# Leetcode – Minimum Operations to Make the Array Increasing Solution

Leetcode is a popular platform for practicing coding problems. Among its wide range of problems, one that stands out for its blend of arrays and optimization techniques is the “Minimum Operations to Make the Array Increasing.” This problem provides an excellent opportunity for programmers to exercise their skills in array manipulation, mathematical reasoning, and algorithmic thinking. This article will walk you through multiple approaches to solving this problem using Python.

## Problem Statement

You are given an integer array nums (0-indexed). In one operation, you can choose an element and increment it by 1.

You are tasked with making all the elements of the array strictly increasing, i.e., nums[i] < nums[i + 1] for all 0 <= i < nums.length - 1. Return the minimum number of operations needed to achieve this condition.

### Constraints

• The array length is between 1 and 10^4.
• Each element in the array will be in the range of 1 to 10^4.

### Example

Input: nums = [1, 1, 1]

Output: 3

Explanation: You can perform the following operations:

1. Increment the second element to 2: [1, 2, 1]
2. Increment the third element to 3: [1, 2, 3]

## Brute Force Approach

One naive approach would be to iterate through the array, and for each pair (nums[i], nums[i+1]), if nums[i] >= nums[i+1], we increment nums[i+1] until it becomes greater than nums[i].

def minOperations(nums):
ops = 0
for i in range(len(nums) - 1):
if nums[i] >= nums[i + 1]:
diff = nums[i] - nums[i + 1] + 1
ops += diff
nums[i + 1] += diff
return ops

### Time Complexity

The time complexity of this approach is O(n), where n is the length of the array. This is because we go through each pair (nums[i], nums[i+1]) exactly once.

### Space Complexity

The space complexity is O(1) as we are not using any additional data structures.

## Optimized Approach: One-Pass Iteration

We can make this algorithm even more efficient by calculating the increment needed for each element in a single pass through the array. While iterating, we can directly compute the required increment for nums[i+1] to be greater than nums[i].

def minOperations(nums):
ops = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i - 1]:
diff = nums[i - 1] - nums[i] + 1
ops += diff
nums[i] += diff
return ops

### Time and Space Complexity

The time complexity remains O(n), and the space complexity remains O(1).

## Unit Testing

Unit tests are crucial for ensuring that your code functions as expected.

def test_minOperations():
assert minOperations([1, 1, 1]) == 3
assert minOperations([1, 5, 2, 4, 1]) == 14
assert minOperations([8, 1, 9, 7]) == 16
print("All test cases pass")

test_minOperations()

## Conclusion

The “Minimum Operations to Make the Array Increasing” problem on Leetcode is an excellent way to fine-tune your skills in array manipulation, mathematical reasoning, and algorithm optimization. We covered a brute-force approach and then refined it to come up with a more optimized solution, both of which have a time complexity of O(n)O(n) and a space complexity of O(1)O(1). Implementing and understanding these solutions provide valuable insights into handling array-related problems in Python effectively.