Hello there! In this unit, we're offering an engaging coding lesson that highlights the performance efficiencies offered by utilizing efficient data structures in Go. We'll address a slice-based problem that requires us to make an optimal choice to minimize the size of our slice. Excited? So am I! Let's get started.

In this unit's task, we'll manipulate a slice of integers. You are required to construct a Go function titled MinimalMaxBlock(). This function should accept a slice as an input and compute an intriguing property related to contiguous blocks within the slice.

More specifically, you must select a particular integer, k, from the slice. Once you've selected k, the function should remove all occurrences of k from the slice, thereby splitting it into several contiguous blocks or remaining sub-slices. A unique feature of k is that it is chosen such that the maximum length among these blocks is minimized.

For instance, consider the slice [1, 2, 2, 3, 1, 4, 4, 4, 1, 2, 5]. If we eliminate all instances of 2 (our k), the remaining blocks would be [1], [3, 1, 4, 4, 4, 1], and [5], with the longest containing 6 elements. Now, if we instead remove all instances of 1, the new remaining blocks would be [2, 2, 3], [4, 4, 4], and [2, 5], the longest of which contains 3 elements. As such, the function should return 1 in this case, as it leads to a minimally maximal block length.

An initial way to approach this problem can be through a brute force method. Each possible value in the slice could be tested in turn by removing it from the slice and then checking the resulting sub-slice sizes. This approach entails iteratively stepping through the slice for each possible value.

package main

import (
    "fmt"
    "math"
)

func MinimalMaxBlockBruteforce(slice []int) int {
    minMaxBlockSize := math.MaxInt32
    minNum := -1

    uniqueElements := make(map[int]bool)
    for _, num := range slice {
        uniqueElements[num] = true
    }

    for num := range uniqueElements {
        indices := []int{-1}
        for i, val := range slice {
            if val == num {
                indices = append(indices, i)
            }
        }
        indices = append(indices, len(slice))

        maxBlockSize := 0
        for i := 1; i < len(indices); i++ {
            maxBlockSize = max(maxBlockSize, indices[i]-indices[i-1]-1)
        }

        if maxBlockSize < minMaxBlockSize {
            minMaxBlockSize = maxBlockSize
            minNum = num
        }
    }

    return minNum
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func main() {
    slice := []int{1, 2, 2, 3, 1, 4, 4, 4, 1, 2, 5}
    fmt.Println(MinimalMaxBlockBruteforce(slice)) // Output: 1
}

This method has a time complexity of O(n^2), as it involves two nested loops: the outer loop cycles through each potential k value and the inner loop sweeps through the slice for each of these k values. However, this approach becomes increasingly inefficient as the size n of the slice grows due to its quadratic time complexity. For larger slices or multiple invocations, the computation time can noticeably increase, demonstrating the need for a more efficient solution.