dfs that takes the current indices i and j of s1 and s2, respectively, and the current difference diff as arguments.i and j have reached the end of their respective strings and diff is zero, return true.i, j, diff) is already computed, return the stored result.i points to a digit in s1, iterate through all possible lengths that the numeric substring starting at i could represent, and recursively call dfs with the new index and adjusted diff.j points to a digit in s2, iterate through all possible lengths that the numeric substring starting at j could represent, and recursively call dfs with the new index and adjusted diff.diff is zero and both i and j point to letters, compare the letters. If they match, recursively call dfs with the next indices.diff is positive, recursively call dfs with the next index of s1 and diff decremented by one.diff is negative, recursively call dfs with the next index of s2 and diff incremented by one.dfs starting with indices 0, 0, and diff 0.