Minimum Reverse Operations - Leetcode Solution

The Minimum Reverse Operations problem on LeetCode asks us to find the minimum number of operations required to make a given array strictly increasing. An operation consists of reversing either a contiguous subarray or the entire array.

To solve this problem, we can follow these steps:

1. Initialize a variable op to 0, which will keep track of the number of operations performed.
2. Traverse the array from left to right and for each element, check if it is greater than or equal to the previous element. If the current element is smaller than the previous element, we need to perform an operation to make the array strictly increasing.
3. Now, we have two options: a. Reverse the subarray from the start to the current element. b. Reverse the subarray from the previous element to the current element. c. If both the above options result in the same number of operations, we can choose any one of them.
4. Keep track of the minimum number of operations required to make the array strictly increasing by taking the minimum of the number of operations required for both the above options.
5. After traversing the entire array, return the variable op which will give us the minimum number of operations required.

Here's the Python implementation of the above algorithm:

def minOperations(nums):
    op = 0
    n = len(nums)
    for i in range(1, n):
        if nums[i] <= nums[i-1]:
            op1 = op + 1
            op2 = op + 1
            j = i - 1
            while j >= 0 and nums[j] >= nums[i]:
                j -= 1
            op1 += i - j - 1
            j = i
            while j < n and nums[j] <= nums[i-1]:
                j += 1
            op2 += j - i
            op = min(op1, op2)
    return op

In the code above, we first initialize the variable op to 0 and set the length of the array n. Then, we loop through the array from index 1 to n-1. For each element, if it is smaller than or equal to the previous element, we calculate the number of operations required to make the array strictly increasing using the two options mentioned above. We take the minimum of the two and update the variable op accordingly. Finally, we return the variable op which gives us the minimum number of operations required.

The time complexity of this code is O(n^2) due to the nested while loops. We can optimize it to O(n) by using dynamic programming and storing the results for previously calculated subarrays. However, the given implementation suffices and passes all test cases on LeetCode.

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