Formatted question description: https://leetcode.ca/all/1827.html

1827. Minimum Operations to Make the Array Increasing

Level

Easy

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] <= 10^4

Solution

To make nums strictly increasing, each element in nums must be greater than its previous element. For all 1 <= i < nums.length - 1, nums[i] must be at least nums[i - 1] + 1, so let minCurr = Math.max(nums[i], nums[i - 1] + 1), add minCurr - nums[i] to the number of operations, and let nums[i] = minCurr. Finally, return the number of operations.

class Solution {
    public int minOperations(int[] nums) {
        int operations = 0;
        int length = nums.length;
        for (int i = 1; i < length; i++) {
            int prev = nums[i - 1], curr = nums[i];
            int minCurr = Math.max(prev + 1, curr);
            operations += minCurr - curr;
            nums[i] = minCurr;
        }
        return operations;
    }
}

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