# BSXOR-Editorial

Setter: Jeevan Jyot Singh
Tester: Nishank Suresh, Satyam
Editorialist: Devendra Singh

Simple

# PREREQUISITES:

bitwise XOR operation

# PROBLEM:

Chef has a binary string S of length N. He wonders if it is possible to divide S into exactly K non-empty substrings such that each S_i belongs to exactly one substring and the \texttt{XOR} of each substring is the same. Can you help Chef to determine if it is possible to do so?

Note: \texttt{XOR} of substring S_{L \cdots R} is defined as: \texttt{XOR} (S_{L \cdots R}) = S_L \oplus S_{L+1} \oplus \cdots \oplus S_R.

Here, \oplus denotes the bitwise XOR operation.

# EXPLANATION:

The bitwise XOR of any binary string is either 0 or 1. Therefore if an answer exists then it is either K non-overlapping substrings having XOR equal to 1 or K non-overlapping substrings having XOR equal to 0. We can iterate over the binary string and check whether we can divide the string into K non-overlapping substrings having XOR equal to 0 or K non-overlapping substrings having XOR equal to 1
We can greedily select the first K-1 substrings having same XOR (0\: or\: 1) and check whether there XOR is equal to the substring formed by the remaining characters. If it is equal output `YES` else `NO`.

We can show that if an answer exists it can always be constructed by this greedy approach.
Suppose there exists an answer such that the first K-1 substrings are not selected greedily, let it be S[1,S_1],S[S_1+1,S_2],....................S[S_{K-1}+1,N], where S_i is the right end of the selected substring. Let there be and index S_0 < S_1 such that XOR of substring S[1,S_1] is same as XOR of substring S[1,S0], we select S[1,S0] as the first substring and append the remaining characters of substring S[1,S_1] to the start of the second string. The XOR of 1st and 2nd substrings remains same as XOR\:of\: S[1,S_0]=S[1,S_1] and XOR\:of\:S[S_0+1,S_2]=S[1,S_2]\oplus S[1,S_0] = S[S_1+1,S_2] as XOR of S[1,S_1]=S[1,S_0] Now move to the next substring and keep on repeating the same procedure until all the K-1 substrings are selected greedily i.e. we select the substring as soon as the XOR becomes equal to what we need (0\: or\: 1).

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Setter's Solution

#ifdef WTSH
#include <wtsh.h>
#else
#include <bits/stdc++.h>
using namespace std;
#define dbg(β¦)
#endif

#define int long long
#define endl β\nβ
#define sz(w) (int)(w.size())
using pii = pair<int, int>;

const long long INF = 1e18;

const int N = 1e6 + 5;

// -------------------- Input Checker Start --------------------

long long readInt(long long l, long long r, char endd)
{
long long x = 0;
int cnt = 0, fi = -1;
bool is_neg = false;
while(true)
{
char g = getchar();
if(g == β-β)
{
assert(fi == -1);
is_neg = true;
continue;
}
if(β0β <= g && g <= β9β)
{
x *= 10;
x += g - β0β;
if(cnt == 0)
fi = g - β0β;
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;
if(!(l <= x && x <= r))
{
cerr << l << β β << r << β β << x << β\nβ;
assert(false);
}
return x;
}
else
{
assert(false);
}
}
}

string readString(int l, int r, char endd)
{
string ret = ββ;
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, β '); }
long long readIntLn(long long l, long long r) { return readInt(l, r, β\nβ); }
string readStringSp(int l, int r) { return readString(l, r, β '); }
void readEOF() { assert(getchar() == EOF); }

vector readVectorInt(int n, long long l, long long r)
{
vector a(n);
for(int i = 0; i < n - 1; i++)
a[n - 1] = readIntLn(l, r);
return a;
}

