vejon
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- vejon in United StatesGiven a number n, make a set of m integers such that the integers lie between [1, n] and the sum of two integers should produce an integer which is in the set. Eg. **Input** n=20, m=5 **Output** [3, 6, 11, 14, 17]
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ThoughtWorks Applications Developer - 1
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AnswerGiven 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.
- vejon in United States
"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 position was already '1')
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:
5 7
00000
1
2 3
1
2 5
1
2 4
1
Example Output:
0
1
1
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Google Software Developer
#include<bits/stdc++.h>
using namespace std;
int a[1000<<2];
int main()
{
long long int i,j,k,m,n;
cin>>n;
for(i=0;i<n;i++)
{
cin>>a[i];
}
for(i=0;i<n;i=i+2)
{
if(i==0)
{
if(a[i]<a[i+1])
{
swap(a[i],a[i+1]);
}
}
if(a[i]<a[i+1])
{
swap(a[i],a[i+1]);
}
if(a[i]<a[i-1])
{
swap(a[i],a[i-1]);
}
}
for(i=0;i<n;i++)
{
cout<<a[i]<<" ";
}
}
#include<bits/stdc++.h>
using namespace std;
int a[1000<<2];
int main()
{
long long int i,j,k,m,n;
cin>>n;
for(i=0;i<n;i++)
{
cin>>a[i];
}
for(i=0;i<n;i=i+2)
{
if(i==0)
{
if(a[i]<a[i+1])
{
swap(a[i],a[i+1]);
}
}
if(a[i]<a[i+1])
{
swap(a[i],a[i+1]);
}
if(a[i]<a[i-1])
{
swap(a[i],a[i-1]);
}
}
for(i=0;i<n;i++)
{
cout<<a[i]<<" ";
}
}
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seat should have screen id too
- vejon June 04, 2019