PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Setter:
Tester: Rahul Dugar
Editorialist: Taranpreet Singh
DIFFICULTY
Simple
PREREQUISITES
None
PROBLEM
Given two binary strings A and B of same length, we need to determine the minimum number of operations to apply on A to convert A into B.
An operation involves picking an odd length substring of A and flipping all bits at odd position within this substring
QUICK EXPLANATION
- Since each operation can affect positions of only same parity modulo 2, we can divide this into two simpler sub-problems, by splitting
- Dividing both A and B into two strings each, odd indexed positions in one string (A_o and B_o) and even indexed positions in other string (A_e and B_e).
- The operation now becomes to flip a consecutive segment, either in A_o or in A_e, in order to make A_o same as B_o and A_e same as B_e
- It can be solved by a single pass over each string, maintaining the number of flips needed to make prefix of A_o and B_o and extending at each step.
EXPLANATION
Let’s consider an example of operation A = 0101101011, we apply operation at substring
A_{3,7} = 11010, we get 0100111111, as only bits at position 3, 5 and 7 are flipped in original string. Similarly, if operation is applied at A_{2,4}, we get 0111111011 as only positions 2 and 4 are flipped.
Crucial Observation
It’s easy to prove that no matter which substring we choose, the operation only affects either odd positioned bits, or even positioned bits.
Let’s split A into two strings A_o containing odd positioned bits of A in original order, and A_e containing even positioned bits of A in original order. Let’s do same for B.
Since our operation always affected consecutive bits of same parity, the operation on A is equivalent to flipping a consecutive sub-string of A_o or A_e, and the goal becomes to make A_o same as B_o and A_e same as B_e
Since both sub-problems are same, Let’s just focus on making A_e same as B_e
Another thing we can prove is that in optimal case, no two operation sub-strings would overlap, since for the overlap interval, flipping it twice gives same sub-string.
Hence, for each position, we can determine whether it needs to be flipped or not (by taking XOR of A_o and A_e). Let’s call this C_o
It’s intuitive to prove that the number of flips needed is the number of blocks consisting of ones in C_o.
We can similarly find C_e and add the operations needed for both odd and even positioned bits separately to get required number of operations.
TIME COMPLEXITY
Time complexity is O(N) per test case.
SOLUTIONS
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long
const int N = 2e5 + 5;
int32_t main()
{
IOS;
int t;
cin >> t;
assert(t >= 1 && t <= 1000);
int sumlen = 0;
while(t--)
{
string a, b;
cin >> a >> b;
assert(a.size() == b.size());
assert(a.size() >= 1 && a.size() <= 1e5);
sumlen += a.size();
assert(sumlen <= 1e5);
int ans = 0;
int n = a.size();
for(int i = 0; i < n; i += 2)
{
if(a[i] == b[i])
continue;
while(i < n && a[i] != b[i])
i += 2;
ans++;
}
for(int i = 1; i < n; i += 2)
{
if(a[i] == b[i])
continue;
while(i < n && a[i] != b[i])
i += 2;
ans++;
}
cout << ans << endl;
}
return 0;
}
Tester'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 readString(int l,int r,char endd){
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){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int sum_n=0;
int holve(string a, string b) {
int d=0,an=0;
for(int i=0; i<sz(a); i++) {
if(d) {
if(a[i]==b[i])
d=0;
} else {
if(a[i]!=b[i]) {
d=1;
an++;
}
}
}
return an;
}
void solve() {
string a=readStringLn(1,100000);
string b=readStringLn(sz(a),sz(a));
int n=sz(a);
sum_n+=n;
assert(sum_n<=100000);
string pa,pb,ppa,ppb;
for(int i=0; i<n; i++) {
if(i&1) {
pa+=a[i];
pb+=b[i];
} else {
ppa+=a[i];
ppb+=b[i];
}
}
cout<<holve(pa,pb)+holve(ppa,ppb)<<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=readIntLn(1,1000);
// 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
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class FLIP{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
String A = n(), B = n();
StringBuilder[] a = new StringBuilder[]{new StringBuilder(), new StringBuilder()};
StringBuilder[] b = new StringBuilder[]{new StringBuilder(), new StringBuilder()};
int N = A.length();
hold(A.length() == B.length());
for(int i = 0; i< N; i++){
a[i%2].append(A.charAt(i));
b[i%2].append(B.charAt(i));
}
pn(calc(a[0].toString(), b[0].toString())+calc(a[1].toString(), b[1].toString()));
}
int calc(String A, String B) throws Exception{
hold(A.length() == B.length());
int flip = 0, ans = 0;
for(int i = 0; i< A.length(); i++){
int d = (A.charAt(i)-'0')^(B.charAt(i)-'0');
if(d != flip){
ans++;
flip ^= 1;
}
}
return (ans+1)/2;
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new FLIP().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
VIDEO EDITORIAL (Hindi):
VIDEO EDITORIAL (English):
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