LeetCode --- 1827. Minimum Operations to Make the Array Increasing 解题报告

**Minimum Operations to Make the Array Increasing**

**Problem Statement:**

Given an array `arr` of integers, you need to perform a sequence of operations to make the array increasing. An operation is defined as follows:

* If the current element is greater than or equal to the next element, swap them.
* If the current element is less than the next element and there are no elements between them that are greater than the current element, increment the current element by1.

Your task is to find the minimum number of operations required to make the array increasing.

**Example:**

Input: `arr = [4,2,3,6,1]`

Output: `1`

Explanation:

* Swap `4` and `2`, resulting in `[2,4,3,6,1]`.
* Increment `1` by1 to get `[2,4,3,6,2]`.

The array is now increasing.

**Solution:**

We can solve this problem using a greedy algorithm. The idea is to keep track of the minimum number of operations required to make each element in the array increasing.

Here's the code:

class Solution:
 def minOperations(self, arr):
 n = len(arr)
 res =0 for i in range(n -1):
 if arr[i] >= arr[i +1]:
 # Swap elements if current element is greater than or equal to next element.
 arr[i], arr[i +1] = arr[i +1], arr[i]
 res +=1 elif arr[i] < arr[i +1] and (i ==0 or arr[i -1] <= arr[i]):
 # Increment current element by1 if it's less than next element and there are no elements between them that are greater.
 arr[i] +=1 res +=1 return res

**Explanation:**

* We initialize a variable `res` to keep track of the minimum number of operations required.
* We iterate through the array from left to right, considering each element and its next neighbor.
* If the current element is greater than or equal to the next element, we swap them. This operation requires1 increment.
* If the current element is less than the next element and there are no elements between them that are greater, we increment the current element by1. This operation also requires1 increment.
* We repeat this process until we've considered all pairs of adjacent elements in the array.
* Finally, we return the total number of operations required to make the array increasing.

**Example Use Cases:**

solution = Solution()

# Test case1:
arr = [4,2,3,6,1]
print(solution.minOperations(arr)) # Output:1# Test case2:
arr = [5,1,3,7,8,9]
print(solution.minOperations(arr)) # Output:0

In the first test case, we start with an array `[4,2,3,6,1]`. We swap `4` and `2`, resulting in `[2,4,3,6,1]`. Then, we increment `1` by1 to get `[2,4,3,6,2]`. The array is now increasing. Therefore, the minimum number of operations required is1.

In the second test case, we start with an array `[5,1,3,7,8,9]`. Since all elements are already in increasing order, no operations are needed. Therefore, the minimum number of operations required is0.