Optimizing Range Minimum Queries | CodeSignal Learn

Optimizing Range Minimum Queries | CodeSignal Learn

Introduction

Welcome to this exciting analysis lesson! In this unit, we're going to take a practical approach and utilize several intriguing techniques, such as Skipping Redundant Cases, Optimization Using Precalculation, and Picking the Best Variable to Iterate Over . Our platform for this unit is an array -based problem, where we'll apply these techniques to formulate an efficient and optimized solution. Ready for the challenge? Then, let's get started!

Task Statement

Our task here involves an array composed of at most 1,000 elements and potentially millions of queries. Each query is a pair of integers denoted as l and r, which correspond to some indices in the array. Your goal is to write a Scala function that, for each query, returns the minimum value in the array between indices l and r (inclusive). The catch is this: rather than directly finding the minimum value for each query one by one, we're required to optimize the process. The idea here is to precalculate the minimum value for each possible l and r, store these values, and then proceed with the queries. This way, we can simplify the problem and enhance the speed of our solution by eliminating redundant computations. The function will accept three parameters: arr, Ls, and Rs . The primary array is arr, while Ls and Rs are arrays that hold the l and r values, respectively, for each query. For instance, let's say you have an array like Array(2, 1, 3, 7, 5) and the following queries: Array(0, 2, 4) and Array(1, 3, 4) . The aim of our function would be to return Array(1, 3, 5) as the minimum values within the ranges of the three queries.

Solution Building: Step 1

We'll begin by declaring our function queryMin . Initially, we'll set n to be the size of the given array and create a two-dimensional array, precalc, filled with zeroes. This array will store all the precalculated minimums for all possible range pairs (l, r). Scala def queryMin(arr: Array[Int], Ls: Array[Int], Rs: Array[Int]): Array[Int] = {val n = arr.length val precalc = Array.ofDim[Int](n, n)

Solution Building: Step 2

Now, we shall precalculate the minimum value for all possible l and r pairs. This is essentially the brute-force approach, but by doing this upfront, we optimize our subsequent queries. We loop through every possible range within arr, updating the minimum value found for that range and storing this minimum value in our precalc two-dimensional array. Scala def queryMin(arr: Array[Int], Ls: Array[Int], Rs: Array[Int]): Array[Int] = {val n = arr.length val precalc = Array.ofDim[Int](n, n) for (l <- 0 until n) {var minVal = arr(l) for (r <- l until n) {minVal = math.min(minVal, arr(r)) precalc(l)(r) = minVal}}

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Solution Building: Step 3

With our precalculation complete, we're ready to process the actual queries (the pairs of l and r). For each query, we use the precalculated values in precalc to quickly find the minimum value for that query's range. These values are then collected into our result array, res . def queryMin(arr: Array[Int], Ls: Array[Int], Rs: Array[Int]): Array[Int] = {val n = arr.length val precalc = Array.ofDim[Int](n, n) for (l <- 0 until n) {var minVal = arr(l) for (r <- l until n) {minVal = math.min(minVal, arr(r)) precalc(l)(r) = minVal}} val res = for ((l, r) <- Ls.zip(Rs)) yield precalc(l)(r) res}

Scala

def queryMin(arr: Array[Int], Ls: Array[Int], Rs: Array[Int]): Array[Int] = {
  val n = arr.length
  val precalc = Array.ofDim[Int](n, n)

  for (l <- 0 until n) {
    var minVal = arr(l)
    for (r <- l until n) {
      minVal = math.min(minVal, arr(r))
      precalc(l)(r) = minVal
    }
  }

  val res = for ((l, r) <- Ls.zip(Rs)) yield precalc(l)(r)
  res
}