Subarray Division
Hazrat Ali
Two children, Lily and Ron, want to share a chocolate bar. Each of the squares has an integer on it.
Lily decides to share a contiguous segment of the bar selected such that:
- The length of the segment matches Ron's birth month, and,
- The sum of the integers on the squares is equal to his birth day.
Determine how many ways she can divide the chocolate.
Example
Lily wants to find segments summing to Ron's birth day, with a length equalling his birth month, . In this case, there are two segments meeting her criteria: and .
Function Description
Complete the birthday function in the editor below.
birthday has the following parameter(s):
- int s[n]: the numbers on each of the squares of chocolate
- int d: Ron's birth day
- int m: Ron's birth month
Returns
- int: the number of ways the bar can be divided
Input Format
The first line contains an integer , the number of squares in the chocolate bar.
The second line contains space-separated integers , the numbers on the chocolate squares where .
The third line contains two space-separated integers, and , Ron's birth day and his birth month.
Constraints
- , where ()
Sample Input 0
5
1 2 1 3 2
3 2
Sample Output 0
2
Explanation 0
Lily wants to give Ron squares summing to . The following two segments meet the criteria:
Sample Input 1
6
1 1 1 1 1 1
3 2
Sample Output 1
0
Explanation 1
Lily only wants to give Ron consecutive squares of chocolate whose integers sum to . There are no possible pieces satisfying these constraints:
Thus, we print as our answer.
Sample Input 2
1
4
4 1
Sample Output 2
1
Solution
#include <bits/stdc++.h>
using namespace std;
string ltrim(const string &);
string rtrim(const string &);
vector<string> split(const string &);
/*
* Complete the 'birthday' function below.
*
* The function is expected to return an INTEGER.
* The function accepts following parameters:
* 1. INTEGER_ARRAY s
* 2. INTEGER d
* 3. INTEGER m
*/
int birthday(vector<int>& s, int& a, int& b) {
int sum = 0, temp = 0, valid = 0;
for (int i = 0; i < s.size(); i++) {
temp++;
sum += s[i];
if (temp == b) {
if (sum == a) ++valid;
temp--;
sum -= s[i - b + 1];
}
}
return valid;
}
int main()
{
ofstream fout(getenv("OUTPUT_PATH"));
string n_temp;
getline(cin, n_temp);
int n = stoi(ltrim(rtrim(n_temp)));
string s_temp_temp;
getline(cin, s_temp_temp);
vector<string> s_temp = split(rtrim(s_temp_temp));
vector<int> s(n);
for (int i = 0; i < n; i++) {
int s_item = stoi(s_temp[i]);
s[i] = s_item;
}
string first_multiple_input_temp;
getline(cin, first_multiple_input_temp);
vector<string> first_multiple_input = split(rtrim(first_multiple_input_temp));
int d = stoi(first_multiple_input[0]);
int m = stoi(first_multiple_input[1]);
int result = birthday(s, d, m);
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;
}