100 chairs

100chairs.js

//Problem: variation of Josephus problem
// Take a second to imagine that you are in a room with 100 chairs arranged in a circle.
//These chairs are numbered sequentially from One to One Hundred.
//At some point in time, the person in chair #1 will be told to leave the room.
//The person in chair #2 will be skipped, and the person in chair #3 will be told to leave.
//Next to go is person in chair #6. In other words, 1 person will be skipped initially, and then 2, 3, 4.. and so on.
//This pattern of skipping will keep going around the circle until there is only one person remaining.. the survivor.
//Note that the chair is removed when the person leaves the room.
//
//Write a program to figure out which chair the survivor is sitting in.

var findLastSurvivingChair = function(noOfChairs) {
  noOfChairs = noOfChairs || 100;

  //create an empty chair array
  var chairArray = [];

  // loop from 1 to number of chairs (100) to create array of chairs #1 - #100
  for (var i = 1; i <= noOfChairs; i++) {
    chairArray.push(i);
  }

  // Skip starts at 1 and increments by 1 each time a chair is removed
  // Continue to remove chairs in while loop until chair array contains one lone surviving chair
  var currentIndex = 0;
  var skip = 0;
  while (chairArray.length > 1) {
    chairArray.splice(currentIndex, 1);
    skip += 1;
    currentIndex += skip;

    //if current index becomes greater than number of remaining chairs,
    //then set current index to remainder from current index / number of chairs
    currentIndex %= chairArray.length;
  }

  return chairArray[0];
};