There are N problems numbered 1..N which you need to complete. You've arranged the problems in increasing difficulty order, and the ith problem has estimated difficulty level i. You have also assigned a rating vi to each problem. Problems with similar vi values are similar in nature. On each day, you will choose a subset of the problems and solve them. You've decided that each subsequent problem solved on the day should be tougher than the previous problem you solved on that day. Also, to make it less boring, consecutive problems you solve should differ in their vi rating by at least K. What is the least number of days in which you can solve all problems? Input Format The first line contains the number of test cases T. T test cases follow. Each case contains an integer N and K on the first line, followed by integers v1,...,vn on the second line. Constraints 1 <= T <= 100 1 <= N <= 300 1 <= vi <= 1000 1 <= K <= 1000 Output Format Output T lines, one for each test case, containing the minimum number of days in which all problems can be solved. Sample Input 2 3 2 5 4 7 5 1 5 3 4 5 6 Sample Output 2 1 Explanation For the first example, you can solve the problems with rating 5 and 7 on the first day and the problem with rating 4 on the next day. Note that the problems with rating 5 and 4 cannot be completed consecutively because the ratings should differ by at least K (which is 2). Also, the problems cannot be completed in order 5,7,4 in one day because the problems solved on a day should be in increasing difficulty level. For the second example, all problems can be solved on the same day. #include using namespace std; string ltrim(const string &); string rtrim(const string &); vector split(const string &); /* * Complete the 'problemSolving' function below. * * The function is expected to return an INTEGER. * The function accepts following parameters: * 1. INTEGER k * 2. INTEGER_ARRAY v */ int problemSolving(int k, vector v) { } int main() { ofstream fout(getenv("OUTPUT_PATH")); string t_temp; getline(cin, t_temp); int t = stoi(ltrim(rtrim(t_temp))); string first_multiple_input_temp; getline(cin, first_multiple_input_temp); vector first_multiple_input = split(rtrim(first_multiple_input_temp)); int n = stoi(first_multiple_input[0]); int k = stoi(first_multiple_input[1]); string v_temp_temp; getline(cin, v_temp_temp); vector v_temp = split(rtrim(v_temp_temp)); vector v(n); for (int i = 0; i < n; i++) { int v_item = stoi(v_temp[i]); v[i] = v_item; } int result = problemSolving(k, v); fout << result << "\n"; fout.close(); return 0; } string ltrim(const string &str) { string s(str); s.erase( s.begin(), find_if(s.begin(), s.end(), not1(ptr_fun(isspace))) ); return s; } string rtrim(const string &str) { string s(str); s.erase( find_if(s.rbegin(), s.rend(), not1(ptr_fun(isspace))).base(), s.end() ); return s; } vector split(const string &str) { vector tokens; string::size_type start = 0; string::size_type end = 0; while ((end = str.find(" ", start)) != string::npos) { tokens.push_back(str.substr(start, end - start)); start = end + 1; } tokens.push_back(str.substr(start)); return tokens; } don't copy old bartleby answer it's wrong

There are N problems numbered 1..N which you need to complete. You've arranged the problems in increasing difficulty order, and the ith problem has estimated difficulty level i. You have also assigned a rating vi to each problem. Problems with similar vi values are similar in nature. On each day, you will choose a subset of the problems and solve them. You've decided that each subsequent problem solved on the day should be tougher than the previous problem you solved on that day. Also, to make it less boring, consecutive problems you solve should differ in their vi rating by at least K. What is the least number of days in which you can solve all problems? Input Format The first line contains the number of test cases T. T test cases follow. Each case contains an integer N and K on the first line, followed by integers v1,...,vn on the second line. Constraints 1 <= T <= 100 1 <= N <= 300 1 <= vi <= 1000 1 <= K <= 1000 Output Format Output T lines, one for each test case, containing the minimum number of days in which all problems can be solved. Sample Input 2 3 2 5 4 7 5 1 5 3 4 5 6 Sample Output 2 1 Explanation For the first example, you can solve the problems with rating 5 and 7 on the first day and the problem with rating 4 on the next day. Note that the problems with rating 5 and 4 cannot be completed consecutively because the ratings should differ by at least K (which is 2). Also, the problems cannot be completed in order 5,7,4 in one day because the problems solved on a day should be in increasing difficulty level. For the second example, all problems can be solved on the same day. #include using namespace std; string ltrim(const string &); string rtrim(const string &); vector split(const string &); /* * Complete the 'problemSolving' function below. * * The function is expected to return an INTEGER. * The function accepts following parameters: * 1. INTEGER k * 2. INTEGER_ARRAY v */ int problemSolving(int k, vector v) { } int main() { ofstream fout(getenv("OUTPUT_PATH")); string t_temp; getline(cin, t_temp); int t = stoi(ltrim(rtrim(t_temp))); string first_multiple_input_temp; getline(cin, first_multiple_input_temp); vector first_multiple_input = split(rtrim(first_multiple_input_temp)); int n = stoi(first_multiple_input[0]); int k = stoi(first_multiple_input[1]); string v_temp_temp; getline(cin, v_temp_temp); vector v_temp = split(rtrim(v_temp_temp)); vector v(n); for (int i = 0; i < n; i++) { int v_item = stoi(v_temp[i]); v[i] = v_item; } int result = problemSolving(k, v); fout << result << "\n"; fout.close(); return 0; } string ltrim(const string &str) { string s(str); s.erase( s.begin(), find_if(s.begin(), s.end(), not1(ptr_fun(isspace))) ); return s; } string rtrim(const string &str) { string s(str); s.erase( find_if(s.rbegin(), s.rend(), not1(ptr_fun(isspace))).base(), s.end() ); return s; } vector split(const string &str) { vector tokens; string::size_type start = 0; string::size_type end = 0; while ((end = str.find(" ", start)) != string::npos) { tokens.push_back(str.substr(start, end - start)); start = end + 1; } tokens.push_back(str.substr(start)); return tokens; } don't copy old bartleby answer it's wrong

