BDGFT - Editorial



Contest: Division 1

Contest: Division 2

Setter: Anik Sarker, Ezio Auditore

Tester: Teja Vardhan Reddy

Editorialist: Taranpreet Singh



Prefix Sums, Sliding Window, brute-force, amortization.


Given a string S of length N consisting of characters ‘0’ and ‘1’ only. Find the number of non-empty substrings of S such that cnt_0 = cnt_1*cnt_1 where cnt_c denote the frequency of character c in substring.


Let us assume cnt_1 = x for some substring. Then cnt_0 = x^2, hence the length of substring should be cnt_0+cnt_1 = x^2+x since we cannot have any other character in substring. Hence, we can surely reject all substrings whose length cannot be written as x^2+x for some integer x.

Now, we can see, the maximum value of x such that x^2+x \leq N is around \sqrt{N}. So, what we can do is to try all possible values of x and for each x, consider all substrings of S with length x^2+x and out of these, count the number of substrings which contain exactly x occurrences of character ‘1’. We can see, that this substring shall contain exactly x^2+x-x = x^2 occurrences of character ‘0’ then, which satisfies our requirement.

Code Snippet
for(int x = 1; x*x+x <= N; x++)
//we need to consider all substrings of length (x^2+x) and count the substrings, which have x occurrences of 1

For doing this, we have two ways, sliding window, and prefix-sums. In the sliding window, we can make a counter variable counting the number of occurrences of character ‘1’ in the current range. Both of these allows us to consider all substrings of a specific length in O(N) time.

It can be seen, that we need to consider \sqrt{N} different lengths of substrings and considering substrings of each length takes O(N) time, we have solved this problem in O(N*\sqrt{N}) time.


The time complexity is O(N*\sqrt{N}) per test case.


Setter 1 Solution
using namespace std;
#define MAX 100005
int C[MAX];

int main(){
	int t;
	assert(1 <= t && t<= 10);

	for(int cs=1;cs<=t;cs++){
	    string s;

	    int n = s.size();
	    assert(1 <= n && n<= 100000);
	    for(int i=1;i<=n;i++){
	        assert(s[i-1] == '0' || s[i-1] == '1');
	        C[i] = C[i-1] + s[i-1] - '0';

	    int Count = 0;
	    for(int i=0;i<n;i++){
	        int x = 1;
	            int j = i + x*x + x;
	            if(j > n) break;

	            int One = C[j] - C[i];
	            if(One == x) Count++;
Setter 2 Solution
using namespace std;
typedef long long ll;
const int maxn = 100009;
int n, m, x, y, z;
string str;
int csum[maxn];
int main()
//    freopen("input05.txt", "r", stdin);
//    freopen("output05.txt", "w", stdout);

	int t, cs = 1;
	cin >> t;
	if(t < 1 || t > 10) assert(false);


	    cin >> str;
	    memset(csum, 0, sizeof(csum));
	    if(str.size() > 100000) assert(false);
	    int n = str.size();
	    str = " " + str;

	    for(int i = 1; i <= n; i++){
	        if(str[i] == '0') csum[i] = csum[i - 1];
	        else if(str[i] == '1') csum[i] = csum[i - 1] + 1;
	    int ans = 0;

	    for(int i = 1; ; i++){

	        int len = i * i + i;

	        if(len > n) break;

	        for(int j = 1; j <= n - len + 1; j++){

	            long long cnt1 = csum[j + len - 1] - csum[j - 1];
	            long long cnt0 = len - cnt1;

	            if(cnt0 == cnt1 * cnt1) ans++;



	    printf("%d\n", ans);

	return 0;
Tester's Solution
#include <bits/stdc++.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <utility>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <iomanip>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp> 
//setbase - cout << setbase (16); cout << 100 << endl; Prints 64
//setfill -   cout << setfill ('x') << setw (5); cout << 77 << endl; prints xxx77
//setprecision - cout << setprecision (14) << f << endl; Prints x.xxxx
//cout.precision(x)  cout<<fixed<<val;  // prints x digits after decimal in val

using namespace std;
using namespace __gnu_pbds;

#define f(i,a,b) for(i=a;i<b;i++)
#define rep(i,n) f(i,0,n)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define vi vector< int >
#define vl vector< ll >
#define ss second
#define ff first
#define ll long long
#define pii pair< int,int >
#define pll pair< ll,ll >
#define sz(a) a.size()
#define inf (1000*1000*1000+5)
#define all(a) a.begin(),a.end()
#define tri pair<int,pii>
#define vii vector<pii>
#define vll vector<pll>
#define viii vector<tri>
#define mod (1000*1000*1000+7)
#define pqueue priority_queue< int >
#define pdqueue priority_queue< int,vi ,greater< int > >
#define flush fflush(stdout) 
#define primeDEN 727999983
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

