1345. Jump Game IV

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

• i + 1 where: i + 1 < arr.length.
• i - 1 where: i - 1 >= 0.
• j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Example 4:

Input: arr = [6,1,9]
Output: 2

Example 5:

Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

Constraints:

• 1 <= arr.length <= 5 * 10^4
• -10^8 <= arr[i] <= 10^8

# @lc code=start
using LeetCode

function min_jumps_1345(arr::Vector{Int})::Int
    # data initialize
    n, step = length(arr), 0
    nums = Dict{Int,Set}(x => Set{Int}() for x in Set{Int}(arr))
    for (i, val) in enumerate(arr)
        push!(nums[val], i)
    end
    # BFS Search
    up_nums, up_inds, inds = Set{Int}([arr[1]]), Set{Int}([1]), Set{Int}([0, 1, n + 1])
    while !(n in up_inds)
        down_inds = Set{Int}()
        # indexs by numbers
        for num in up_nums
            union!(down_inds, nums[num])
        end
        # nearby indexs
        for i in up_inds
            push!(down_inds, i - 1)
            push!(down_inds, i + 1)
        end
        # remove extra indexs
        for i in down_inds
            i in inds && delete!(down_inds, i)
        end
        up_inds, up_nums = down_inds, Set{Int}(arr[i] for i in down_inds)
        union!(inds, up_inds)
        step += 1
    end
    return step
end

# @lc code=end

min_jumps_1345 (generic function with 1 method)

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