LeetCode #789 — MEDIUM

Escape The Ghosts

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are playing a simplified PAC-MAN game on an infinite 2-D grid. You start at the point [0, 0], and you are given a destination point target = [x_target, y_target] that you are trying to get to. There are several ghosts on the map with their starting positions given as a 2D array ghosts, where ghosts[i] = [x_i, y_i] represents the starting position of the i^th ghost. All inputs are integral coordinates.

Each turn, you and all the ghosts may independently choose to either move 1 unit in any of the four cardinal directions: north, east, south, or west, or stay still. All actions happen simultaneously.

You escape if and only if you can reach the target before any ghost reaches you. If you reach any square (including the target) at the same time as a ghost, it does not count as an escape.

Return true if it is possible to escape regardless of how the ghosts move, otherwise return false.

Example 1:

Input: ghosts = [[1,0],[0,3]], target = [0,1]
Output: true
Explanation: You can reach the destination (0, 1) after 1 turn, while the ghosts located at (1, 0) and (0, 3) cannot catch up with you.

Example 2:

Input: ghosts = [[1,0]], target = [2,0]
Output: false
Explanation: You need to reach the destination (2, 0), but the ghost at (1, 0) lies between you and the destination.

Example 3:

Input: ghosts = [[2,0]], target = [1,0]
Output: false
Explanation: The ghost can reach the target at the same time as you.

Constraints:

1 <= ghosts.length <= 100
ghosts[i].length == 2
-10⁴ <= x_i, y_i <= 10⁴
There can be multiple ghosts in the same location.
target.length == 2
-10⁴ <= x_target, y_target <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are playing a simplified PAC-MAN game on an infinite 2-D grid. You start at the point [0, 0], and you are given a destination point target = [xtarget, ytarget] that you are trying to get to. There are several ghosts on the map with their starting positions given as a 2D array ghosts, where ghosts[i] = [xi, yi] represents the starting position of the ith ghost. All inputs are integral coordinates. Each turn, you and all the ghosts may independently choose to either move 1 unit in any of the four cardinal directions: north, east, south, or west, or stay still. All actions happen simultaneously. You escape if and only if you can reach the target before any ghost reaches you. If you reach any square (including the target) at the same time as a ghost, it does not count as an escape. Return true if it is possible to escape regardless of how the ghosts move, otherwise return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[[1,0],[0,3]]
[0,1]

Example 2

[[1,0]]
[2,0]

Example 3

[[2,0]]
[1,0]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #789: Escape The Ghosts
class Solution {
    public boolean escapeGhosts(int[][] ghosts, int[] target) {
        int tx = target[0], ty = target[1];
        for (var g : ghosts) {
            int x = g[0], y = g[1];
            if (Math.abs(tx - x) + Math.abs(ty - y) <= Math.abs(tx) + Math.abs(ty)) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #789: Escape The Ghosts
func escapeGhosts(ghosts [][]int, target []int) bool {
	tx, ty := target[0], target[1]
	for _, g := range ghosts {
		x, y := g[0], g[1]
		if abs(tx-x)+abs(ty-y) <= abs(tx)+abs(ty) {
			return false
		}
	}
	return true
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #789: Escape The Ghosts
class Solution:
    def escapeGhosts(self, ghosts: List[List[int]], target: List[int]) -> bool:
        tx, ty = target
        return all(abs(tx - x) + abs(ty - y) > abs(tx) + abs(ty) for x, y in ghosts)

// Accepted solution for LeetCode #789: Escape The Ghosts
use std::cell::RefCell;
use std::rc::Rc;

type OptNode = Option<Rc<RefCell<TreeNode>>>;

impl Solution {
    pub fn min_diff_in_bst(root: OptNode) -> i32 {
        let mut min = i32::MAX;
        let mut prev = None;

        Self::dfs(root, &mut prev, &mut min);
        min
    }

    fn dfs(root: Option<Rc<RefCell<TreeNode>>>, prev: &mut Option<i32>, min: &mut i32) {
        if let Some(node) = root {
            let node = node.borrow();

            Self::dfs(node.left.clone(), prev, min);

            if let Some(prev) = prev {
                *min = (*min).min(node.val - *prev)
            }
            *prev = Some(node.val);

            Self::dfs(node.right.clone(), prev, min);
        }
    }
}

// Accepted solution for LeetCode #789: Escape The Ghosts
function escapeGhosts(ghosts: number[][], target: number[]): boolean {
    const [tx, ty] = target;
    for (const [x, y] of ghosts) {
        if (Math.abs(tx - x) + Math.abs(ty - y) <= Math.abs(tx) + Math.abs(ty)) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.