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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

Related Topics:
Array, Greedy

Solution 1.

// OJ: https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minOperations(vector<int>& A) {
        int ans = 0;
        for (int i = 1; i < A.size(); ++i) {
            int tmp = A[i];
            A[i] = max(A[i], A[i - 1] + 1);
            ans += A[i] - tmp;
        }
        return ans;
    }
};