Service Lane

A driver is driving on the freeway. The check engine light of his vehicle is on, and the driver wants to get service immediately. Luckily, a service lane runs parallel to the highway. It varies in width along its length.

Paradise Highway

You will be given an array of widths at points along the road (indices), then a list of the indices of entry and exit points. Considering each entry and exit point pair, calculate the maximum size vehicle that can travel that segment of the service lane safely.

Example

If the entry index, and the exit, , there are two segment widths of and respectively. The widest vehicle that can fit through both is . If and , the widths are which limits vehicle width to .

Function Description

Complete the serviceLane function in the editor below.

serviceLane has the following parameter(s):

int n: the size of the array
int cases[t][2]: each element contains the starting and ending indices for a segment to consider, inclusive

Returns

int[t]: the maximum width vehicle that can pass through each segment of the service lane described

Input Format

The first line of input contains two integers, and , where denotes the number of width measurements and , the number of test cases. The next line has space-separated integers which represent the array .

The next lines contain two integers, and , where is the start index and is the end index of the segment to check.

Constraints

Sample Input

STDIN               Function
-----               --------
8 5                 n = 8, t = 5
2 3 1 2 3 2 3 3     width = [2, 3, 1, 2, 3, 2, 3, 3]
0 3                 cases = [[0, 3], [4, 6], [6, 7], [3, 5], [0, 7]]
4 6
6 7
3 5
0 7

Sample Output

1
2
3
2
1

Solution

#include <bits/stdc++.h>

using namespace std;

vector<string> split_string(string);

// Complete the serviceLane function below.

vector<int> serviceLane(int n, vector<vector<int>> cases, vector<int> width) {

cout << "Value of n: " << n << endl;

cout << "Number of cases: " << cases.size() << endl;

cout << "Number of widths: " << width.size() << endl;

int i, j, k, smallest;

vector<int> ans(cases.size());

for (int x = 0; x < cases.size(); x++) {

i = cases[x][0];

j = cases[x][1];

k = i;

smallest = INT_MAX;

while (k <= j) {

smallest = min(width[k], smallest);

k++;

}

ans[x] = smallest;

}

return ans;

}

int main()

{

ofstream fout(getenv("OUTPUT_PATH"));

string nt_temp;

getline(cin, nt_temp);

vector<string> nt = split_string(nt_temp);

int n = stoi(nt[0]);

int t = stoi(nt[1]);

string width_temp_temp;

getline(cin, width_temp_temp);

vector<string> width_temp = split_string(width_temp_temp);

vector<int> width(n);

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

int width_item = stoi(width_temp[i]);

width[i] = width_item;

}

vector<vector<int>> cases(t);

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

cases[i].resize(2);

for (int j = 0; j < 2; j++) {

cin >> cases[i][j];

}

cin.ignore(numeric_limits<streamsize>::max(), '\n');

}

vector<int> result = serviceLane(n, cases, width);

for (int i = 0; i < result.size(); i++) {

fout << result[i];

if (i != result.size() - 1) {

fout << "\n";

}

fout << "\n";

fout.close();

return 0;

}

vector<string> split_string(string input_string) {

string::iterator new_end = unique(input_string.begin(), input_string.end(), [](const char& x, const char& y) {

return x == y and x == ' ';

});

input_string.erase(new_end, input_string.end());

while (input_string[input_string.length() - 1] == ' ') {

input_string.pop_back();

}

vector<string> splits;

char delimiter = ' ';

size_t i = 0;

size_t pos = input_string.find(delimiter);

while (pos != string::npos) {

splits.push_back(input_string.substr(i, pos - i));

i = pos + 1;

pos = input_string.find(delimiter, i);

}

splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));

return splits;

}

Comments