Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using core interview patterns strategy.
There is an n x 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:
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.
Example 1:
Input: n = 3, startPos = [0,1], s = "RRDDLU" Output: [1,5,4,3,1,0] Explanation: Starting from startPos and beginning execution from the ith instruction: - 0th: "RRDDLU". Only one instruction "R" can be executed before it moves off the grid. - 1st: "RDDLU". All five instructions can be executed while it stays in the grid and ends at (1, 1). - 2nd: "DDLU". All four instructions can be executed while it stays in the grid and ends at (1, 0). - 3rd: "DLU". All three instructions can be executed while it stays in the grid and ends at (0, 0). - 4th: "LU". Only one instruction "L" can be executed before it moves off the grid. - 5th: "U". If moving up, it would move off the grid.
Example 2:
Input: n = 2, startPos = [1,1], s = "LURD" Output: [4,1,0,0] Explanation: - 0th: "LURD". - 1st: "URD". - 2nd: "RD". - 3rd: "D".
Example 3:
Input: n = 1, startPos = [0,0], s = "LRUD" Output: [0,0,0,0] Explanation: No matter which instruction the robot begins execution from, it would move off the grid.
Constraints:
m == s.length1 <= n, m <= 500startPos.length == 20 <= startrow, startcol < ns consists of 'L', 'R', 'U', and 'D'.Problem summary: There is an n x 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
3 [0,1] "RRDDLU"
2 [1,1] "LURD"
1 [0,0] "LRUD"
out-of-boundary-paths)robot-return-to-origin)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2120: Execution of All Suffix Instructions Staying in a Grid
class Solution {
public int[] executeInstructions(int n, int[] startPos, String s) {
int m = s.length();
int[] ans = new int[m];
Map<Character, int[]> mp = new HashMap<>(4);
mp.put('L', new int[] {0, -1});
mp.put('R', new int[] {0, 1});
mp.put('U', new int[] {-1, 0});
mp.put('D', new int[] {1, 0});
for (int i = 0; i < m; ++i) {
int x = startPos[0], y = startPos[1];
int t = 0;
for (int j = i; j < m; ++j) {
char c = s.charAt(j);
int a = mp.get(c)[0], b = mp.get(c)[1];
if (0 <= x + a && x + a < n && 0 <= y + b && y + b < n) {
x += a;
y += b;
++t;
} else {
break;
}
}
ans[i] = t;
}
return ans;
}
}
// Accepted solution for LeetCode #2120: Execution of All Suffix Instructions Staying in a Grid
func executeInstructions(n int, startPos []int, s string) []int {
m := len(s)
mp := make(map[byte][]int)
mp['L'] = []int{0, -1}
mp['R'] = []int{0, 1}
mp['U'] = []int{-1, 0}
mp['D'] = []int{1, 0}
ans := make([]int, m)
for i := 0; i < m; i++ {
x, y := startPos[0], startPos[1]
t := 0
for j := i; j < m; j++ {
a, b := mp[s[j]][0], mp[s[j]][1]
if 0 <= x+a && x+a < n && 0 <= y+b && y+b < n {
x += a
y += b
t++
} else {
break
}
}
ans[i] = t
}
return ans
}
# Accepted solution for LeetCode #2120: Execution of All Suffix Instructions Staying in a Grid
class Solution:
def executeInstructions(self, n: int, startPos: List[int], s: str) -> List[int]:
ans = []
m = len(s)
mp = {"L": [0, -1], "R": [0, 1], "U": [-1, 0], "D": [1, 0]}
for i in range(m):
x, y = startPos
t = 0
for j in range(i, m):
a, b = mp[s[j]]
if 0 <= x + a < n and 0 <= y + b < n:
x, y, t = x + a, y + b, t + 1
else:
break
ans.append(t)
return ans
// Accepted solution for LeetCode #2120: Execution of All Suffix Instructions Staying in a Grid
impl Solution {
pub fn execute_instructions(n: i32, start_pos: Vec<i32>, s: String) -> Vec<i32> {
let s = s.as_bytes();
let m = s.len();
let mut ans = vec![0; m];
for i in 0..m {
let mut y = start_pos[0];
let mut x = start_pos[1];
let mut j = i;
while j < m {
match s[j] {
b'U' => {
y -= 1;
}
b'D' => {
y += 1;
}
b'L' => {
x -= 1;
}
_ => {
x += 1;
}
}
if y == -1 || y == n || x == -1 || x == n {
break;
}
j += 1;
}
ans[i] = (j - i) as i32;
}
ans
}
}
// Accepted solution for LeetCode #2120: Execution of All Suffix Instructions Staying in a Grid
function executeInstructions(n: number, startPos: number[], s: string): number[] {
const m = s.length;
const ans = new Array(m);
for (let i = 0; i < m; i++) {
let [y, x] = startPos;
let j: number;
for (j = i; j < m; j++) {
const c = s[j];
if (c === 'U') {
y--;
} else if (c === 'D') {
y++;
} else if (c === 'L') {
x--;
} else {
x++;
}
if (y === -1 || y === n || x === -1 || x === n) {
break;
}
}
ans[i] = j - i;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.