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

There are N problems numbered 1..N which you need to complete. You've arranged the problems in increasing difficulty order, and the i^th problem has estimated difficulty level i. You have also assigned a rating vi to each problem. Problems with similar vi values are similar in nature. On each day, you will choose a subset of the problems and solve them. You've decided that each subsequent problem solved on the day should be tougher than the previous problem you solved on that day. Also, to make it less boring, consecutive problems you solve should differ in their vi rating by at least K. What is the least number of days in which you can solve all problems?

Input Format

The first line contains the number of test cases T. T test cases follow. Each case contains an integer N and K on the first line, followed by integers v1,...,vn on the second line.

Constraints

1 <= T <= 100
1 <= N <= 300
1 <= vi <= 1000
1 <= K <= 1000

Output Format

Output T lines, one for each test case, containing the minimum number of days in which all problems can be solved.

Sample Input

2 3 2 5 4 7 5 1 5 3 4 5 6

Sample Output

2 1

Explanation

For the first example, you can solve the problems with rating 5 and 7 on the first day and the problem with rating 4 on the next day. Note that the problems with rating 5 and 4 cannot be completed consecutively because the ratings should differ by at least K (which is 2). Also, the problems cannot be completed in order 5,7,4 in one day because the problems solved on a day should be in increasing difficulty level.

For the second example, all problems can be solved on the same day.

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);

string rtrim(const string &);

vector<string> split(const string &);

* Complete the 'problemSolving' function below.

* The function is expected to return an INTEGER.

* The function accepts following parameters:

* 1. INTEGER k

* 2. INTEGER_ARRAY v

int problemSolving(int k, vector<int> v) {

}

int main()

{

ofstream fout(getenv("OUTPUT_PATH"));

string t_temp;

getline(cin, t_temp);

int t = stoi(ltrim(rtrim(t_temp)));

string first_multiple_input_temp;

getline(cin, first_multiple_input_temp);

vector<string> first_multiple_input = split(rtrim(first_multiple_input_temp));

int n = stoi(first_multiple_input[0]);

int k = stoi(first_multiple_input[1]);

string v_temp_temp;

getline(cin, v_temp_temp);

vector<string> v_temp = split(rtrim(v_temp_temp));

vector<int> v(n);

for (int i = 0; i < n; i++) {

int v_item = stoi(v_temp[i]);

v[i] = v_item;

}

int result = problemSolving(k, v);

fout << result << "\n";

fout.close();

return 0;

}

string ltrim(const string &str) {

string s(str);

s.erase(

s.begin(),

find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))

);

return s;

}

string rtrim(const string &str) {

string s(str);

s.erase(

find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),

s.end()

);

return s;

}

vector<string> split(const string &str) {

vector<string> tokens;

string::size_type start = 0;

string::size_type end = 0;

while ((end = str.find(" ", start)) != string::npos) {

tokens.push_back(str.substr(start, end - start));

start = end + 1;

}

tokens.push_back(str.substr(start));

return tokens;

}

don't copy old bartleby answer it's wrong

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