Longest Mirror Digit Sequence
Autumn Tournament (2008)
Submission name: pal.cpp
Problem summary
Select and rearrange some of the given file sizes 1,2,5,10. Concatenate their decimal representations without spaces to obtain a palindrome. Maximize the digit-string length; among maximum-length answers choose the lexicographically smallest. Print No when no nonempty palindrome can be formed.
Source clarification. The short statement does not explicitly spell out rearrangement or character-level comparison. Its samples and the official archive’s count-based reference solution establish this intended interpretation; the tokens may be reordered, and 10 contributes two decimal characters. Original test data: https://infos.infosbg.com/files/Contests/ETI/2008/C/problems/pal.zip
Input
N, followed by N values from {1,2,5,10}.
Output
The optimal palindrome as a digit string, or No.
Constraints
1<N<10000.
Worked example
Original statement sample.
Input
8 5 1 5 2 5 2 1 2
Output
1252521
Solution analysis
- The order may be rearranged: count the four token types. A 10 contributes a zero preceded by a one. In a palindrome, that zero must also be followed by a one. Hence using any 10 requires at least one standalone 1; with only tens no palindrome exists.
- Try keeping each available token count or one fewer. Removing more than one of a type cannot help repair parity: restoring a pair increases length while preserving palindrome feasibility. Also try using no tens, which is necessary when no standalone 1 is available. For each choice, the digit counts are ones+ tens, tens, twos, fives; at most one may be odd.
- For a feasible choice, construct its first half greedily in character order 0,1,2,5. A zero must follow a one, and the next character after a zero must be one. Before accepting a character, check that the remaining ones can separate the remaining zeros and satisfy the center boundary. Each future zero needs a preceding and following one; consecutive 101 chains share their separating ones. Other digits may have to go before such a chain when there is no spare one. The helper canFinish implements this constant-time count test.
- In the sample the counts are two 1s, three 2s and three 5s. Drop one 5, giving first half 125 and center 2, so the answer is 1252521. Keeping center 5 instead would produce the lexicographically larger 1255521.
- In canFinish, let Z be remaining half-zeros and O remaining half-ones. If the prefix ends in zero, a one is mandatory next. Otherwise Z zeros require Z ones, plus one when the previous character is not one, minus one if center1 permits the half to end in zero. If the prefix ends in one and there are other digits, a tight alternating chain needs an extra separating one when the center is 0 or 1. With no zeros left, only a zero center still requires the half to finish in one. These conditions are both necessary and constructible by placing the other digits before or after the alternating chain.
Why it works
For fixed counts, a decimal palindrome requires at most one odd digit count. If any tens are used, at least one standalone one is necessary and, together with parity, sufficient for the construction described. Retaining at most one fewer token of each type suffices because any further removed pair can be restored without breaking these conditions; the only exceptional reduction is dropping all tens when no one exists. For fixed counts, the completion test counts precisely the ones needed to surround all remaining zeros, accounting for whether the previous character already supplies a one, whether center1 allows a final zero, and whether other digits require a break in an otherwise tight alternating chain. Greedily choosing the smallest character with a feasible completion minimizes the first half, and hence the complete palindrome. Enumerating these choices and comparing length then lexicographic order yields the optimum.
Watch out. The palindrome is tested on decimal characters, not on the sequence of numeric tokens: a single 10 is not a palindrome. Treat 10 as an indivisible input token. Do not mistake this for longest palindromic subsequence in the original input order.
Commented C++17 solution
An independently written complete program with standard headers.
#include<algorithm>
#include<array>
#include<iostream>
#include<string>
using namespace std;
bool canFinish(array<int,4>remaining,char previous,char center){
int count=remaining[0]+remaining[1]+remaining[2]+remaining[3];
if(count==0){
return (previous!='0'||center=='1')&&(center!='0'||previous=='1');
}
if(previous=='0'){
if(remaining[1]==0){
return false;
}
remaining[1]--;
return canFinish(remaining,'1',center);
}
int zeros=remaining[0];
int ones=remaining[1];
int others=remaining[2]+remaining[3];
if(zeros==0){
return center!='0'||ones>0;
}
int needed=zeros+(previous!='1')-(center=='1');
// Without a spare one, other digits cannot interrupt an alternating chain.
if(previous=='1'&&others>0&&(center=='0'||center=='1')){
needed++;
}
return ones>=needed;
}
int main(){
int n;
cin>>n;
array<int,4>count{};
for(int i=0;i<n;i++){
int value;
cin>>value;
int index=value==1?0:value==2?1:value==5?2:3;
count[index]++;
}
string best;
for(int mask=0;mask<16;mask++){
for(int dropTens=0;dropTens<2;dropTens++){
auto use=count;
bool valid=true;
for(int i=0;i<4;i++){
use[i]-=(mask>>i)&1;
if(use[i]<0){
valid=false;
}
}
if(dropTens){
use[3]=0;
}
if(!valid||(use[3]>0&&use[0]==0)){
continue;
}
array<int,4>digits{use[3],use[0]+use[3],use[1],use[2]};
string symbols="0125";
string middle;
for(int i=0;i<4;i++){
if(digits[i]%2==1){
middle+=symbols[i];
}
}
if(middle.size()>1){
continue;
}
array<int,4>remaining;
for(int i=0;i<4;i++){
remaining[i]=digits[i]/2;
}
string half;
char previous='?';
int halfLength=remaining[0]+remaining[1]+remaining[2]+remaining[3];
for(int position=0;position<halfLength;position++){
for(int i=0;i<4;i++){
char next=symbols[i];
if(remaining[i]==0||(next=='0'&&previous!='1')||
(previous=='0'&&next!='1')){
continue;
}
remaining[i]--;
if(canFinish(remaining,next,middle.empty()?'?':middle[0])){
half+=next;
previous=next;
break;
}
remaining[i]++;
}
}
string candidate=half+middle;
reverse(half.begin(),half.end());
candidate+=half;
if(candidate.size()>best.size()||(candidate.size()==best.size()&&candidate<best)){
best=candidate;
}
}
}
cout<<(best.empty()?"No":best)<<'\n';
}