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Given a string of size n consisting of 0s and/or 1s.you have to perform k queries and there are two types of queries possible.

"1"(without quotes): Print length of the longest substring with all '1'.
"2 X"(without quotes): where X is an Integer between 1 to n.In this query, you will
change character at Xth position to '1' (it is possible that the character at ith

Input Format:
* First Line of input contains n and k, where n is string length and k is the number of queries.
* Next line contains a string of 0's and/or 1's of length n.
* Each of next k lines contains query of any one type (i.e 1 or 2).

Output Format:
For each query of type 1, print in new line the maximum size of subarray with all 1's.

```Example Input:                        Example Output:
5 7                                   0
00000                                 1
1                                     1
2 3                                   3
1
2 5
1
2 4
1```

What I have tried:

My Solution: O(k*n) (if most of the queries of type 1):

```if(type==1){
int count=0, ans=0;
for(int i=0;i<str.size();i++){  //longest len substring
if(str[i]=='1'){
count++;
}else{
ans=max(ans,count);
count=0;
}
}

printf("%d\n",ans);
}else{
int xth;
scanf("%d",&xth);
str[xth-1]='1';   //update
}```

I am not able to figure out an efficient solution, as for 'type 1' query only solution I could think of is to iterate through string every time and maintain a "count" variable with all 1's consecutively and finally update "ans" variable when ith str becomes '0'.

I am thinking of using segment tree but don't know how to do. As required good solution should be O(k*log(n)) (doing "type1" query in log(n) time)
Posted
PIEBALDconsult 2-Dec-17 10:34am
Define and implement a data structure that will maintain the information you need.
I would imagine a form of lookup table -- indexed by character (0,1) -- of maps that contain a start index and a length.