// Key points that helped me figure it out
// - If the index (-1 because array is 0 based) and the number are the same
//   then the person has not bribed or been bribed
// - We can count how many bribes a person has done by checking how many numbers
//   are after it that are greater than his/her number.

function minimumBribes(q: number[]): void {
    // Write your code here
    let total_bribes = 0
    
    for(let i=0;i<q.length;i++){ // Loop through the persons
        // Write your code here
    
        // If i (index) is lower than the person's number minus the 2 positions 
        // and minus 1 because the array is 0 based (or just -3), then it means the  
        // person bribed more than 2 times
        if(i < q[i]-3){
            process.stdout.write("Too chaotic \n")
            return // Return to break out
        }
            
        // Check how many numbers greater than the person's current
        // one exist down the line
       let current_bribes = 0       
        for(let j=i;j<q.length;j++){  // Loop through the rest of persons from current        
            if(q[i]>q[j])
              current_bribes++ // +1 bribe
              
            // We already know that no one bribed 3 times,
            // so we can break on the 2nd one for optimization
            if(current_bribes === 2){  
                break
            }
        
       }
       
       total_bribes += current_bribes // Add person's bribes to the total
    }
    process.stdout.write(total_bribes+' \n') // Return total bribes

}