Enum CoinFlips { Heads, Tails }

int NumberOfFlipsToMakeCoinFlipsAlternate(CoinFlips[] coinFlips){

CoinFlip previousCoinFlip;

int numberOfFlipsNeededToMakeSequenceAlternate = 0;

for(var i =0; i< listOfCoinFlips.length; i++){ if(i==0) { previousCoinFlip = listOfCoinFlips[i]; continue; // deals with the first coin case while in a single for loop, as well as dealing with a trivial sequence of one coin. }

var latestCoinFlip = listOfCoinFlips[i];

if(latestCoinFlip ! = previousCoinFlip) { numberOfFlipsNeededToMakeSequenceAlternate ++; }

return numberOfFlipsNeededToMakeSequenceAlternate; }

If you have HH, you need to flip the second coin. If you have HHT... or HHH..., you still needed to flip that second coin to get a valid sequence. Given that flip has happened, you then decide whether or not you need to flip the third coin (yes and no respectively), and so on. Given the last k are correct, you need to ensure k+1 is.

You can make optimisations, but this can be guaranteed to be done in a single iteration over the array. If it's O(n) it's good enough.