1827. Minimum Operations to Make the Array Increasing

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

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] <= 104

------

----

	class Solution {
	public int minOperations(int[] nums) {
	int ans = 0;
	for (int i = 1; i < nums.length; i++) {
	if (nums[i] <= nums[i - 1]) {
	ans += nums[i - 1] + 1 - nums[i];
	nums[i] = nums[i - 1] + 1;
	}
	}
	return ans;
	}
	}

view raw 1827.MinimumOperationsToMakeTheArrayIncreasing.java hosted with ❤ by GitHub

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

view raw 1827.MinimumOperationsToMakeTheArrayIncreasing.py hosted with ❤ by GitHub