How do you solve Minimum Operations to Make the Array Increasing on LeetCode?

Minimum Operations to Make the Array Increasing (LeetCode #1827) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each element must be strictly greater than the previous one.

What is the time complexity of Minimum Operations to Make the Array Increasing?

The optimal solution for Minimum Operations to Make the Array Increasing runs in O(n) time and uses O(1) space. This complexity is optimal because we only make a single pass through the array, checking each element once.

What is the brute force solution for Minimum Operations to Make the Array Increasing?

The brute force approach for Minimum Operations to Make the Array Increasing is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to increment elements to make the array strictly increasing. This means for each element, we might need to check all previous elements to ensure the condition is satisfied, leading to a lot of unnecessary calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make the Array Increasing?

Minimum Operations to Make the Array Increasing uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Minimum Operations to Make the Array Increasing?

Always think about edge cases, such as arrays with all equal elements. Explain your thought process clearly to the interviewer as you write code. Consider both time and space complexity when proposing solutions.

What are common mistakes when solving Minimum Operations to Make the Array Increasing?

Not updating the current element after increments, leading to incorrect calculations. Overcomplicating the logic instead of using a straightforward greedy approach.

#1827

Minimum Operations to Make the Array Increasing

Easy

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a greedy strategy to ensure each element is at least one greater than the previous element. This avoids unnecessary increments and ensures we only make the minimum number of operations needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable 'operations' to 0 and set 'prev' to the first element.
2Step 2: Loop through the array starting from the second element.
3Step 3: If the current element is less than or equal to 'prev', calculate the increments needed to make it 'prev + 1', update 'operations', and set 'prev' to this new value. Otherwise, just update 'prev' to the current element.

solution.py10 lines

1def minOperations(nums):
2    operations = 0
3    prev = nums[0]
4    for i in range(1, len(nums)):
5        if nums[i] <= prev:
6            operations += (prev - nums[i] + 1)
7            prev += 1
8        else:
9            prev = nums[i]
10    return operations

ℹ

Complexity note: This complexity is optimal because we only make a single pass through the array, checking each element once.

1Each element must be strictly greater than the previous one.
2Incrementing an element affects only the current and subsequent elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.