LeetCode in Kotlin

1827. Minimum Operations to Make the Array Increasing

Easy

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

Solution

class Solution {
    fun minOperations(nums: IntArray): Int {
        var minsOps = 0
        for (i in 1 until nums.size) {
            if (nums[i] <= nums[i - 1]) {
                minsOps += nums[i - 1] - nums[i] + 1
                nums[i] = nums[i - 1] + 1
            }
        }
        return minsOps
    }
}

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