My solution: http://www.codechef.com/viewsolution/3769202

I followed the ‘maintaining count of each digit’ logic as in the editorial yet I get a TLE for the above solution. What could I be missing out?

you need to store number of occurrence of digit 0-9 at every index…

And the given X is the index not the value at that index.

check My solution

you should not compute from scratch for every query you need to store the values of your previous computation and use it for next computation

But it isn’t given that the queries are in ascending order. So how will use previous computation for next one?

for(i=1;i<=n;++i)

{

for(j=0;j<10;++j)

{

count_till_now[i][j]=count_till_now[i-1][j];

}

count_till_now[i][given[i]]++;

}

@letsctheworld 2d array is not necessary.you just need to have a single counter[10].@sandy999 firstly do the computation before evaluating the queries. 1,2,…n and store the values at each index in some array(say answer[n]).and after that you can answer queries directly through answer array.

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