To solve this problem, you can use a binary search algorithm. Here are the steps:

1. Initialize two pointers, `start` and `end`, to the beginning and end of the array, respectively.

2. While `start` is less than or equal to `end`:

a. Calculate the middle index `mid` as `(start + end) / 2`.

b. If `arr[mid]` is less than `x`, update `start` to `mid + 1`.

c. Otherwise, update `end` to `mid - 1`.

3. After the loop, `start` will be the lower bound of `x`. If `start` is equal to the size of the array, it means all numbers in the array are smaller than `x`.

Here is a Python code snippet that implements the above steps:

```python
def lower_bound(arr, x):
start, end = 0, len(arr) - 1
while start

Question

To solve this problem, you can use a binary search algorithm. Here are the steps:

1. Initialize two pointers, `start` and `end`, to the beginning and end of the array, respectively.

2. While `start` is less than or equal to `end`:

a. Calculate the middle index `mid` as `(start + end) / 2`.

b. If `arr[mid]` is less than `x`, update `start` to `mid + 1`.

c. Otherwise, update `end` to `mid - 1`.

3. After the loop, `start` will be the lower bound of `x`. If `start` is equal to the size of the array, it means all numbers in the array are smaller than `x`.

Here is a Python code snippet that implements the above steps:

```python
def lower_bound(arr, x):
    start, end = 0, len(arr) - 1
    while start <= end:
        mid = (start + end) // 2
        if arr[mid] < x:
            start = mid + 1
        else:
            end = mid - 1
    return start
```

This function takes an array `arr` and a number `x` as input, and returns the index of the lower bound of `x` in `arr`. If all numbers in `arr` are smaller than `x`, it returns the size of `arr`.

Knowee AI · Accepted Answer