Regarding linear time implementation of shakutori

I introduced this problem before: https://pekempey.hatenablog.com/entry/2018/09/18/085432

This technique can be applied to shakutori (also known as two pointers or sliding window). Its implementation is a little complicated but it might be efficient.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cassert>

using namespace std;

int gcd(int x, int y) {
  return y == 0 ? x : gcd(y, x % y);
}

int f(vector<int> a) {
  const int n = a.size();
  int res = 0;
  for (int i = 0; i < n; i++) {
    int g = 0;
    for (int j = i; j < n; j++) {
      g = gcd(g, a[j]);
      if (g != 1) {
        res = max(res, j - i + 1);
      }
    }
  }
  return res;
}

int g(vector<int> a) {
  const int n = a.size();
  vector<pair<int, int>> L, R;
  L.emplace_back(0, 0);
  R.emplace_back(0, 0);
  int j = 0;
  int res = 0;
  for (int i = 0; i < n; i++) {
    while (j < n && gcd(a[j], gcd(L.back().first, R.back().first)) != 1) {
      L.emplace_back(gcd(L.back().first, a[j]), a[j]);
      j++;
    }
    res = max(res, j - i);
    if (i == j) j++;
    else {
      if (R.size() == 1) {
        while (L.size() >= 2) {
          R.emplace_back(gcd(R.back().first, L.back().second), L.back().second);
          L.pop_back();
        }
      }
      R.pop_back();
    }
  }
  return res;
}

int main() {
  assert(f({1,2,1,2,6,1}) == 2);
  assert(f({1,10,1}) == 1);
  assert(f({1,1,1}) == 0);
  assert(g({1,2,1,2,6,1}) == 2);
  assert(g({1,10,1}) == 1);
  assert(g({1,1,1}) == 0);
}

This technique is described in [1] : http://hirzels.com/martin/papers/debs17-tutorial.pdf

[1] M. Hirzel, S. Schneider, K. Tangwongsan. Sliding-Window Aggregation Algorithms: Tutorial, DEBS'17 Proceedings of the 11th ACM International Conference on Distributed and Event-based Systems, pp.11--14, 2017.