How do you solve Minimum Moves to Reach Target in Grid on LeetCode?

Minimum Moves to Reach Target in Grid (LeetCode #3609) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(max(tx, ty))) time complexity and O(1) space complexity. Key insight: Working backwards simplifies the problem.

What is the time complexity of Minimum Moves to Reach Target in Grid?

The optimal solution for Minimum Moves to Reach Target in Grid runs in O(log(max(tx, ty))) time and uses O(1) space. The complexity is logarithmic because we effectively halve the larger coordinate each step, reducing the problem size quickly.

What is the brute force solution for Minimum Moves to Reach Target in Grid?

The brute force approach for Minimum Moves to Reach Target in Grid is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach explores all possible paths from the start to the target, checking every possible move. It's inefficient as it doesn't leverage the structure of the problem. The optimal Optimal Solution approach improves this to O(log(max(tx, ty))).

What algorithm or data structure is used to solve Minimum Moves to Reach Target in Grid?

Minimum Moves to Reach Target in Grid uses the following concepts: Math. The recommended patterns to study are: Greedy, Backtracking.

What are the interview tips for Minimum Moves to Reach Target in Grid?

Explain your thought process clearly. Use diagrams to illustrate your approach. Practice similar problems to build intuition.

What are common mistakes when solving Minimum Moves to Reach Target in Grid?

Forgetting to check bounds before moving. Not considering the case where coordinates match.

#3609

Minimum Moves to Reach Target in Grid

Hard

MathGreedyBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(max(tx, ty)))
Space	O(1)	O(1)

💡

Intuition

Time O(log(max(tx, ty)))Space O(1)

Instead of moving forward, we work backwards from the target to the start. This allows us to efficiently determine the minimum moves by undoing the operations.

⚙️

Algorithm

3 steps

1Step 1: While (tx, ty) is greater than (sx, sy), check if tx > ty or ty > tx.
2Step 2: If the larger coordinate is at least double the smaller, halve the larger coordinate.
3Step 3: Otherwise, subtract the smaller coordinate from the larger until reaching (sx, sy) or impossible.

solution.py17 lines

1def minMoves(sx, sy, tx, ty):
2    moves = 0
3    while tx >= sx and ty >= sy:
4        if tx == sx and ty == sy:
5            return moves
6        if tx > ty:
7            if ty > sy:
8                tx -= ty // 2
9            else:
10                tx -= ty
11        else:
12            if tx > sx:
13                ty -= tx // 2
14            else:
15                ty -= tx
16        moves += 1
17    return -1

ℹ

Complexity note: The complexity is logarithmic because we effectively halve the larger coordinate each step, reducing the problem size quickly.

1Working backwards simplifies the problem.
2Halving the larger coordinate reduces the number of moves.

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