How can I find number of consecutive sequences of various lengths satisfy a particular property?

Question

I am given a array A[] having N elements which are positive integers .I have to find the number of sequences of lengths 1,2,3,..,N that satisfy a particular property? I have built an interval tree with O(nlogn) complexity.Now I want to count the number of sequences that satisfy a certain property ?

All the properties required for the problem are related to sum of the sequences

Note an array will have N*(N+1)/2 sequences. How can I iterate over all of them in O(nlogn) or O(n) ?

Answer 1

If we let k be the moving index from 0 to N(elements), we will run an algorithm that is essentially looking for the MIN R that satisfies the condition (lets say I), then every other subset for L = k also is satisfied for R >= I (this is your short circuit). After you find I, simply return an output for (L=k, R>=I). This of course assumes that all numerics in your set are >= 0.

To find I, for every k, begin at element k + (Nk)/2. Figure out if this defined subset from (L=k, R=k+(Nk)/2) satisfies your condition. If it does, then decrement R until your condition is NOT met, then R=1 is your MIN (your could choose to print these results as you go, but they results in these cases would be essentially printed backwards). If (L=k, R=k+(Nk)/2) does not satisfy your condition, then INCREMENT R until it does, and this becomes your MIN for that L=k. This degrades your search space for each L=k by a factor of 2. As k increases and approaches N, your search space continuously decreases.

// This declaration wont work unless N is either a constant or MACRO defined above
unsigned int myVals[N];
unsigned int Ndiv2 = N / 2;
unsigned int R;

for(unsigned int k; k < N; k++){

    if(TRUE == TESTVALS(myVals, k, Ndiv2)){ // It Passes
        for(I = NDiv2; I>=k; I--){
            if(FALSE == TESTVALS(myVals, k, I)){
                I++;
                break;
            } 
        }
    }else{                                  // It Didnt Pass
        for(I = NDiv2; I>=k; I++){
            if(TRUE == TESTVALS(myVals, k, I)){
                break;
            } 
        }
    }

    // PRINT ALL PAIRS from L=k, from R=I to R=N-1


    if((k & 0x00000001) == 0) Ndiv2++;

} // END --> for(unsigned int k; k < N; k++)

The complexity of the algorithm above is O(N^2). This is because for each k in N(ie N iterations / tests) there is no greater than N/2 values for each that need testing. Big O notation isnt concerned about the N/2 nor the fact that truly N gets smaller as k grows, it is concerned with really only the gross magnitude. Thus it would say N tests for every N values thus O(N^2)

There is an Alternative approach which would be FASTER. That approach would be to whenever you wish to move within the secondary (inner) for loops, you could perform a move have the distance algorithm. This would get you to your O(nlogn) set of steps. For each k in N (which would all have to be tested), you run this half distance approach to find your MIN R value in logN time. As an example, lets say you have a 1000 element array. when k = 0, we essentially begin the search for MIN R at index 500. If the test passes, instead of linearly moving downward from 500 to 0, we test 250. Lets say the actual MIN R for k = 0 is 300. Then the tests to find MIN R would look as follows:

R=500 R=250 R=375 R=312 R=280 R=296 R=304 R=300

While this is oversimplified, your are most likely going to have to optimize, and test 301 as well 299 to make sure youre in the sweet spot. Another not is to be careful when dividing by 2 when you have to move in the same direction more than once in a row.

Answer 2

@user1907531: First of all , if you are participating in an online contest of such importance at national level , you should refrain from doing this cheap tricks and methodologies to get ahead of other deserving guys. Second, a cheater like you is always a cheater but all this hampers the hard work of those who have put in making the questions and the competitors who are unlike you. Thirdly, if @trumetlicks asks you why haven't you tagged the ques as homework , you tell another lie there.And finally, I don't know how could so many people answer this question this cheater asked without knowing the origin/website/source of this question. This surely can't be given by a teacher for homework in any Indian school. To tell everyone this cheater has asked you the complete solution of a running collegiate contest in India 6 hours before the contest ended and he has surely got a lot of direct helps and top of that invited 100's others to cheat from the answers given here. So, good luck to all these cheaters .