 Solution

Maintain one array index per input and a candidate value. Cycle through the arrays, counting consecutive agreements with that candidate; when a larger head is encountered, start counting agreements again. p is an auxiliary array of $m$ indices, not a copy of any input.The loop invariant has three parts. All indices satisfy $0\le p[t]\le n[t]$; no discarded entry is common to every input; and every original common value is at least candidate. Additionally, the agree cyclically consecutive arrays immediately preceding index j have heads equal to candidate. These facts are true initially, with an empty agreement run.

A head below candidate is not globally common, by the lower-bound clause, so advancing its pointer is safe. A head above candidate gives a new lower bound: every globally common value lies in this array's unprocessed suffix and so is at least its head. Only this array is yet certified to agree with the new candidate. An equal head extends the run by one. An exhausted suffix proves absence; a run of $m$ agreements covers every array exactly once and proves that candidate is common, and the lower-bound clause proves minimality. The exhaustion test precedes each dereference, including after the last entry of a suffix is discarded.

For algorithm termination and time complexity, charge comparisons to the pair (j,p[j]). A fixed entry is discarded at most once, can raise the candidate at most once, and can be counted as an agreement at most once: before the same array is revisited, either the algorithm has finished or the candidate has strictly increased. Thus there are at most three comparisons per input entry, besides the $O(m)$ initialization. The result is the smallest common value or None in $O(m+{\sum }_{t}n[t])$ time and $O(m)$ auxiliary space. For $m=1$, the first head immediately gives the answer.