1827.Minimum_Operations_to_Make_the_Array_Increasing(Easy).md

1827. Minimum Operations to Make the Array Increasing (Easy)

Date and Time: Sep 27, 2024, 22:32 (EST)

Link: https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/

Question:

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.

Example 1:

Input: nums = [1,1,1]

Output: 3

Explanation: You can do the following operations:
1) Increment nums[2], so nums becomes [1,1,2].
2) Increment nums[1], so nums becomes [1,2,2].
3) Increment nums[2], so nums becomes [1,2,3].

Example 2:

Input: nums = [1,5,2,4,1]

Output: 14

Example 3:

Input: nums = [8]

Output: 0

Constraints:

• 1 <= nums.length <= 5000

• 1 <= nums[i] <= 10^4

Walk-through:

Start looping nums from the second element, and we want to compare nums[i] with nums[i-1], if nums[i] <= nums[i-1], we can record how many operations it take to make it greater than nums[i-1] by nums[i-1]+1 - nums[i] and we also update nums[i] with this value nums[i-1]+1.

Python Solution:

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        # Loop over nums, if nums[i] <= nums[i-1], update
        res = 0
        for i in range(1, len(nums)):
            if nums[i] <= nums[i-1]:
                res += (nums[i-1]+1 - nums[i])
                nums[i] = nums[i-1]+1
        return res

Time Complexity: $O(n)$, n is the length of nums.
Space Complexity: $O(1)$