Optimizing Array Block Splits

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

In this unit's task, we'll manipulate an array of integers. You are required to construct a Scala function named minimalMaxBlock. This function should accept an array as input and compute an intriguing property related to contiguous blocks within the array.

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

For instance, consider the array Array(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 Array(1), Array(3, 1, 4, 4, 4, 1), and Array(5), with the longest containing 6 elements. Now, if we instead remove all instances of 1, the new remaining blocks would be Array(2, 2, 3), Array(4, 4, 4), and Array(2, 5), the longest of which contains 3 elements. As such, the function should return 1 in this case, as it leads to a minimal maximum block length.

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

def minimalMaxBlockBruteforce(arr: Array[Int]): Int = {
  var minMaxBlockSize = Int.MaxValue
  var minNum = arr(0)

  for (num <- arr.distinct) { // Avoid duplicates
    val indices = arr.zipWithIndex.collect { case (x, i) if x == num => i }
    val extendedIndices = (-1 +: indices) :+ arr.length // Add artificial indices at the ends
    val blockSizes = for (i <- 1 until extendedIndices.length)
      yield extendedIndices(i) - extendedIndices(i - 1) - 1
    val maxBlockSize = if (blockSizes.isEmpty) 0 else blockSizes.max

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

  minNum
}

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 array for each of these k values.

However, this approach becomes increasingly inefficient as the size n of the array grows, due to its quadratic time complexity. For larger arrays or multiple invocations, the computation time can noticeably increase, demonstrating the need for a more efficient solution.