Extract a certain sequence of numbers from a vector

Question

I am trying to find a way to extract from a series of numbers the longest sequence that meets the condition :each number must be the prefix of the number that follows it.

Ex: For the series : 523,742,7421,12,123,1234,87 it should display 12,123,1234 .

I thought about storing the series in a vector and then iterate it and move the numbers that meet the condition in another vector. However, I got stuck at choosing the longest sequence ( 12,123,1234 instead of 742, 7421 in the above example).Here is the code I wrote so far:

       bool prefix(int a, int b){
           if ((b / 10 - ((b % 10)) / 10) == a)
           return 1;
           else return 0;
          }

       vector<int> choose_sequence(vector<int> &series){
          vector<int> right_sequence;
           int count = 0;
              for (int i = 0; i < series.size();){
                for (int j = i + 1; j < series.size();){
                   if (prefix(series.at(i), series.at(j))){
                  right_sequence.push_back(series.at(i));
                  right_sequence.push_back(series.at(j));
                        i=j;
                        j++;

                   }
                 else 
                 i++;
            }
              }
          return right_sequence;
       }

Any suggestion or correction is welcomed and most appreciate.Also,if you know a better way to do this using another data type than vectors,please share.

Answer 1

I think the classic way of solving this would be to create a second vector that will contain, for each position, the length of the sequence that ends at that position and the position of the previous element in the sequence (-1 if there is no previous element). For your example, the vectors would be something like: count: 1, 1, 2, 1, 2, 3, 1 prev: -1, -1, 1, -1, 3, 4, -1

You will chose the maximum value from count and determine the sequence using prev.

So, our maximum is 3, the position is 5 and the corresponding element is 1234. The previous element is at position 4 in the original vector and the value is 123. The previous element is at position 3 and the value is 12. The previous element is at position -1 which means there is no previous element so we have our sequence: 12, 123, 1234.

Another approach would be trying to set the length of the sequence that starts at a certain position and the position of the next element. You would have: count: 1, 2, 1, 3, 2, 1, 1 next: -1, 2, -1, 4, 5, -1, -1

Our maximum is again 3, but now we have the first element of the sequence (position 3, value 12). The next element is on position 4 (the value is 123). The next element is on position 5 (the value is 1234). The next element is in position -1 which means we reached the end of the sequence.

The advantage of this approach is that the sequence is generated in the "correct" order.

Answer 2

You can do this without any additional memory and only for one traversal of the input vector.

std::vector<int> longestSequence(const std::vector<int>& numbers)
{
    std::vector<int> result;
    if (numbers.empty())
        return result;
    size_t longestStart = 0, longestLength = 0;
    size_t start = 0;
    for (size_t i = 1, imax = numbers.size(); i < imax; ++i) {
        if (numbers[i] / 10 != numbers[i - 1]) {
            if (i - start > longestLength) {
                longestStart = start;
                longestLength = i - start;
            }
            start = i;
        }
    }
    if (numbers.size() - start > longestLength) {
        longestStart = start;
        longestLength = numbers.size() - start;
    }
    result.assign(begin(numbers) + longestStart, begin(numbers) + longestStart + longestLength);
    return result;
}

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