function minimumBribes(q: number[]): void {
    /**
     * Idea: Reverse thinking, complete the numbers behind first, count the number 
     * of permutations of the argument.
     */

    // 0. fn that swap number in array
    const swap = (s: number[], i: number) => { [s[i], s[i + 1]] = [s[i + 1], s[i]] }

    // 1. create fresh new array to be start, a hash to couting swaped
    let s = Array(q.length).fill(0).map((_, i) => i + 1)
    let swap_count = new Map<number, number>()

    // 2. loop [q], find and swap to complete right side to left side, counting too.
    mainloop: for (let qi = q.length - 1; qi >= 0; qi--) {
        while (true) {
            let si = s.indexOf(q[qi])
            if (si < qi) {
                let swap_num = s[si + 1]

                swap(s, si)
                swap_count.set(swap_num, (swap_count.get(swap_num) || 0) + 1)
                continue
            }
            break;
        }
    }

    // 3. calc swaping is chaos or not
    let counts = Array.from(swap_count.values())
    let chaos = counts.some(e => e > 2)
    let sum = counts.reduce((p, v) => p + v, 0)

    if (chaos) console.log('Too chaotic')
    else console.log(sum)
}