An array A consists of n integers in locations A[0], A[1] ....A[n-1]. It is…

Concept: This is the in-place "cycle-following" (juggling) algorithm for a cyclic left rotation by k, done in O(n) time with no auxiliary array. Under the map j -> (j+k) mod n, the n index positions split into gcd(n,k) disjoint cycles. Each cycle can be resolved by walking it once starting from any unvisited index: save the value at the start in temp, keep overwriting A[j] with the value k slots ahead until the walk returns to the start, then drop temp into the last freed slot. A running pointer (min) is used to know when every index has been swept into some cycle, so the outer loop can keep launching a fresh cycle from the next unvisited start.

Application to this fragment:

1. Outer loop condition: i starts at 0 and min starts at n (out of range), and the outer loop must keep starting new cycles until every index has been visited, i.e. until i catches up with min. That test is i < min, which fixes the first blank and rules out any option built on i > min (that condition is false from the very first check, so the loop would never run at all).

2. Inner loop condition: inside a cycle we walk j -> (j+k) mod n starting at j = i. The index right before the walk wraps back to i is the predecessor of i under this map, namely (n + i - k) mod n. So the inner loop must keep going while j != (n + i - k) mod n. This rules out the option whose inner test is built on (n + i) mod n: that expression simplifies to i, and since j is set equal to i on the previous line, the test is false immediately, so that inner loop body would never execute even once.

3. Assignment inside the loop: each element must be moved back k slots with the index wrapped at the array boundary, i.e. A[j] = A[(j + k) mod n]. Dropping the mod n (writing A[j + k] or A[(j + k)] without the wrap) lets the index run past n-1 once j + k reaches or exceeds n, which is out of bounds rather than a rotation.

4. After the inner loop ends, the one value that was saved in temp before the cycle got overwritten must be written into the last freed slot, which is exactly A[(n + i - k) mod n] = temp.

5. Finally i = i + 1 advances the outer loop, and here is precisely why that is safe. Inside a cycle starting at index i, the code compares j against min only AFTER advancing it to (j+k) mod n - the cycle's own starting index i is never itself compared. For the very first cycle (i = 0), the values that DO get compared are every other member of index 0's residue class modulo g = gcd(n,k), namely g, 2g, ..., n-g; the smallest of these is g, so min becomes exactly g by the time that cycle finishes. Every later cycle (starting at i = 1, 2, ...) walks a different residue class mod g, and apart from that cycle's own start (again never compared), every member of it is at least i + g > g, so none of them can pull min below g. min therefore settles at exactly gcd(n,k), and the outer test i < min lets i run through 0, 1, ..., gcd(n,k)-1 - one fresh cycle per value, since these are exactly the residue classes' representatives - before stopping exactly at i = gcd(n,k), by which point every index has been covered exactly once.

Cross-check: take n = 5, k = 2 (gcd(5,2) = 1, so all 5 indices form one cycle). Starting at i = 0, j walks 0 -> 2 -> 4 -> 1 -> 3 -> back to 0 under j = (j+k) mod n. The inner-loop stop index is (n+i-k) mod n = (5+0-2) mod 5 = 3, so the walk executes for j = 0, 2, 4, 1 (four moves for a 5-element cycle) and then A[3] = temp closes it. Since gcd(5,2) = 1, the outer loop runs only once (i = 0), then min ends at 1 and the outer test 1 < 1 is false. A second cross-check with n = 6, k = 4 (gcd = 2) confirms the general rule: the outer loop runs for i = 0 and i = 1 (2 = gcd(6,4) cycles), covering all 6 indices exactly once with none left out or double-counted.

Conclusion: the fragment i < min; j != (n+i-k) mod n; A[(j+k) mod n]; temp; i + 1 is the only one consistent with the cycle-following algorithm on every count above.