// find_by_order()  // order_of_key
typedef tree<

#define int ll
int a[123456];
	std::ios::sync_with_stdio(false); cin.tie(NULL);
	int t;
		int i;
		string s;
		int cnt=0;
		int ans=0;
		//map<int,int> mapi;
		int n=s.length(),j;
		//int ans=0;
	return 0;   
Editorialist's Solution
import java.util.*;
import java.text.*;
class BDGFT{
	//Into the Hardware Mode
	void pre() throws Exception{}
	void solve(int TC) throws Exception{
	    String s = n();int n = s.length();
	    int[] pre = new int[1+n];
	    for(int i = 1; i<= n; i++){
	        pre[i] = pre[i-1];
	        if(s.charAt(i-1) == '1')pre[i]++;
	    long ans = 0;
	    for(int i = 1; i<= n; i++)
	        //Iterating over all values of cnt1 till it exceed length of substring S[i, n]
	        for(int cnt1 = 1; i-1+cnt1+cnt1*cnt1 <= n; cnt1++)
	            //Checking if substring of length cnt1+cnt1*cnt1 has cnt1 ones.
	            if(pre[i-1+cnt1+cnt1*cnt1]-pre[i-1] == cnt1)
	void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
	long IINF = (long)1e18, mod = (long)1e9+7;
	final int INF = (int)1e9, MX = (int)2e5+5;
	DecimalFormat df = new DecimalFormat("0.00000000000");
	double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-6;
	static boolean multipleTC = true, memory = false, fileIO = false;
	FastReader in;PrintWriter out;
	void run() throws Exception{
	        in = new FastReader("input.txt");
	        out = new PrintWriter("output.txt");
	    }else {
	        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);
	public static void main(String[] args) throws Exception{
	    if(memory)new Thread(null, new Runnable() {public void run(){try{new BDGFT().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
	    else new BDGFT().run();
	long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
	int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
	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;}
	String nln()throws Exception{return in.nextLine();}
	int ni()throws Exception{return Integer.parseInt(;}
	long nl()throws Exception{return Long.parseLong(;}
	double nd()throws Exception{return Double.parseDouble(;}
	class FastReader{
	    BufferedReader br;
	    StringTokenizer st;
	    public FastReader(){
	        br = new BufferedReader(new InputStreamReader(;
	    public FastReader(String s) throws Exception{
	        br = new BufferedReader(new FileReader(s));
	    String next() throws Exception{
	        while (st == null || !st.hasMoreElements()){
	                st = new StringTokenizer(br.readLine());
	            }catch (IOException  e){
	                throw new Exception(e.toString());
	        return st.nextToken();
	    String nextLine() throws Exception{
	        String str = "";
	            str = br.readLine();
	        }catch (IOException e){
	            throw new Exception(e.toString());
	        return str;

Feel free to share your approach, if you want to. (even if its same :stuck_out_tongue: ) . Suggestions are welcomed as always had been. :slight_smile:


My solution using same approach as the editorial…

1 Like

Can anyone solve this question using sparse table…Plzx?

mine o(tnsqrt(n)) in python is not working, it;s giving tle, will you check my code and approach ? it’s different from what u guys have done.

@macdod try submitting in pypy

Kindly tell the error in my code.My code is similar to TESTER’S code.


same algorithm implemented ( prefix table )
but giving t.l.e in python .
whats wrong with python if same algo gives tle in python and ac in c++
getting accepted in .2 secs in c++
getting tle in pypy and the only soln getting accepted takes 0.98secs.
Why dont u reset the time limit for python languages


still timeout, this is my code, plz check what am i doing wrong there.

any idea how sparse table is used in this question

amazing formula for calculating the size of window

1 Like

I have not read your whole code but from line 20 to 22, each time you add an element to the dictionary and copy the whole dictionary to the list k or is it doing something else. If the first one i.e. copying the whole dictionary then I think that is where you are doing wrong(as it will be O(n^2). Please correct if wrong!

Is there any way to solve this problem in o(n) or o(nlog(n))?

my ans is not showing any output
import java.util.;
import java.lang.*;

class cd125

public static void main(String args[])throws IOException
	int i,j;
    BufferedReader br = new BufferedReader (new InputStreamReader(;
	int T = Integer.parseInt(br.readLine());

		String a  = br.readLine();
		int count0 = 0;
		int count1 = 0;
		int dp[] = new int[a.length()];
		int ans = -1;
			dp[i] = 0;
				dp[i] = dp[i-1]+1;
			ans =  Math.max(dp[i],dp[i-1]);
	}catch(Exception e)



your for loop is starting from i=0,which gives runtime error when you access the index
“dp[i-1]” because “0-1” is index out of bounds which is wrong.
hope this helps :slight_smile:

I used the same approach yesterday but in python. I tried to submit my code in Python 3 , PYPY3 and finally in CPYTHON(2.7) but all gave TLE.
Link to Python Solutions:,bindalmukul99

Today I wrote the same code in c++ and it worked (to my surprise in 0.13 sec)
Link to C++ solution:

Can anybody explain what was that? Any help would be appreciated.


I have solved the problem using sliding Window but I am getting a TLE
Here is the link to my Solution
link to Solution


i cannot understand the approach can any body tell me??

I would be helpful if you can explain the conditions in your second loop

I am surprised here :frowning_face:
Let me see if anything is wrong. This is what it might happen if author keep TL tight. People complains that wrong solutions are passing if he keeps it lenient. But I am more sad when I see this type of solution not passing ( if I am not wrong that this is correct approach) .
Here is your solution with AC. Just did a small change. ( removed multiplication since its a costly operation).
You should feel like “lesson learnt, don’t use multiplication unless necessary in such ques”.


Thanks for the help and a very well lesson learned indeed
But the TL was very tight
Also in the editorial it has been mentioned that the problem can be solved either by using sliding window or by prefix table but none of the solutions either setter’s or tester’s is using sliding window otherwise that would have helped.