# SDSTRING - Editorial

Contest

Setter : Jeevan Jyot Singh
Tester : Rahul Dugar
Editorialist : Rajarshi Basu

None

Simple

# PROBLEM

A binary string is called a self-destructing string if it can reduced to an empty string by performing the following operation some number of times (possibly zero): Choose a valid integer i such that the i-th character of the current string is different from the i+1-th character, and remove these two characters from the string.

You are given a binary string s. Your task is to convert s to a self-destructing string. To do that, you may perform the following operation any number of times (possibly zero): Choose an integer i (1 \le i \le |s|-1) such that the i-th character of s is different from the i+1-th character, and invert one of these characters (inverting a character means changing β0β to β1β or β1β to β0β, e.g. the string β01β can be changed to β00β).

Find the smallest number of operations required to convert s to a self-destructing string or determine that it is impossible.

### Constraints

• 1 \le T \le 1,000
• 1 \le |s| \le 10^6
• s contains only characters β0β and β1β
• the sum of |s| over all test cases does not exceed 10^6

# EXPLANATION

The following are the cases :

• the string has odd number of characters. Then, it is obviously impossible since every deletion operation doesnβt change the parity of length.
• If the string has only 0βs or 1βs then we cannot do a change operation, and it is also non reducible, so its impossible.
• Otherwise it is just the difference between number of 1βs and 0βs, divided by 2.

# SOLUTION

Testerβs Solution
``````#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#ifndef rd
#define trace(...)
#define endl '\n'
#endif
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double f80;
#define double long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define sz(x) ((long long)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
const ll infl=0x3f3f3f3f3f3f3f3fLL;
const int infi=0x3f3f3f3f;
//const int mod=998244353;
const int mod=1000000007;
typedef vector<int> vi;
typedef vector<ll> vl;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
auto clk=clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
uniform_int_distribution<int> uid(0,lim-1);
return uid(rang);
}
int powm(int a, int b) {
int res=1;
while(b) {
if(b&1)
res=(res*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return res;
}

long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
assert(l<=x && x<=r);
return x;
} else {
assert(false);
}
}
}
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
}
long long readIntLn(long long l,long long r){
}
}
}

int suss=0;
void solve() {
for(char i:s)
if(i!='0'&&i!='1')
assert(0);
//	cin>>s;
//	suss+=sz(s);
//	assert(suss<=1000000);
if(sz(s)&1) {
cout<<-1<<endl;
} else {
int ones=accumulate(all(s),0LL)-(sz(s)*'0');
if(ones==sz(s)||ones==0)
cout<<-1<<endl;
else
cout<<max(ones-sz(s)/2,sz(s)/2-ones)<<endl;
}
}

signed main() {
ios_base::sync_with_stdio(0),cin.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
cout<<fixed<<setprecision(7);
//	int t=1;
//	cin>>t;
fr(i,1,t)
solve();
//	assert(getchar()==EOF);
#ifdef rd
//	cerr<<endl<<endl<<endl<<"Time Elapsed: "<<((double)(clock()-clk))/CLOCKS_PER_SEC<<endl;
#endif
}
``````

# VIDEO EDITORIAL (Hindi):

why is this condition important??

• If the string has only 0βs0βs or 1βs1βs then we cannot do a change operation, and it is also non reducible, so its impossible.

authors @rajarshi_basu plz reply to this query

1 Like