We can naively simulate the insertion for each pair, adding the left character and the new character. For example, the rules NN -> C, NC -> B for NNC would give NC + NB (+ C) = NCNBC.
Instead of storing the entire string, you can instead count the number of times each pair occurs. Following the above example, the next step for NNC would increase the count for the pairs NC, CN, NB, BC. Afterward, we can sum the leftmost character of each pair and the rightmost character of the initial string to get the total times each letter occurs.
Similar idea to Day 6, where we optimized the code for part 2 by storing a different state.