Problem Overview
Codeforces Problem 1006B - Polycarp's Practice is a problem from the Codeforces contest that involves sorting elements and maximizing the sum of specific indices. The goal is to find the optimal way to select certain indices, calculate the sum of their values, and compute the distance between these indices for output.

In this problem, we are given an array of integers and tasked with selecting the largest k elements. After selecting them, we must calculate the sum of those elements and determine the distances between their positions in the original array. This requires sorting the elements, selecting the k largest values, and calculating the resulting indices' gaps.

Problem Breakdown
Given two integers n and k, and an array a of size n, we need to perform the following tasks:

  1. Find the k largest elements in the array a.
  2. Compute the sum of these k largest elements.
  3. Output the sum.
  4. Output the distances between the positions of these k largest elements.

Input:

  • The first line contains two integers n and k.
  • The second line contains n integers, the elements of the array a.

Output:

  • Print the sum of the k largest elements.
  • For each pair of consecutive largest elements, print the distance between their positions in the array.

Detailed Explanation and Solution

Let's walk through the solution step-by-step.

Step 1: Parse Input and Prepare Arrays

We are given n and k. We first need to read the array a. We will also create an array b which will hold a copy of a for sorting, and an array c to store the indices of the largest elements.

int n, i, j, k, mx = -1, ans;
cin >> n >> k;
int a[n + 10], b[n + 10], c[k + 10];

Here, a is the original array, b is the sorted copy of a, and c will store the indices of the largest k elements.

Step 2: Find the Largest k Elements

To find the largest k elements, we sort the array b and then use the sorted array to determine the indices of the largest elements in the original array a.

 
for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
}

 

Step 3: Sorting and Finding Indices

After sorting the array b, we use a greedy approach to pick the largest element, mark its position, and then exclude it from future selections. We repeat this process k times.

 
for (i = 0; i < k; i++) {
    mx = -1;
    for (j = 0; j < n; j++) {
        if (b[j] > mx) {
            mx = b[j];
            ans = j;
        }
    }
    b[ans] = 0;  // Mark this element as selected
    c[i] = ans;  // Store the index
}

 

Step 4: Sorting the Indices

Once we have selected the k largest elements and their indices, we sort the array c to ensure that the indices are in increasing order. This helps when calculating the distances between them later.

 
sort(c, c + k);

 

Step 5: Calculate the Sum of the Largest Elements

The sum of the largest k elements is calculated by adding the corresponding elements from the original array a using the indices stored in c.

 
ans = 0;
for (i = 0; i < k; i++)
    ans += a[c[i]];
cout << ans << endl;

 

Step 6: Output the Distances Between Indices

For each pair of consecutive largest elements, we compute the distance between their positions and output it.

 
if (k == 1)
    cout << n << endl;
else {
    for (i = 0; i < k - 1; i++) {
        if (i == 0)
            ans = c[i] + (c[i + 1] - c[i]);
        else
            ans = c[i + 1] - c[i];
        cout << ans << " ";
    }
    ans = n - c[k - 1];
    cout << ans << endl;
}

 

Complete Solution Code:

 
#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n, i, j, k, mx = -1, ans;
    cin >> n >> k;
    int a[n + 10], b[n + 10], c[k + 10];
    for (i = 0; i < n; i++) {
        cin >> a[i];
        b[i] = a[i];
    }
    for (i = 0; i < k; i++) {
        mx = -1;
        for (j = 0; j < n; j++) {
            if (b[j] > mx) {
                mx = b[j];
                ans = j;
            }
        }
        b[ans] = 0;
        c[i] = ans;
    }
    sort(c, c + k);
    ans = 0;
    for (i = 0; i < k; i++)
        ans += a[c[i]];
    cout << ans << endl;
    if (k == 1)
        cout << n << endl;
    else {
        for (i = 0; i < k - 1; i++) {
            if (i == 0)
                ans = c[i] + (c[i + 1] - c[i]);
            else
                ans = c[i + 1] - c[i];
            cout << ans << " ";
        }
        ans = n - c[k - 1];
        cout << ans << endl;
    }

    return 0;
}

 

Time and Space Complexity Analysis

  • Time Complexity:

    • The time complexity is dominated by the sorting operation. Sorting the array b takes O(nlog⁡n)O(n \log n)O(nlogn). After sorting, we find the indices of the k largest elements in a greedy manner, which is O(k⋅n)O(k \cdot n)O(k⋅n). The overall time complexity of the solution is O(nlog⁡n+k⋅n)O(n \log n + k \cdot n)O(nlogn+k⋅n), which simplifies to O(n2)O(n^2)O(n2) in the worst case.

  • Space Complexity:

    • We are using additional arrays a, b, and c, each of size n or k. Hence, the space complexity is O(n)O(n)O(n).

Common Mistakes to Avoid

  • Not Sorting Correctly: It's important to sort the array of indices correctly after selecting the largest elements so that the distances between the selected positions are accurate.
  • Misunderstanding the Problem Statement: Always ensure you understand the requirement of selecting indices and calculating the sum and distances. This is a common trap.
  • Incorrect Indexing: While working with indices in the array, ensure that you're using zero-based indexing correctly (or one-based if required) throughout the solution.

Related Problems to Practice

  1. Codeforces 1040B - Combination Lock: This problem involves similar logic of selecting the largest or smallest elements and calculating the gaps between their indices.
  2. Codeforces 1351B - Caisa and Pylons: Involves sorting and calculating sums based on selected indices.
  3. LeetCode 215 - Kth Largest Element in an Array: This problem focuses on finding the k largest elements in an array, which is quite similar to this problem.

Conclusion
In this problem, we successfully implemented a solution that sorts the array, selects the k largest elements, computes the sum, and calculates the distances between their indices in the original array. This approach works efficiently by sorting the array and then performing simple array manipulations. The time complexity of this solution is primarily dominated by the sorting step, which is O(nlog⁡n)O(n \log n)O(nlogn), making it efficient for the given problem constraints.

This problem helps improve understanding of sorting, greedy algorithms, and array manipulation. By focusing on these techniques, you can efficiently solve a variety of problems involving sorting and selecting specific elements.