Lesson 2

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!

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 C++ 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 vectors that hold the `l`

and `r`

values respectively for each query. For instance, let's say you have an array like `{2, 1, 3, 7, 5}`

and the following queries: `{0, 2, 4}`

and `{1, 3, 4}`

. The aim of our function would be to return `{1, 3, 5}`

as the minimum values within the ranges of the three queries.

We'll begin by declaring our function `query_min`

.

Initially, we'll set `n`

to be the size of the given array and create a two-dimensional vector, `precalc`

, filled with zeroes. This vector will store all the precalculated minimums for all possible range pairs (`l`

, `r`

).

C++`1#include <vector> 2#include <algorithm> // for std::min 3 4std::vector<int> query_min(const std::vector<int>& arr, const std::vector<int>& Ls, const std::vector<int>& Rs) { 5 int n = arr.size(); 6 std::vector<std::vector<int>> precalc(n, std::vector<int>(n, 0));`

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 store this minimum value in our `precalc`

two-dimensional vector.

C++`1#include <vector> 2#include <algorithm> // for std::min 3 4std::vector<int> query_min(const std::vector<int>& arr, const std::vector<int>& Ls, const std::vector<int>& Rs) { 5 int n = arr.size(); 6 std::vector<std::vector<int>> precalc(n, std::vector<int>(n, 0)); 7 for (int l = 0; l < n; ++l) { 8 int min_val = arr[l]; 9 for (int r = l; r < n; ++r) { 10 min_val = std::min(min_val, arr[r]); 11 precalc[l][r] = min_val; 12 } 13 }`

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 added to our result vector, `res`

.

C++`1#include <vector> 2#include <algorithm> // for std::min 3 4std::vector<int> query_min(const std::vector<int>& arr, const std::vector<int>& Ls, const std::vector<int>& Rs) { 5 int n = arr.size(); 6 std::vector<std::vector<int>> precalc(n, std::vector<int>(n, 0)); 7 for (int l = 0; l < n; ++l) { 8 int min_val = arr[l]; 9 for (int r = l; r < n; ++r) { 10 min_val = std::min(min_val, arr[r]); 11 precalc[l][r] = min_val; 12 } 13 } 14 std::vector<int> res; // Initialize the result vector 15 16 for (size_t i = 0; i < Ls.size(); ++i) { 17 int l = Ls[i]; 18 int r = Rs[i]; 19 res.push_back(precalc[l][r]); 20 } 21 22 return res; 23}`

Congratulations on reaching the end of this analysis lesson! You've learned to utilize techniques such as *Skipping Redundant Cases*, *Optimization Using Precalculation*, and *Picking the Best Variable to Iterate Over* to solve a complex array-based problem. Mastering these techniques is unquestionably advantageous when tackling larger datasets or equivalent challenges with stringent time constraints. Now is the time for you to consolidate what you've learned. I encourage you to explore more problems where these concepts can be applied and try to create optimized solutions using the principles you learned today. Happy coding!