Ceil

We will declare the 2 pointers and an ‘ans’ variable initialized to -1(If we don’t find any index, we will return -1).

Place the 2 pointers i.e. low and high: Initially, we will place the pointers like this: low will point to the first index and high will point to the last index.
Calculate the ‘mid’: Now, we will calculate the value of mid using the following formula: mid = (low+high) // 2 ( ‘//’ refers to integer division)
Compare arr[mid] with x: With comparing arr[mid] to x, we can observe 2 different cases:

Case 1 - If arr[mid] >= x: This condition means that the index arr[mid] may be an answer. So, we will update the ‘ans’ variable with arr[mid] and search in the left half if there is any smaller number that satisfies the same condition. Here, we are eliminating the right half.
Case 2 - If arr[mid] < x: In this case, arr[mid] cannot be our answer and we need to find some bigger element. So, we will eliminate the left half and search in the right half for the answer.

Floor

We will declare the 2 pointers and an ‘ans’ variable initialized to -1(If we don’t find any index, we will return -1).

Place the 2 pointers i.e. low and high: Initially, we will place the pointers like this: low will point to the first index and high will point to the last index.
Calculate the ‘mid’: Now, we will calculate the value of mid using the following formula: mid = (low+high) // 2 ( ‘//’ refers to integer division)
Compare arr[mid] with x: With comparing arr[mid] to x, we can observe 2 different cases:

Case 1 - If arr[mid] <= x: The index arr[mid] is a possible answer. So, we will store it and will try to find a larger number that satisfies the same condition. That is why we will remove the left half and try to find the number in the right half.
Case 2 - If arr[mid] > x: arr[mid] is definitely not the answer and we need a smaller number. So, we will reduce the search space to the left half by removing the right half.

Code