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1827. 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

Solutions

Solution 1: Single Pass

We use a variable $mx$ to record the maximum value of the current strictly increasing array, initially $mx = 0$.

Traverse the array nums from left to right. For the current element $v$, if $v \lt mx + 1$, we need to increase it to $mx + 1$ to ensure the array is strictly increasing. Therefore, the number of operations we need to perform this time is $max(0, mx + 1 - v)$, which is added to the answer, and then we update $mx=max(mx + 1, v)$. Continue to traverse the next element until the entire array is traversed.

The time complexity is $O(n)$, where $n$ is the length of the array nums. The space complexity is $O(1)$.

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

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

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


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

  • function minOperations(nums: number[]): number {
    let ans = 0;
    let max = 0;
    for (const v of nums) {
        ans += Math.max(0, max + 1 - v);
        max = Math.max(max + 1, v);
    }
    return ans;
}


  • public class Solution {
    public int MinOperations(int[] nums) {
        int ans = 0, mx = 0;
        foreach (int v in nums) {
            ans += Math.Max(0, mx + 1 - v);
            mx = Math.Max(mx + 1, v);
        }
        return ans;
    }
}


  • impl Solution {
    pub fn min_operations(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut max = 0;
        for &v in nums.iter() {
            ans += (0).max(max + 1 - v);
            max = v.max(max + 1);
        }
        ans
    }
}

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