function lcss_len(x::Vector{T}, y::Vector{T}, i=1, j=1, a=nothing) where {T}
    a = isnothing(a) ? zeros(T, min(length(x), length(y))) : a
    if i > length(x) || j > length(y)
        0
    elseif x[i] == y[j]
        n = 1 + lcss_len(x, y, i+1, j+1, a)
        a[n] = x[i]
        n
    else
        max(lcss_len(x, y, i+1, j, a), lcss_len(x, y, i, j+1, a))
    end
end

function lcss_diff(x::Vector{T}, y::Vector{T}) where {T}
    lcss = longest_common_subsequence(x, y)
    i, j, k = 1, 1, 1
    while i <= length(lcss) && j <= length(x) && k <= length(y)
        if x[j] == lcss[i] && y[k] == lcss[i]
            i += 1
            j += 1
            k += 1
        elseif x[j] == lcss[i] && y[k] != lcss[i]
            println("Added: $i $(y[k])")
            k += 1
        elseif y[k] == lcss[i] && x[j] != lcss[i]
            println("Removed: $i $(x[j])")
            j += 1
        else
            println("Changed: $(x[j]) -> $(y[k])")
            j += 1
            k += 1
        end
    end
end

function lrss_len(x::Vector{T}, i=1, j=1) where {T}
    if max(i, j) > length(x)
        0
    elseif x[i] == x[j] && i != j
        1 + lrss_len(x, i+1, j+1)
    else
        max(lrss_len(x, i+1, j), lrss_len(x, i, j+1))
    end
end

# This has a bug: e.g. [1,2,3,4,5,6] vs [1,2,4,5,6]. Sequences need to have same length
function longest_common_subsequence(x::Vector{T}, y::Vector{T})::Vector{T} where {T}
    m, n = length(x), length(y)
    C = zeros(Int, m+1, n+1)
    # Not necessary:
    C[:,1] .= 0
    C[1,:] .= 0
    for i = 1:m
        for j = 1:n
            if x[i] == y[j]
                C[i+1,j+1] = C[i, j] + 1
            else
                C[i+1,j+1] = max(C[i, j+1], C[i+1, j])
            end
        end
    end

    display(C)
    # Reconstruct sequence
    i, j = m, n
    k = C[m+1, n+1]
    seq = Vector{T}(undef, k)
    while k > 0
        if C[i,j] + 1 == C[i+1,j+1]
            seq[k] = x[i]
            k -= 1
            i -= 1
            j -= 1
        elseif C[i,j+1] == C[i+1,j+1]
            i -= 1
        elseif C[i+1,j] == C[i+1,j+1]
            j -= 1
        end
    end
    seq
end