algoadvance

You are given a 0-indexed integer array nums. You are allowed to remove exactly one element from the array. Return true if it can be made strictly increasing after removing exactly one element, otherwise return false.

An array nums is strictly increasing if nums[i] < nums[i+1] for every index i where 0 <= i < nums.length - 1.

Clarifying Questions

1. What is the size of the input array?
   • The size may vary but usually falls within the constraints such that 1 <= nums.length <= 10^4.
2. What should be returned if the array is already strictly increasing?
   • If the array is already strictly increasing, we should return true since removing any element will still maintain the order.
3. Are there any edge cases we should consider?
   • Yes, edge cases like when the array length is 1 or 2, as they are trivially strictly increasing after removing one element.

Strategy

1. Initial Check for Strictly Increasing Array
   • Check if the array is already strictly increasing. If it is, return true.
2. Validation of Increasing Properties on Removal
   • Traverse the array and identify points where the strict increasing property is violated.
   • For each violation, attempt two things:
     1. Remove the element that violates the property.
     2. Remove the element right after the violation.
   • For each removal scenario, check if the remaining elements maintain the strictly increasing property.
3. Helper Function
   • Create a helper function is_strictly_increasing(arr) to check if a given subarray is strictly increasing.
4. Edge Case Handling
   • Naturally, removing one element from arrays of size 1 and 2 should always return true since they can’t violate the strictly increasing property.

Code

def can_be_increasing(nums):
    def is_strictly_increasing(arr):
        for i in range(len(arr) - 1):
            if arr[i] >= arr[i + 1]:
                return False
        return True
    
    n = len(nums)
    
    if n <= 2:
        return True
    
    for i in range(n - 1):
        if nums[i] >= nums[i + 1]:
            # Try removing nums[i]
            if is_strictly_increasing(nums[:i] + nums[i+1:]):
                return True
            # Try removing nums[i+1]
            if is_strictly_increasing(nums[:i+1] + nums[i+2:]):
                return True
            return False
    
    return True

# Example Test Cases
print(can_be_increasing([1,2,10,5,7]))  # Expected True
print(can_be_increasing([2,3,1,2]))     # Expected False
print(can_be_increasing([1,1,1]))       # Expected False
print(can_be_increasing([1,2,3]))       # Expected True

Time Complexity

• Time Complexity: The is_strictly_increasing function checks the array in O(n) time. As we might call this function twice in one traversal for an array of length n, the overall complexity is O(n).
• Space Complexity: The space complexity is O(1) additional space since we are only creating a few variables and not using any extra space that scales with input size.

This strategy efficiently identifies whether the array can be made strictly increasing by removing one element, ensuring a linear runtime.

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