Maximum Digit
Autumn Tournament (2009)
Submission name: maxdigit.cpp
Problem summary
Given three three-digit positive integers B, C, D and a position K counted from the left, find the largest of their K-th digits that satisfies 9 + (A − 5) > 8. Print No if none does.
Input
B, C, D, K on one line.
Output
The largest eligible digit, or No.
Constraints
100 ≤ B, C, D ≤ 999; 1 ≤ K ≤ 3.
Worked example
Original statement sample.
Input
235 567 123 3
Output
7
Solution analysis
- The inequality simplifies to A + 4 > 8, or A > 4. Thus only digits 5 through 9 qualify.
- Use divisor 100, 10, or 1 for K equal to 1, 2, or 3. The selected digit is (number/divisor)%10. Find the maximum of the three selected digits and test it against 4.
- For 235, 567, 123 at position 3, the selected digits are 5, 7, 3. Their maximum 7 satisfies the inequality.
Why it works
The divisor formula extracts the requested position from every number. If the maximum exceeds 4 it is eligible and at least as large as every other candidate. If it does not, no candidate exceeds 4, so none satisfies the inequality.
Complexity: O(1) time and O(1) extra space.
Watch out. Positions are counted from the left. The inequality is strict: digit 4 is ineligible.
Commented C++17 solution
An independently written complete program with standard headers.
// Key idea: The divisor formula extracts the requested position from every number. If the maximum
// exceeds 4 it is eligible and at least as large as every other candidate.
#include<algorithm>
#include<iostream>
#include<limits>
#include<string>
using namespace std;
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int b,c,d,position;
cin>>b>>c>>d>>position;
int divisor=1;
for(int step=position;step<3;step++){
divisor*=10;
}
int largest=max({(b/divisor)%10,(c/divisor)%10,(d/divisor)%10});
if(largest>4){
cout<<largest<<'\n';
}
else{
cout<<"No\n";
}
}