LeetCode #1824 — MEDIUM

Minimum Sideway Jumps

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

The Problem

Problem Statement

There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts at point 0 in the second lane and wants to jump to point n. However, there could be obstacles along the way.

You are given an array obstacles of length n + 1 where each obstacles[i] (ranging from 0 to 3) describes an obstacle on the lane obstacles[i] at point i. If obstacles[i] == 0, there are no obstacles at point i. There will be at most one obstacle in the 3 lanes at each point.

For example, if obstacles[2] == 1, then there is an obstacle on lane 1 at point 2.

The frog can only travel from point i to point i + 1 on the same lane if there is not an obstacle on the lane at point i + 1. To avoid obstacles, the frog can also perform a side jump to jump to another lane (even if they are not adjacent) at the same point if there is no obstacle on the new lane.

For example, the frog can jump from lane 3 at point 3 to lane 1 at point 3.

Return the minimum number of side jumps the frog needs to reach any lane at point n starting from lane 2 at point 0.

Note: There will be no obstacles on points 0 and n.

Example 1:

Input: obstacles = [0,1,2,3,0]
Output: 2 
Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).
Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).

Example 2:

Input: obstacles = [0,1,1,3,3,0]
Output: 0
Explanation: There are no obstacles on lane 2. No side jumps are required.

Example 3:

Input: obstacles = [0,2,1,0,3,0]
Output: 2
Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps.

Constraints:

obstacles.length == n + 1
1 <= n <= 5 * 10⁵
0 <= obstacles[i] <= 3
obstacles[0] == obstacles[n] == 0

Step 01

Brute Force Baseline

Problem summary: There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts at point 0 in the second lane and wants to jump to point n. However, there could be obstacles along the way. You are given an array obstacles of length n + 1 where each obstacles[i] (ranging from 0 to 3) describes an obstacle on the lane obstacles[i] at point i. If obstacles[i] == 0, there are no obstacles at point i. There will be at most one obstacle in the 3 lanes at each point. For example, if obstacles[2] == 1, then there is an obstacle on lane 1 at point 2. The frog can only travel from point i to point i + 1 on the same lane if there is not an obstacle on the lane at point i + 1. To avoid obstacles, the frog can also perform a side jump to jump to another lane (even if they are not adjacent) at the same point if there is no obstacle on the new lane. For example, the frog can jump

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

[0,1,2,3,0]

Example 2

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

Example 3

[0,2,1,0,3,0]

Core Insight

What unlocks the optimal approach

At a given point, there are only 3 possible states for where the frog can be.
Check all the ways to move from one point to the next and update the minimum side jumps for each lane.

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 #1824: Minimum Sideway Jumps
class Solution {
    public int minSideJumps(int[] obstacles) {
        final int inf = 1 << 30;
        int[] f = {1, 0, 1};
        for (int i = 1; i < obstacles.length; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (obstacles[i] == j + 1) {
                    f[j] = inf;
                    break;
                }
            }
            int x = Math.min(f[0], Math.min(f[1], f[2])) + 1;
            for (int j = 0; j < 3; ++j) {
                if (obstacles[i] != j + 1) {
                    f[j] = Math.min(f[j], x);
                }
            }
        }
        return Math.min(f[0], Math.min(f[1], f[2]));
    }
}

// Accepted solution for LeetCode #1824: Minimum Sideway Jumps
func minSideJumps(obstacles []int) int {
	f := [3]int{1, 0, 1}
	const inf = 1 << 30
	for _, v := range obstacles[1:] {
		for j := 0; j < 3; j++ {
			if v == j+1 {
				f[j] = inf
				break
			}
		}
		x := min(f[0], min(f[1], f[2])) + 1
		for j := 0; j < 3; j++ {
			if v != j+1 {
				f[j] = min(f[j], x)
			}
		}
	}
	return min(f[0], min(f[1], f[2]))
}

# Accepted solution for LeetCode #1824: Minimum Sideway Jumps
class Solution:
    def minSideJumps(self, obstacles: List[int]) -> int:
        f = [1, 0, 1]
        for v in obstacles[1:]:
            for j in range(3):
                if v == j + 1:
                    f[j] = inf
                    break
            x = min(f) + 1
            for j in range(3):
                if v != j + 1:
                    f[j] = min(f[j], x)
        return min(f)

// Accepted solution for LeetCode #1824: Minimum Sideway Jumps
struct Solution;

impl Solution {
    fn min_side_jumps(obstacles: Vec<i32>) -> i32 {
        let n = obstacles.len();
        let mut dp: Vec<Vec<i32>> = vec![vec![0, 0, 0]; n];
        dp[0][0] = 1;
        dp[0][2] = 1;
        for i in 1..n {
            for j in 0..3 {
                if obstacles[i] == (j + 1) as i32 {
                    dp[i][j] = std::i32::MAX;
                } else {
                    let mut min = std::i32::MAX;
                    for k in 0..3 {
                        if !(obstacles[i - 1] == (k + 1) as i32 || obstacles[i] == (k + 1) as i32) {
                            min = min.min(dp[i - 1][k] + if k == j { 0 } else { 1 });
                        }
                    }
                    dp[i][j] = min;
                }
            }
        }
        *dp[n - 1].iter().min().unwrap()
    }
}

#[test]
fn test() {
    let obstacles = vec![0, 1, 2, 3, 0];
    let res = 2;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
    let obstacles = vec![0, 1, 1, 3, 3, 0];
    let res = 0;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
    let obstacles = vec![0, 2, 1, 0, 3, 0];
    let res = 2;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
}

// Accepted solution for LeetCode #1824: Minimum Sideway Jumps
function minSideJumps(obstacles: number[]): number {
    const inf = 1 << 30;
    const f = [1, 0, 1];
    for (let i = 1; i < obstacles.length; ++i) {
        for (let j = 0; j < 3; ++j) {
            if (obstacles[i] == j + 1) {
                f[j] = inf;
                break;
            }
        }
        const x = Math.min(...f) + 1;
        for (let j = 0; j < 3; ++j) {
            if (obstacles[i] != j + 1) {
                f[j] = Math.min(f[j], x);
            }
        }
    }
    return Math.min(...f);
}

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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.