[Practice] Get the count of Ones in a sorted array of 0s and 1s

[lpatmo]

Given a sorted array of 0s and 1s, count the number of ones in the array.

Requirement: Challenge: perform this in O(log(N)) time complexity.

console.log(findOnes([0,0,1,1,1,1,1,1,1])); //7
console.log(findOnes([0,1])); //1
console.log(findOnes([0,0,0,1,1,1,1,1,1,1,1]));  //8
console.log(findOnes([0,0,0]));  //0
console.log(findOnes([1,1,1,1,1,1]));  //6
console.log(findOnes([0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1]));  //6

[lpatmo]

function findOnes(arr) {
  //choose an arbitrary midpoint (1/2 of array)
  //base case: midpoint = 1, and left number is 0
  //base case: last element is 0
  //base case: first element is 1
  //if midpoint is 0, move to right by arr up to midpoint divided by 2
  // if midpoint is 1, move to left by arr up to midpoint divided by 2... 
  let start = 0;
  let end = arr.length -1;
  let midpoint = Math.floor((start+end)/2);
 // console.log(midpoint)
  if (arr[midpoint] === 1 && arr[midpoint-1] === 0) {
    return arr.length - midpoint;
  }
  if (arr[arr.length-1] === 0) {
    return 0;
  }
  if (arr[0] === 1) {
    return arr.length;
  }
  //console.log('midpiont initial', midpoint)
  //Basically practicing binary search below: 
  while (start < end) {
    if (arr[midpoint] === 1 && arr[midpoint-1] === 0) {
       return arr.length - midpoint; //not happy with this, but it's a necessary check
    }
    if (arr[midpoint] === 0) {
      start = midpoint + 1;
    } else {
      end = midpoint - 1;
    }
    midpoint = Math.floor((start+end)/2);
  }
  return arr.length - midpoint;
}