Formatted question description: https://leetcode.ca/all/1827.html

# 1827. Minimum Operations to Make the Array Increasing

Easy

## 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] <= 10^4

## Solution

To make nums strictly increasing, each element in nums must be greater than its previous element. For all 1 <= i < nums.length - 1, nums[i] must be at least nums[i - 1] + 1, so let minCurr = Math.max(nums[i], nums[i - 1] + 1), add minCurr - nums[i] to the number of operations, and let nums[i] = minCurr. Finally, return the number of operations.

class Solution {
public int minOperations(int[] nums) {
int operations = 0;
int length = nums.length;
for (int i = 1; i < length; i++) {
int prev = nums[i - 1], curr = nums[i];
int minCurr = Math.max(prev + 1, curr);
operations += minCurr - curr;
nums[i] = minCurr;
}
return operations;
}
}