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
EXPLANATION:
一道easy的题目, 直接一个for循环就可以做出来了. 因为每个数只需要比前一个数大1即可. 将结果累加就行.
SOLUTION:
class Solution {
func minOperations(_ nums: [Int]) -> Int {
var result:Int = 0
var arr = nums
for index in stride(from: 1, to: nums.count, by: 1) {
if (arr[index] <= arr[index - 1]) {
result += arr[index-1] - arr[index] + 1
arr[index] = arr[index-1]+1
}
}
return result
}
}