LeetCode #3609 — HARD

Minimum Moves to Reach Target in Grid

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given four integers sx, sy, tx, and ty, representing two points (sx, sy) and (tx, ty) on an infinitely large 2D grid.

You start at (sx, sy).

At any point (x, y), define m = max(x, y). You can either:

Move to (x + m, y), or
Move to (x, y + m).

Return the minimum number of moves required to reach (tx, ty). If it is impossible to reach the target, return -1.

Example 1:

Input: sx = 1, sy = 2, tx = 5, ty = 4

Output: 2

Explanation:

The optimal path is:

Move 1: max(1, 2) = 2. Increase the y-coordinate by 2, moving from (1, 2) to (1, 2 + 2) = (1, 4).
Move 2: max(1, 4) = 4. Increase the x-coordinate by 4, moving from (1, 4) to (1 + 4, 4) = (5, 4).

Thus, the minimum number of moves to reach (5, 4) is 2.

Example 2:

Input: sx = 0, sy = 1, tx = 2, ty = 3

Output: 3

Explanation:

The optimal path is:

Move 1: max(0, 1) = 1. Increase the x-coordinate by 1, moving from (0, 1) to (0 + 1, 1) = (1, 1).
Move 2: max(1, 1) = 1. Increase the x-coordinate by 1, moving from (1, 1) to (1 + 1, 1) = (2, 1).
Move 3: max(2, 1) = 2. Increase the y-coordinate by 2, moving from (2, 1) to (2, 1 + 2) = (2, 3).

Thus, the minimum number of moves to reach (2, 3) is 3.

Example 3:

Input: sx = 1, sy = 1, tx = 2, ty = 2

Output: -1

Explanation:

It is impossible to reach (2, 2) from (1, 1) using the allowed moves. Thus, the answer is -1.

Constraints:

0 <= sx <= tx <= 10⁹
0 <= sy <= ty <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given four integers sx, sy, tx, and ty, representing two points (sx, sy) and (tx, ty) on an infinitely large 2D grid. You start at (sx, sy). At any point (x, y), define m = max(x, y). You can either: Move to (x + m, y), or Move to (x, y + m). Return the minimum number of moves required to reach (tx, ty). If it is impossible to reach the target, return -1.

Baseline thinking

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

Pattern signal: Math

Example 1

Example 2

Example 3

Step 02

Core Insight

What unlocks the optimal approach

Work backwards from <code>(tx, ty)</code> to <code>(sx, sy)</code>, undoing one move at each step.
If the larger coordinate >= 2 × (the smaller), undo by halving the larger; otherwise undo by subtracting the smaller from the larger.
Count these undo-steps until you hit <code>(sx, sy)</code> (return the count), or return -1 if you drop below or get stuck.

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

Largest constraint values

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 #3609: Minimum Moves to Reach Target in Grid
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minMoves(sx, sy, x, y int) (ans int) {
// 	for ; x != sx || y != sy; ans++ {
// 		if x < sx || y < sy {
// 			return -1
// 		}
// 		if x == y {
// 			if sy > 0 {
// 				x = 0
// 			} else {
// 				y = 0
// 			}
// 			continue
// 		}
// 		// 保证 x > y
// 		if x < y {
// 			x, y = y, x
// 			sx, sy = sy, sx
// 		}
// 		if x >= y*2 {
// 			if x%2 > 0 {
// 				return -1
// 			}
// 			x /= 2
// 		} else {
// 			x -= y
// 		}
// 	}
// 	return
// }

// Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
package main

// https://space.bilibili.com/206214
func minMoves(sx, sy, x, y int) (ans int) {
	for ; x != sx || y != sy; ans++ {
		if x < sx || y < sy {
			return -1
		}
		if x == y {
			if sy > 0 {
				x = 0
			} else {
				y = 0
			}
			continue
		}
		// 保证 x > y
		if x < y {
			x, y = y, x
			sx, sy = sy, sx
		}
		if x >= y*2 {
			if x%2 > 0 {
				return -1
			}
			x /= 2
		} else {
			x -= y
		}
	}
	return
}

# Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
import math

class Solution:
    def minMoves(self, sx: int, sy: int, tx: int, ty: int) -> int:
        moves = 0
        
        # Helper to check power of two
        def is_power_of_two(n):
            return n > 0 and (n & (n - 1)) == 0
        
        while tx >= sx and ty >= sy:
            if tx == sx and ty == sy:
                return moves
            
            if tx > ty:
                # If we are in a state where y matches sy, and sx is larger than ty,
                # we are purely in the doubling regime for x. We can fast-forward.
                if ty == sy and sx > ty:
                    if tx % sx == 0 and is_power_of_two(tx // sx):
                        return moves + (tx // sx).bit_length() - 1
                    else:
                        return -1

                if tx >= 2 * ty:
                    # Must have come from doubling x
                    if tx % 2 != 0:
                        return -1
                    tx //= 2
                    moves += 1
                else:
                    # Must have come from x + y
                    tx -= ty
                    moves += 1
            elif ty > tx:
                # Symmetric logic for y
                if tx == sx and sy > tx:
                    if ty % sy == 0 and is_power_of_two(ty // sy):
                        return moves + (ty // sy).bit_length() - 1
                    else:
                        return -1

                if ty >= 2 * tx:
                    if ty % 2 != 0:
                        return -1
                    ty //= 2
                    moves += 1
                else:
                    ty -= tx
                    moves += 1
            else:
                # tx == ty
                # Can only reach equal coordinates from a state where one coordinate is 0.
                # e.g. (0, y) -> (y, y) or (x, 0) -> (x, x)
                res = float('inf')
                
                # Check possibility coming from x=0
                if sx == 0:
                    # We need to check distance from (0, sy) to (0, ty)
                    # This path is purely doubling y
                    if ty == sy:
                         res = min(res, moves + 1)
                    elif sy > 0 and ty % sy == 0 and is_power_of_two(ty // sy):
                         res = min(res, moves + 1 + (ty // sy).bit_length() - 1)
                    # If sy == 0, we can only reach (0, ty) if ty == 0, but we are at tx=ty>0 here usually
                    # (unless tx=ty=0 which is caught at start)
                    
                # Check possibility coming from y=0
                if sy == 0:
                    # Check distance from (sx, 0) to (tx, 0)
                    if tx == sx:
                        res = min(res, moves + 1)
                    elif sx > 0 and tx % sx == 0 and is_power_of_two(tx // sx):
                        res = min(res, moves + 1 + (tx // sx).bit_length() - 1)
                
                if res != float('inf'):
                    return res
                else:
                    return -1
                    
        return -1

// Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minMoves(sx, sy, x, y int) (ans int) {
// 	for ; x != sx || y != sy; ans++ {
// 		if x < sx || y < sy {
// 			return -1
// 		}
// 		if x == y {
// 			if sy > 0 {
// 				x = 0
// 			} else {
// 				y = 0
// 			}
// 			continue
// 		}
// 		// 保证 x > y
// 		if x < y {
// 			x, y = y, x
// 			sx, sy = sy, sx
// 		}
// 		if x >= y*2 {
// 			if x%2 > 0 {
// 				return -1
// 			}
// 			x /= 2
// 		} else {
// 			x -= y
// 		}
// 	}
// 	return
// }

// Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3609: Minimum Moves to Reach Target in Grid
// package main
// 
// // https://space.bilibili.com/206214
// func minMoves(sx, sy, x, y int) (ans int) {
// 	for ; x != sx || y != sy; ans++ {
// 		if x < sx || y < sy {
// 			return -1
// 		}
// 		if x == y {
// 			if sy > 0 {
// 				x = 0
// 			} else {
// 				y = 0
// 			}
// 			continue
// 		}
// 		// 保证 x > y
// 		if x < y {
// 			x, y = y, x
// 			sx, sy = sy, sx
// 		}
// 		if x >= y*2 {
// 			if x%2 > 0 {
// 				return -1
// 			}
// 			x /= 2
// 		} else {
// 			x -= y
// 		}
// 	}
// 	return
// }

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.