README_EN.md

1827. Minimum Operations to Make the Array Increasing

中文文档

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.

<li>For example, if <code>nums = [1,2,3]</code>, you can choose to increment <code>nums[1]</code> to make <code>nums = [1,<u><b>3</b></u>,3]</code>.</li>

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:

<li><code>1 &lt;= nums.length &lt;= 5000</code></li>

<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>

Solutions

Python3

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        mx = ans = 0
        for v in nums:
            ans += max(0, mx + 1 - v)
            mx = max(mx + 1, v)
        return ans

Java

class Solution {
    public int minOperations(int[] nums) {
        int ans = 0;
        int mx = 0;
        for (int v : nums) {
            ans += Math.max(0, mx + 1 - v);
            mx = Math.max(mx + 1, v);
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minOperations(vector<int>& nums) {
        int ans = 0;
        int mx = 0;
        for (int& v : nums) {
            ans += max(0, mx + 1 - v);
            mx = max(mx + 1, v);
        }
        return ans;
    }
};

Go

func minOperations(nums []int) int {
	ans, mx := 0, 0
	for _, v := range nums {
		ans += max(0, mx+1-v)
		mx = max(mx+1, v)
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

...