Given an array of 0 and 1, find minimum no. of swaps to bring all 1s together (only adjacent swaps allowed)

Question

If given an array of 1\'s and 0\'s, what\'s good algorithm to show the minimum number of adjacent swaps needed to group all of the 1\'s together. The 1\'s don\'t need to be

Answer 1

UPDATED:

The algorithm determines center by just getting an array of all indices of 1's. The center of that array will always hold the center index. Much faster.

oneIndices = array of indices of all 1's in the input
middleOfOnesIndices = round(oneIndices.length/2)-1    // index to the center index
minimumSwaps = 0
foreach index i of oneIndices
    minimumSwaps += aboluteValue(oneIndices[middleOfOneIndices]-oneIndices[i])-absoluteValue(middleOfOneIndices-i);

Here's a fiddle to see it in action:

https://jsfiddle.net/3pmwrk0d/6/

This was a fun one. Thanks for the question.