Minimum Operations
This post is part of the Algorithms Problem Solving series.
Problem description
This is the Minimum Operations problem. The description looks like this:
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 incrementnums[1]
to makenums = [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.
Examples
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
Solution
First solution counting and update the array in place.
def min_operations(nums):
previous_num = nums[0]
operations = 0
for index, num in enumerate(nums[1:]):
if num <= previous_num:
new_num = previous_num - num + 1
operations += new_num
nums[index + 1] += new_num
previous_num = nums[index + 1]
return operations
Second and more optimized solution without the need to update the array.
def min_operations(nums):
prev, ops = 0, 0
for num in nums:
if num <= prev:
prev += 1
ops += prev - num
else:
prev = num
return ops
Resources
- Learning Python: From Zero to Hero
- Algorithms Problem Solving Series
- Stack Data Structure
- Queue Data Structure
- Linked List
- Tree Data Structure
Have fun, keep learning, and always keep coding!