The problem “Remove One Element to Make the Array Strictly Increasing” is a fascinating exercise in array manipulation and subarray analysis. It not only tests your understanding of arrays but also introduces you to the importance of maintaining certain properties in an array, such as strict increasing order. In this exhaustive guide, we’ll dissect the problem statement, explore multiple approaches to solve it, and delve into time and space complexities for each.

## 1. Problem Statement

Given a 0-indexed integer array `nums`

, you have to determine if it’s possible to remove exactly one element from the array such that it becomes strictly increasing. Return `True`

if it can be done and `False`

otherwise.

## 2. Brute-Force Approach

One simple way to solve this problem is to iterate through the array and remove each element once, then check if the resulting array is strictly increasing. This approach works but is not the most efficient.

Python Code:

```
def canBeIncreasing(nums):
for i in range(len(nums)):
temp = nums[:i] + nums[i+1:]
if all(temp[j] < temp[j+1] for j in range(len(temp) - 1)):
return True
return False
```

## 3. Optimized Approaches

#### 3.1 Two-Pass Method

Instead of removing an element each time, you can traverse the array twice: once from the beginning and once from the end. Keep track of the positions where the array fails to be strictly increasing and then evaluate if removing any of these would make the array strictly increasing.

Python Code:

```
def canBeIncreasing(nums):
n = len(nums)
left, right = [0]*n, [0]*n
left[0], right[-1] = nums[0], nums[-1]
for i in range(1, n):
left[i] = max(left[i-1], nums[i])
for i in range(n-2, -1, -1):
right[i] = min(right[i+1], nums[i])
for i in range(1, n-1):
if left[i-1] < nums[i+1] or right[i+1] > nums[i-1]:
return True
return False
```

#### 3.2 One-Pass Method

We can optimize it further by using a single pass through the array and using variables to keep track of the elements to be considered for removal.

Python Code:

```
def canBeIncreasing(nums):
cnt = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i-1]:
cnt += 1
if cnt > 1:
return False
if i >= 2 and nums[i] <= nums[i-2]:
nums[i] = nums[i-1]
return True
```

## 4. Time and Space Complexity Analysis

**Brute-Force Method**- Time Complexity: O(n^2)
- Space Complexity: O(n)

**Two-Pass Method**- Time Complexity: O(n)
- Space Complexity: O(n)

**One-Pass Method**- Time Complexity: O(n)
- Space Complexity: O(1)

## 5. Conclusion

This problem serves as a great learning experience, not only for understanding array manipulations but also for getting a grasp on optimizing solutions. The techniques you learn from this problem could be widely applied to many other array-related problems.