0_ceiling-of-a-number.js

/**
 *
 * Problem:
 * Given an array of numbers sorted in an ascending order, find the ceiling of a given
 * number 'key'. The ceiling of the 'key' will be the smallest element in the given array
 * greater than or equal to the 'key'. Write a function to return the index of the
 * ceiling of the 'key'. If there isn’t any ceiling return -1.
 *
 * Example 1:
 * Input: [4, 6, 10], key = 6
 * Output: 1
 * Explanation: The smallest number greater than or equal to '6' is '6' having index '1'.
 *
 * Example 2:
 * Input: [1, 3, 8, 10, 15], key = 12
 * Output: 4
 * Explanation: The smallest number greater than or equal to '12' is '15' having index
 * '4'.
 *
 *
 * Time: O(log n)
 * Space: O(1)
 *
 * @param {number[]} arr
 * @param {number} key
 * @return {number}
 */
function ceilingOfNumber(arr, key) {
  let start = 0;
  let end = arr.length;
  while (start < end) {
    const middle = start + Math.floor((end - start) / 2);
    if (arr[middle] >= key) {
      end = middle;
    } else {
      start = middle + 1;
    }
  }
  if (start < arr.length) {
    return start;
  }
  return -1;
}

// Test
console.log(ceilingOfNumber([4, 6, 10], 6)); // 1
console.log(ceilingOfNumber([1, 3, 8, 10, 15], 12)); // 4
console.log(ceilingOfNumber([4, 6, 10], 17)); // -1
console.log(ceilingOfNumber([4, 6, 10], -1)); // 0