Reading only 'm' entries in a 'n x m' matrix

Question

Given two dimensional k x ((26^k)-1) array, containing every possible string of capital letters of length k , except one of them. How can we know the missing string while reading only theta(26^k) entries of the array, and not theta(kx (26^k)) entries?

We have thought of using 'k' pointers for all the 26^k columns, but it will still be the same as looking kx 26^k entries, we also considered checking for a[:0], a[:1], a[:2], . . . a[:(26^k)-1] but it is still the same as looking for kx 26^k values as slicing also counts as looking at those entries.

Answer 1

Here's one of the ways to do it in Θ(26^k) time complexity by using an additional O(26^k) space.

Let's look at the intuition first -

1. Each row of the matrix would have (26^k)-1 entries. In the first row, all letters would have the same frequency except one whose frequency would be one lesser than the other 25. This letter would be the first letter of our missing string.
2. Now we already know the first letter of our missing string, so for the second row, we don't need to look at all the entries and instead, we can look at the 25*26^(k-1) cols for which the corresponding first-row value was the missing letter.(but now in order to know which cols to look into we would need to store those cols somewhere else using additional space). And we would only have to look into 26^(k-1) - 1 entries in the second row.
3. For a given row, we can count the frequency of letters in given entries in the O(no. of entries) and constant space(26 letters). Once we know the missing letter in a given row, we can go through the row again and store indices of all the cols for which the letter is the missing letter and store those indices in a separate array using additional space O(26^(k-1-i)) for ith row.
4. On subsequent rows, we only visit the cols which are maintained in our previous row's array which contains indices of cols for which the letter was the missing letter. eg for the 3rd row, we can do away with looking at just 26^(k-2) - 1 entries for indices which we stored in our array when we looked at 2nd row.

So the total no. of entries we would have to look at in each row(after summing up the geometric series) would be -

sum{(26 - 1), (26^2 - 1), (26^3 - 1) ... (26^k -1)} = (26*(1 - (26^k))/(1-26) - k

which is Θ(26^k) time and space complexity.