// -------------------- Input Checker End --------------------

void solve()
{
bool ok = false;
for(int x: {0, 1})
{
int cur = 0, cnt = 0, i = 0;
for(; i < n and cnt < k - 1; i++)
{
cur ^= s[i] - β0β;
if(cur == x)
cnt++, cur = 0;
}
if(i == n)
continue;
for(; i < n; i++)
cur ^= s[i] - β0β;
ok |= (cur == x);
}
cout << (ok ? βYESβ : βNOβ) << endl;
}

int32_t main()
{
ios::sync_with_stdio(0);
cin.tie(0);
for(int tc = 1; tc <= T; tc++)
{
// cout << βCase #β << tc << ": ";
solve();
}
return 0;
}

Tester-1's Solution(Python)
``````for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
zeros, ones, pref = 0, 0, 0
for c in s:
if c == '1':
pref ^= 1
zeros += pref == 0
ones += pref != (ones%2)
print('YES' if (pref == 0 and zeros >= k) or (ones >= k and ones%2 == k%2) else 'NO')
``````
Tester-2's Solution
``````#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline;
#else
#define debug(x);
#endif

/*
------------------------Input Checker----------------------------------
*/

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;
}

if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}

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){
}
}
}

/*
------------------------Main code starts here----------------------------------
*/
int MAX=100000;
int check_bin(string s){
for(auto it:s){
if((it!='0')&&(it!='1')){
return 0;
}
}
return 1;
}
int sum_cases=0;
void solve(){
sum_cases+=n;
assert(check_bin(s));
int now=0;
vector<int> track; int zro=0;
for(int i=0;i<n;i++){
now+=s[i]-'0'; now=now%2;
track.push_back(now); zro+=now==0;
}
int freq=0;
if((k%2==0)&&(now)){
cout<<"NO\n";
return;
}
if(k%2==0){
if(zro>=k){
cout<<"YES\n";
return;
}
now=1;
}
else{
now=track.back();
}
int us=now;
for(int i=0;i<n;i++){
if(track[i]==now){
now^=us;
freq++;
}
}
if(freq>=k){
cout<<"YES\n";
}
else{
cout<<"NO\n";
}
return;
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("error.txt", "w", stderr);
#endif
while(test_cases--){
solve();
}
assert(getchar()==-1);
assert(sum_cases<=2*MAX);
return 0;
}

``````
Editorialist's Solution
``````#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
void sol(void)
{
int n, k;
cin >> n >> k;
string s;
cin >> s;
int x = 0, cnt = 0;
for (int i = 0; i < n; i++)
{
if (s[i] == '1')
x ^= 1;
if (x == 1 && cnt<k-1 && i!=n-1)
cnt++, x = 0;
}
if (cnt == k-1 && x==1)
{
cout << "YES\n";
return;
}
x = cnt = 0;
for (int i = 0; i < n; i++)
{
if (s[i] == '1')
x ^= 1;
if (x == 0 && cnt<k-1 && i!=n-1)
cnt++;
}
if (cnt == k-1 && x==0)
{
cout << "YES\n";
return;
}
else
cout << "NO\n";
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}

``````
1 Like

i have a doubt in test case of 111, why no soln? canβt we split it as [1, 2] and [3,3] substring? as of the first substring 1 ^ 1 = 0 and for second also 1 ^ 1 = 0 , what i have missed??

[1, 2] and [3, 3] will be 11 and 1.
The second sub-string have only one character, so for first sub-string 1 ^ 1 = 0, and for second sub-string 1 = 1.
Thereβs no way to split 111 into two sub-strings with same xor.

3 Likes

Oh, i was confused whether we have to take single element xor with itself, thanks got it

Could anyone point out the three test cases I missed?

https://www.codechef.com/viewsolution/62820376

1 Like

Can someone tell me what is wrong with this approach ?
https://www.codechef.com/viewsolution/62874598

Can I know tasks 1 and 2 in subtask 1 to know where my solution went wrong

https://www.codechef.com/viewsolution/62812907
Any idea why this solution is giving WA ?

Hey @parzival18s
there is the problem of index of bound in else as when w becomes n in if condition it will point null value.

1 Like

I see. That was a terrible blunder on my side. Thanks for pointing it out.

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