What is the time complexity of Execution of All Suffix Instructions Staying in a Grid?

The optimal solution for Execution of All Suffix Instructions Staying in a Grid runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only make a single pass through the instruction string, and the space complexity is O(n) due to the answer array.

What is the brute force solution for Execution of All Suffix Instructions Staying in a Grid?

The brute force approach for Execution of All Suffix Instructions Staying in a Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the robot's movement for each possible starting instruction in the string. For each instruction, we will execute the commands until we either run out of commands or move off the grid. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Execution of All Suffix Instructions Staying in a Grid?

Execution of All Suffix Instructions Staying in a Grid uses the following concepts: String, Simulation. The recommended patterns to study are: Simulation, Dynamic Programming.

What are the interview tips for Execution of All Suffix Instructions Staying in a Grid?

Always clarify the boundaries of the grid and initial position. Think about how you can optimize a brute-force solution. Practice simulating movements in a grid to build intuition.

What are common mistakes when solving Execution of All Suffix Instructions Staying in a Grid?

Not considering edge cases where the robot might immediately go off the grid. Failing to update the position correctly based on the instruction.

#2120

Execution of All Suffix Instructions Staying in a Grid

Medium

StringSimulationSimulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a single pass from the end of the instruction string to the beginning, calculating how many instructions can be executed from each index. This way, we can build the answer array in linear time by leveraging previously computed results.

⚙️

Algorithm

3 steps

1Step 1: Initialize an answer array of the same length as the instruction string, filled with zeros.
2Step 2: Start from the last instruction and move backwards to the first instruction.
3Step 3: For each instruction, check if the robot can move in the specified direction and update the position and count accordingly.

solution.py23 lines

1# Full working Python code
2
3def executeInstructions(n, startPos, s):
4    m = len(s)
5    answer = [0] * m
6    x, y = startPos
7    for i in range(m - 1, -1, -1):
8        if i < m - 1:
9            answer[i] = answer[i + 1]
10        if s[i] == 'L':
11            y -= 1
12        elif s[i] == 'R':
13            y += 1
14        elif s[i] == 'U':
15            x -= 1
16        elif s[i] == 'D':
17            x += 1
18        if 0 <= x < n and 0 <= y < n:
19            answer[i] += 1
20        else:
21            break
22    return answer
23

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the instruction string, and the space complexity is O(n) due to the answer array.

1The robot's movement can be efficiently simulated by processing the instructions in reverse.
2Using previously computed results can help avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.