LeetCode 1824 - Minimum Sideway Jumps

This problem describes a frog navigating a 3-lane road of length n. The frog starts at point 0 in lane 2 and wants to reach point n. Each point along the road may have at most one obstacle in one of the three lanes, represented by the obstacles array.

leetcodemediumarraydynamic-programminggreedy

LeetCode Problem 1824

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Greedy

Solution

Problem Understanding

This problem describes a frog navigating a 3-lane road of length n. The frog starts at point 0 in lane 2 and wants to reach point n. Each point along the road may have at most one obstacle in one of the three lanes, represented by the obstacles array. The frog can move forward in the same lane if there is no obstacle, or perform a side jump to switch to a different lane at the same point if the target lane has no obstacle.

The input is an array obstacles of length n + 1, where obstacles[i] indicates which lane has an obstacle at point i (0 means no obstacle). The output is the minimum number of side jumps required for the frog to reach any lane at the final point n starting from lane 2 at point 0.

Key points include:

  • The frog always starts at lane 2, point 0.
  • Obstacles are never present at points 0 or n.
  • The frog may jump forward if the lane is clear or side jump if blocked.
  • Only one obstacle per point ensures we never have to consider multiple conflicts simultaneously.

Important edge cases:

  • obstacles contains no obstacles except at lanes other than lane 2; the frog may not need to side jump.
  • Obstacles may force immediate side jumps at early points.
  • Very large n up to 5 * 10^5 requires an efficient O(n) solution.

Approaches

The brute-force approach is to simulate all possible frog movements recursively. At each point, the frog could move forward in its current lane if there is no obstacle or jump to any other lane without an obstacle. This approach tries all combinations, recursively exploring all side jumps. It is correct but extremely inefficient because the number of states grows exponentially with n, making it infeasible for n up to 5 * 10^5.

The key observation for a better solution is that we can track the minimum side jumps required to reach each lane at each point. Since there are only 3 lanes, we can use dynamic programming or even a rolling array to keep track of the current minimum side jumps for lanes 1, 2, and 3 at each point. At each step, the frog either moves forward in the same lane if unobstructed or computes the minimum among side jumps to the other lanes.

This leads to an O(n) time and O(1) space solution using a simple DP array of size 3.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Recursively explores all possible moves; too slow for large n
Optimal O(n) O(1) Track minimum side jumps for each lane using a rolling DP array of size 3

Algorithm Walkthrough

  1. Initialize a DP array dp of size 3 to store the minimum side jumps needed to be at lanes 1, 2, and 3 at the current point. Since the frog starts at lane 2, dp = [1, 0, 1]. This is because starting at lane 2 requires 0 jumps, and being at lanes 1 or 3 requires one side jump initially.
  2. Iterate through each point i from 1 to n:

a. For each lane, if there is an obstacle at point i, set dp[lane] to infinity (or a very large number), as that lane cannot be occupied.

b. After marking blocked lanes, check each lane l. If dp[l] is not blocked, update it as the minimum of its current value and dp[other_lane] + 1 for the two other lanes. This simulates the possibility of performing a side jump to reach lane l. 3. After processing all points, the answer is the minimum value in dp, representing the minimum side jumps needed to reach the end in any lane.

Why it works: The DP array always maintains the minimum number of side jumps to reach each lane at the current point. By updating with the minimum among possible side jumps at each step, we ensure that the frog's path is globally optimal. The rolling array ensures that space is constant while correctly propagating the minimum jumps forward.

Python Solution

from typing import List

class Solution:
    def minSideJumps(self, obstacles: List[int]) -> int:
        n = len(obstacles)
        # Initialize DP array: dp[lane] = min side jumps to reach current point in that lane
        dp = [1, 0, 1]  # lanes 1,2,3 at point 0

        for i in range(1, n):
            # Set blocked lanes to infinity
            for lane in range(3):
                if obstacles[i] == lane + 1:
                    dp[lane] = float('inf')
            
            # Update dp for non-blocked lanes using side jumps
            for lane in range(3):
                if obstacles[i] != lane + 1:
                    dp[lane] = min(dp[lane], min(dp[(lane + 1) % 3], dp[(lane + 2) % 3]) + 1)
        
        return min(dp)

Explanation: The initialization [1,0,1] accounts for the starting position at lane 2. For each point, blocked lanes are marked as inf. Then, for lanes not blocked, we calculate the minimum side jumps considering side jumps from the other two lanes. The final answer is the smallest value in the DP array.

Go Solution

import "math"

func minSideJumps(obstacles []int) int {
    n := len(obstacles)
    dp := []int{1, 0, 1} // lanes 1,2,3 at point 0

    for i := 1; i < n; i++ {
        for lane := 0; lane < 3; lane++ {
            if obstacles[i] == lane+1 {
                dp[lane] = math.MaxInt32
            }
        }
        for lane := 0; lane < 3; lane++ {
            if obstacles[i] != lane+1 {
                minSide := min(dp[(lane+1)%3], dp[(lane+2)%3]) + 1
                if dp[lane] > minSide {
                    dp[lane] = minSide
                }
            }
        }
    }
    return min(dp[0], min(dp[1], dp[2]))
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

Go-specific notes: Go does not have inf for integers, so we use math.MaxInt32 to mark blocked lanes. We use slices for the DP array and a helper function min to simplify comparisons.

Worked Examples

Example 1: obstacles = [0,1,2,3,0]

Point Obstacles dp (lane1,lane2,lane3)
0 0 [1,0,1]
1 1 [inf,0,2]
2 2 [1,inf,2]
3 3 [1,2,inf]
4 0 [2,2,2]

Minimum = 2 side jumps.

Example 2: obstacles = [0,1,1,3,3,0]

Point Obstacles dp
0 0 [1,0,1]
1 1 [inf,0,1]
2 1 [inf,0,1]
3 3 [0,0,inf]
4 3 [0,0,inf]
5 0 [0,0,1]

Minimum = 0 side jumps.

Complexity Analysis

Measure Complexity Explanation
Time O(n) We iterate through all points once and update 3 lanes at each point
Space O(1) We use a fixed-size array of 3 elements, independent of n

The algorithm scales linearly with the road length and uses constant space since we only track the 3 lanes at the current point.

Test Cases

# Provided examples
assert Solution().minSideJumps([0,1,2,3,0]) == 2  # example 1
assert Solution().minSideJumps([0,1,1,3,3,0]) == 0  # example 2
assert Solution().minSideJumps([0,2,1,0,3,0]) == 2  # example 3

# Edge cases
assert Solution().minSideJumps([0,0,0,0,0]) == 0  # no obstacles
assert Solution().minSideJumps([0,1,0,1,0]) == 1  # alternating obstacles
assert Solution().minSide