LeetCode 1827. Increment array with minimal operation

1827. The least operation increments the array

Give you an array of integers nums （ Subscript from 0 Start ）. In every operation , You can select an element in the array , And add it 1 .

• For example , If nums = [1,2,3] , You can choose to increase it nums[1] obtain nums = [1,**3**,3] .

Please return to make nums Strictly increasing Of least Operating frequency .

We call it an array nums yes Strictly increasing , When it satisfies for all 0 <= i < nums.length - 1 There are nums[i] < nums[i+1] . A length of 1 Is a special case of strictly incrementing .

Example 1：

 Input ：nums = [1,1,1]
 Output ：3
 explain ： You can do the following ：
1)  increase  nums[2] , The array becomes  [1,1,2] .
2)  increase  nums[1] , The array becomes  [1,2,2] .
3)  increase  nums[2] , The array becomes  [1,2,3] .

Example 2：

 Input ：nums = [1,5,2,4,1]
 Output ：14

Example 3：

 Input ：nums = [8]
 Output ：0

Tips ：

• 1 <= nums.length <= 5000
• 1 <= nums[i] <= 104

Two 、 Method 1

simulation

class Solution {
    
    public int minOperations(int[] nums) {
    
        if (nums.length == 1) {
    
            return 0;
        }

        int res = 0;
        for (int i = 0; i < nums.length; i++) {
    
            if (i + 1 < nums.length && nums[i + 1] <= nums[i]) {
    
                res += (nums[i] + 1 - nums[i + 1]);
                nums[i + 1] = nums[i] + 1;
            }
        }
        return res;
    }
}

Complexity analysis

• Time complexity ：O(n).

• Spatial complexity ：O(1).