Can't shake the feeling that other solutions and explanations are just too complicated.

Observations: - This is a que of people, normally always starts 1,2,3,4... - Since person can skip at most 2 positions, it means that if we look at que segment of size 3, we known that for arbirary position i only i, i+1 and i+2 can be there - So for initial segment [p1,p2,p3] at position p1 can only be 1 (the original), 2 (if it bribed) or 3 (if it bribed twice)

So if normally the queue was supposed to start with [p1, p2, p3] = [1, 2, 3] , then by scanning the queue from left to right (via var x) and keeping track of p1, p2, p3 we can encounter the following cases

1) x == p1 // x is at it's place and did not bribe anyone
p1, p2, p3 = p2, p3, p3+1 // shift the window to the right

2) x == p2 // x bribed once and skipepd 1 ahead bribes += 1 p2, p3 = p3, p3+1 // expected p1 is the same, p2 and p3 move to the right

3) x == p3 // x bribed twice and skipped 2 ahead bribes += 2 p3 = p3+1 // we are still expecting the same p1 and p2, p3 moves to the right

4) x is outside the window thus it's chaos

Code in golang

func minimumBribes(q []int32) {
	// Write your code here
	var bribes int32
	// window of size 3
	// starts by expecting the right order
	var p1, p2, p3 int32 = 1, 2, 3

	for _, x := range q {
		switch {
		// x = 1;  1 2 3 -> 2 3 4
		case x == p1:
			p1, p2, p3 = p2, p3, p3+1
		// x = 2; 2 1 3 -> 1 3 4
		case x == p2:
			bribes += 1
			p2, p3 = p3, p3+1
		// x = 3; 3 1 2 -> 1 2 4
		case x == p3:
			bribes += 2
			p3 = p3 + 1
		default:
			fmt.Println("Too chaotic")
			return
		}
	}

	fmt.Println(bribes)
}