: R→[0,2], next R would go out of bounds = 1 step. Index 1: R→[0,2], D→[1,2], D→[2,2], L→[2,1], U→[1,1] = 5 steps. Index 2: D→[1,1], D→[2,1], L→[2,0], U→[1,0] = 4 steps. Index 3: D→[1,1], L→[1,0], U→[0,0] = 3 steps. Index 4: L→[0,0], next U would go out of bounds = 1 step. Index 5: U would go out of bounds immediately = 0 steps.

Example 2 — Smaller Grid

$ Input: n = 2, startPos = [1,1], s = "LURD"

› Output: [4,1,1,1]

💡 Note: From [1,1] in 2x2 grid: Index 0 LURD: L→[1,0], U→[0,0], R→[0,1], D→[1,1] all valid = 4 steps. Index 1 URD: U→[0,1], R→would go to [0,2] out of bounds = 1 step.

Example 3 — Edge Position

$ Input: n = 1, startPos = [0,0], s = "LRUD"

› Output: [0,0,0,0]

💡 Note: In 1x1 grid at [0,0], any movement L,R,U,D goes out of bounds immediately, so 0 steps for all starting positions.

Constraints

1 ≤ n ≤ 1000
0 ≤ startPos[0], startPos[1] < n
1 ≤ s.length ≤ 1000
s consists of 'L', 'R', 'U', and 'D'

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to simulate robot movement for each possible starting instruction position. For each suffix of instructions, we track the robot's position and count valid moves until it hits a boundary. The optimal approach uses direction mapping for cleaner code. Time: O(m²), Space: O(1).

Common Approaches

✓ Optimized Simulation with Early Termination

⏱️ Time: O(m²) Space: O(1)

Uses the same simulation strategy but with more efficient direction handling using coordinate deltas. The time complexity remains the same, but the code is cleaner and slightly more efficient in practice.

Simulation for Each Suffix

⏱️ Time: O(m²) Space: O(1)

For each starting position i in the instruction string, simulate the robot's movement from that point until it either goes out of bounds or reaches the end of instructions. Count how many instructions were executed successfully.

Optimized Simulation with Early Termination — Algorithm Steps

Step 1: Create direction mapping for L,R,U,D movements
Step 2: For each starting index, simulate with delta coordinates
Step 3: Use early termination when bounds are exceeded

Visualization

Tap to expand

Step-by-Step Walkthrough

Direction Map

L:(-1,0), R:(1,0), U:(0,-1), D:(0,1)

Delta Movement

Add direction deltas to current position

Bounds Check

Validate new position within grid

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int executeInstructions(int n, int startRow, int startCol, char* s, int start) {
    int row = startRow;
    int col = startCol;
    int count = 0;
    int len = strlen(s);
    
    for (int i = start; i < len; i++) {
        int newRow = row;
        int newCol = col;
        
        switch (s[i]) {
            case 'L':
                newCol = col - 1;
                break;
            case 'R':
                newCol = col + 1;
                break;
            case 'U':
                newRow = row - 1;
                break;
            case 'D':
                newRow = row + 1;
                break;
        }
        
        // Check if new position is out of bounds
        if (newRow < 0 || newRow >= n || newCol < 0 || newCol >= n) {
            break;
        }
        
        // Update position and increment count
        row = newRow;
        col = newCol;
        count++;
    }
    
    return count;
}

int main() {
    int n;
    int startRow, startCol;
    char s[1001];
    
    // Read n
    scanf("%d", &n);
    
    // Read startPos [startRow, startCol]
    char startPosLine[100];
    scanf("%s", startPosLine);
    
    // Parse [startRow, startCol]
    char* ptr = startPosLine;
    while (*ptr && *ptr != '[') ptr++;
    if (*ptr == '[') ptr++;
    
    startRow = 0;
    while (*ptr && *ptr >= '0' && *ptr <= '9') {
        startRow = startRow * 10 + (*ptr - '0');
        ptr++;
    }
    
    while (*ptr && (*ptr == ',' || *ptr == ' ')) ptr++;
    
    startCol = 0;
    while (*ptr && *ptr >= '0' && *ptr <= '9') {
        startCol = startCol * 10 + (*ptr - '0');
        ptr++;
    }
    
    // Read instructions string
    scanf("%s", s);
    
    int len = strlen(s);
    
    // Calculate result for each suffix
    printf("[");
    for (int i = 0; i < len; i++) {
        if (i > 0) printf(",");
        int count = executeInstructions(n, startRow, startCol, s, i);
        printf("%d", count);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m²)

Same as brute force - for each of m positions, potentially execute m instructions

✓ Linear Growth

Space Complexity

O(1)

Only using direction deltas and position variables, no extra data structures

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

856 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Execution of All Suffix Instructions Staying in a Grid - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized Simulation with Early Termination — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler