Approach #1: Interval Stabbing [Accepted]


Say N = 10 and A[2] = 5. Then there are 5 rotations that are bad for this number: rotation indexes 0, 1, 2, 8, 9 - these rotations will cause this number to not get 1 point later.

In general, for each number in the array, we can map out what rotation indexes will be bad for this number. It will always be a region of one interval, possibly two if the interval wraps around (eg. 8, 9, 0, 1, 2 wraps around, to become [8, 9] and [0, 1, 2].)

At the end of plotting these intervals, we need to know which rotation index has the least intervals overlapping it - this one is the answer.


First, an element like A[2] = 5 will not get score in (up to) 5 posiitons: when the 5 is at final index 0, 1, 2, 3, or 4. When we shift by 2, we'll get final index 0. If we shift 5-1 = 4 before this, this element will end up at final index 4. In general (modulo N), a shift of i - A[i] + 1 to i will be the rotation indexes that will make A[i] not score a point.

If we are trying to plot an interval like [2, 3, 4], then instead of doing bad[2]--; bad[3]--; bad[4]--;, what we will do instead is keep track of the cumulative total: bad[2]--; bad[5]++. For "wrap-around" intervals like [8, 9, 0, 1, 2], we will keep track of this as two separate intervals: bad[8]--, bad[10]++, bad[0]--, bad[3]++. (Actually, because of our implementation, we don't need to remember the bad[10]++ part.)

At the end, we want to find a rotation index with the least intervals overlapping. We'll maintain a cumulative total cur representing how many intervals are currently overlapping our current rotation index, then update it as we step through each rotation index.

Complexity Analysis

  • Time Complexity: , where is the length of A.

  • Space Complexity: .

Analysis written by: @awice.