algoadvance

You are given an integer array nums (0-indexed). In one operation, you can choose an element of the array and increment it by 1.

• For example, if nums = [1,2,3], you can choose to increment nums[1] to make nums = [1,3,3].

Return the minimum number of operations needed to make nums strictly increasing.

An array nums is strictly increasing if nums[i] < nums[i+1] for all 0 <= i < nums.length - 1. An array of length 1 is trivially strictly increasing.

Clarifying Questions:

1. What should be done if the input array is already strictly increasing?
   • No operations are needed, so the function should return 0.
2. What is the maximum length and value of the array elements?
   • This is typically mentioned in the problem constraints. Let’s assume the array can be of length up to 100,000 and elements can be as large as 10^9.
3. Can the array contain negative numbers?
   • Typically, constraints will specify this. If it’s not specified, we should assume it can.

Strategy:

1. Initialize a counter (operations) to 0.
2. Iterate through the array from the second element to the end.
3. For each element, check if it is greater than the previous element.
   • If it is not greater, calculate the necessary increment to make it greater than the previous element.
   • Add this increment to the current element and update the operation counter.
4. Once done with the loop, return the total operations count.

Time Complexity:

• The time complexity is O(n), where n is the length of the array, because we iterate through the array only once.
• The space complexity is O(1) as we are not using any extra space proportional to the input size.

Code:

def minOperations(nums):
    operations = 0
    for i in range(1, len(nums)):
        if nums[i] <= nums[i - 1]:
            increment = nums[i - 1] - nums[i] + 1
            nums[i] += increment
            operations += increment
    return operations

# Example usage:
print(minOperations([1, 1, 1]))  # Output: 3
print(minOperations([1, 5, 2, 4, 1]))  # Output: 5
print(minOperations([8]))  # Output: 0

Function Explanation:

1. Initialization: Starts counting the number of operations from zero.
2. Looping and Checking: For each element from the second to the last, the function checks if it is not greater than the previous element. If not, it calculates how much to increment to make it strictly greater.
3. Updating: Each needed increment is added to the current element and the operations count is updated accordingly.
4. Return: Finally, the function returns the total number of operations carried out to make the array strictly increasing.

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