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Minimum number of sets with numbers less than Y in C++
Problem statement
Given a string of consecutive digits and a number Y, the task is to find the number of minimum sets such that every set follows the below rule −
- Set should contain consecutive numbers
- No digit can be used more than once.
- The number in the set should not be more than Y.
Example
If str = “1234” and Y = 20 then answer is 3 as below sets are created −
{12} {3} and {4}
Algorithm
- Convert string to number
- If the number is not greater than Y, then mark f = 1
- If the number exceeds Y, then increase count if f = 1 and re-initialize f as 0 and also initialise num as s[i]-‘0’ or num as 0
- After iterating in the string completely, then increase count if f is 1
Example
#include <iostream>
#include <string>
using namespace std;
int getMinSets(string str, int y) {
int cnt = 0;
int num = 0;
int l = str.length();
int f = 0;
for (int i = 0; i < l; ++i) {
num = num * 10 + str[i] - 48;
if (num <= y) {
f = 1;
continue;
}
if (f) {
++cnt;
}
num = str[i] - '0';
f = 0;
if (num <= y) {
f = 1;
} else {
num = 0;
}
}
if (f) {
++cnt;
}
return cnt;
}
int main() {
string str = "1234";
int y = 20;
cout << "Minimum sets = " << getMinSets(str, y) << endl;
return 0;
}Output
When you compile and execute above program. It generates following output −
Minimum sets = 3
Updated on: 22-Nov-2019
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