35. Search Insert Position

Problem:

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5
Output: 2

Example 2:

Input: [1,3,5,6], 2
Output: 1

Example 3:

Input: [1,3,5,6], 7
Output: 4

Example 4:

Input: [1,3,5,6], 0
Output: 0

Solution:

Same as simple binary search except it returns the start index when does not find a match.

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
let searchInsert = function (nums, target) {
  let s = 0;
  let e = nums.length - 1;

  while (s <= e) {
    const p = ((s + e) / 2) | 0;
    const diff = nums[p] - target;
    if (diff === 0) {
      return p;
    } else if (diff < 0) {
      s = p + 1;
    } else {
      e = p - 1;
    }
  }

  return s;
};

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