LeetCode Meditations — Chapter 5: Binary Search

Binary search is one of the most well-known algorithms. It's also a divide-and-conquer algorithm, where we break the problem into smaller components.

The crux of binary search is to find a target element in a given sorted array.
We have two pointers: high to point to the largest element, and low to point to the smallest element. We first initialize them for the whole array itself, with high being the last index and low being the first index. Then, we calculate the midpoint. If the target is greater than the midpoint, then we adjust our low pointer to point to the mid + 1, otherwise if the target is less than the midpoint, we adjust high to be mid - 1. With each iteration, we eliminate half the array until the midpoint equals target or the low pointer passes high.

If we find the index of the target, we can return it as soon as we find it; otherwise, we can just return -1 to indicate that the target doesn't exist in the array.

For example, if we have a nums array [-1, 0, 3, 5, 9, 12] and our target is 9, the operation looks like this:

We can write it in TypeScript like this:

function search(nums: number[], target: number): number {
  let high = nums.length - 1;
  let low = 0;

  while (high >= low) {
    let mid = Math.floor((high + low) / 2);

    if (target > nums[mid]) {
      low = mid + 1;
    } else if (target < nums[mid]) {
      high = mid - 1;
    } else {
      return mid;
    }
  }

  return -1;
};

Time and space complexity

The time complexity of a binary search algorithm is $O(log \ n)$ in the worst case. (For example, if the target is not in the array, we'll be halving the array until there is one element left.) The space complexity is $O(1)$ as we don't need extra space.

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The first problem in this chapter will be Find Minimum in Rotated Sorted Array, until then, happy coding.