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No need to swap values, no need to loop backwards, no need to loop more than once. Just loop through each person in the queue and check two things: 1. Has this person moved more than two positions forward?
2. How many times has this person been bribed?

In Python3:

defminimumBribes(Q):## initialize the number of movesmoves=0## decrease Q by 1 to make index-matching more intuitive# so that our values go from 0 to N-1, just like our# indices. (Not necessary but makes it easier to# understand.)Q=[P-1forPinQ]## Loop through each person (P) in the queue (Q)fori,Pinenumerate(Q):# i is the current position of P, while P is the# original position of P.## First check if any P is more than two ahead of # its original positionifP-i>2:print("Too chaotic")return## From here on out, we don't care if P has moved# forwards, it is better to count how many times# P has RECEIVED a bribe, by looking at who is# ahead of P. P's original position is the value# of P.# Anyone who bribed P cannot get to higher than# one position in front if P's original position,# so we need to look from one position in front# of P's original position to one in front of P's# current position, and see how many of those # positions in Q contain a number large than P.# In other words we will look from P-1 to i-1,# which in Python is range(P-1,i-1+1), or simply# range(P-1,i). To make sure we don't try an# index less than zero, replace P-1 with# max(P-1,0)forjinrange(max(P-1,0),i):ifQ[j]>P:moves+=1print(moves)

## New Year Chaos

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No need to swap values, no need to loop backwards, no need to loop more than once. Just loop through each person in the queue and check two things: 1. Has this person moved more than two positions forward? 2. How many times has this person been bribed?

In Python3: