Problem Description

You have an integer array nums (0-indexed) and you can perform operations on it. In each operation, you can select any element from the array and increase its value by 1.

For example, if nums = [1,2,3], you could choose to increment nums[1] to get nums = [1,3,3].

Your goal is to make the array strictly increasing, which means every element must be smaller than the next element: nums[i] < nums[i+1] for all valid indices i.

The task is to find the minimum number of operations needed to achieve this strictly increasing property.

Note that an array with only one element is considered strictly increasing by default.

For instance:

• If nums = [1,1,1], you would need to make it something like [1,2,3], requiring 3 operations total (increment the second element once and the third element twice)
• If nums = [1,5,2,4,1], you would need to transform it to ensure each element is larger than the previous one

The solution tracks the maximum value that must be maintained at each position to ensure the strictly increasing property, and counts the total operations needed to adjust elements that don't meet this requirement.

Intuition

When we traverse the array from left to right, at each position we need to ensure the current element is larger than the previous element. The key insight is that we want to perform the minimum number of operations, so we should only increase an element to the smallest value that maintains the strictly increasing property.

Think about it this way: if we're at position i, the current element must be at least previous_element + 1. If it's already larger than that, great - we don't need any operations. But if it's smaller, we need to increase it to exactly previous_element + 1 (no more, no less, since we want minimum operations).

As we move through the array, we keep track of what the "previous element" effectively becomes. This is captured by the variable mx in the solution. After processing each element, mx represents the minimum value that the current position must have to maintain the strictly increasing property.

For each element v:

• If v < mx + 1, we need (mx + 1 - v) operations to bring it up to the required minimum
• If v >= mx + 1, we don't need any operations for this element

After processing each element, we update mx to be the maximum of either:

• mx + 1 (the minimum required value for strict increasing)
• v (the actual value, if it's already larger)

This greedy approach works because increasing any element more than necessary would just waste operations without providing any benefit. By always choosing the minimum valid value at each position, we ensure the total number of operations is minimized.

Learn more about Greedy patterns.

Solution Approach

The solution uses a single-pass greedy algorithm with two key variables:

• ans: accumulates the total number of operations needed
• mx: tracks the minimum value required at the current position to maintain strict increasing order

Let's walk through the implementation step by step:

1. Initialize variables: Start with ans = 0 (no operations yet) and mx = 0 (representing the value before the first element conceptually).

2. Traverse the array: For each element v in nums:

   • Calculate operations needed: max(0, mx + 1 - v)
     • If v < mx + 1, we need (mx + 1 - v) operations to increase v to mx + 1
     • If v >= mx + 1, we need 0 operations (the element is already large enough)
   • Add these operations to ans
   • Update mx to max(mx + 1, v):
     • This represents the value at the current position after any necessary operations
     • If we performed operations, the value becomes mx + 1
     • If no operations were needed, the value remains v

3. Return the result: After processing all elements, ans contains the minimum total operations.

Example walkthrough with nums = [1, 1, 1]:

• Initially: ans = 0, mx = 0
• Process nums[0] = 1:
  • Operations: max(0, 0 + 1 - 1) = 0
  • Update: ans = 0, mx = max(1, 1) = 1
• Process nums[1] = 1:
  • Operations: max(0, 1 + 1 - 1) = 1 (need to increase to 2)
  • Update: ans = 1, mx = max(2, 1) = 2
• Process nums[2] = 1:
  • Operations: max(0, 2 + 1 - 1) = 2 (need to increase to 3)
  • Update: ans = 3, mx = max(3, 1) = 3
• Result: 3 operations total

The time complexity is O(n) where n is the length of the array, and space complexity is O(1) as we only use constant extra space.

Example Walkthrough

Let's walk through the solution with nums = [1, 5, 2, 4, 1]:

Initial State:

• ans = 0 (total operations counter)
• mx = 0 (tracks minimum required value)

Step 1: Process nums[0] = 1

• Required minimum: mx + 1 = 0 + 1 = 1
• Current value: 1
• Operations needed: max(0, 1 - 1) = 0
• Update ans = 0 + 0 = 0
• Update mx = max(1, 1) = 1
• Array state: [1, 5, 2, 4, 1]

Step 2: Process nums[1] = 5

• Required minimum: mx + 1 = 1 + 1 = 2
• Current value: 5
• Operations needed: max(0, 2 - 5) = 0 (already larger than required)
• Update ans = 0 + 0 = 0
• Update mx = max(2, 5) = 5
• Array state: [1, 5, 2, 4, 1]

Step 3: Process nums[2] = 2

• Required minimum: mx + 1 = 5 + 1 = 6
• Current value: 2
• Operations needed: max(0, 6 - 2) = 4 (need to increase from 2 to 6)
• Update ans = 0 + 4 = 4
• Update mx = max(6, 2) = 6
• Array state conceptually: [1, 5, 6, 4, 1]

Step 4: Process nums[3] = 4

• Required minimum: mx + 1 = 6 + 1 = 7
• Current value: 4
• Operations needed: max(0, 7 - 4) = 3 (need to increase from 4 to 7)
• Update ans = 4 + 3 = 7
• Update mx = max(7, 4) = 7
• Array state conceptually: [1, 5, 6, 7, 1]

Step 5: Process nums[4] = 1

• Required minimum: mx + 1 = 7 + 1 = 8
• Current value: 1
• Operations needed: max(0, 8 - 1) = 7 (need to increase from 1 to 8)
• Update ans = 7 + 7 = 14
• Update mx = max(8, 1) = 8
• Final array conceptually: [1, 5, 6, 7, 8]

Result: Minimum operations needed = 14

The key insight: at each position, we only increase the element to the exact minimum needed (mx + 1) to maintain strict increasing order, never more. This greedy choice ensures we use the minimum total operations.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the array nums. The algorithm iterates through the array exactly once with a single for loop. Inside the loop, all operations (max(), addition, subtraction) are constant time O(1) operations. Therefore, the overall time complexity is linear with respect to the input size.

Space Complexity: O(1). The algorithm only uses a fixed amount of extra space regardless of the input size. It maintains three variables: ans for accumulating the result, mx for tracking the maximum value needed, and v for iterating through the array. Since these variables don't grow with the input size, the space complexity is constant.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Updating the Maximum Value

One common mistake is updating mx (or current_max) before calculating the operations needed, or using the wrong value for the update.

Incorrect Implementation:

def minOperations(self, nums: List[int]) -> int:
    total_operations = 0
    current_max = 0
  
    for num in nums:
        current_max = max(current_max + 1, num)  # WRONG: Updated before calculation
        operations_needed = max(0, current_max - num)
        total_operations += operations_needed
  
    return total_operations

Why it's wrong: By updating current_max first, we lose track of what the minimum required value should be at this position, leading to incorrect operation counts.

Correct Approach: Always calculate operations first based on the current state, then update the tracking variable.

2. Off-by-One Error in Strict Increasing Condition

Another pitfall is forgetting that "strictly increasing" means each element must be strictly greater than the previous, not just greater than or equal.

Incorrect Implementation:

def minOperations(self, nums: List[int]) -> int:
    total_operations = 0
    current_max = 0
  
    for num in nums:
        operations_needed = max(0, current_max - num)  # WRONG: Missing +1
        total_operations += operations_needed
        current_max = max(current_max, num)  # WRONG: Should be max(current_max + 1, num)
  
    return total_operations

Why it's wrong: This would make elements equal to the previous maximum, creating a non-decreasing array instead of a strictly increasing one. For [1,1,1], this would incorrectly return 0 operations instead of 3.

Correct Approach: Always ensure the next element is at least current_max + 1 to maintain strict inequality.

3. Integer Overflow in Large Arrays

For very large arrays with large values, the accumulated operations count could potentially overflow in languages with fixed integer sizes.

Prevention: In Python, this isn't an issue due to arbitrary precision integers, but in other languages, consider using appropriate data types (like long long in C++) when dealing with potentially large operation counts.

4. Misunderstanding Initial State

Some might incorrectly initialize current_max to nums[0] and start iterating from index 1.

Incorrect Implementation:

def minOperations(self, nums: List[int]) -> int:
    if len(nums) <= 1:
        return 0
  
    total_operations = 0
    current_max = nums[0]  # WRONG: Assumes first element doesn't need operations
  
    for i in range(1, len(nums)):
        operations_needed = max(0, current_max + 1 - nums[i])
        total_operations += operations_needed
        current_max = max(current_max + 1, nums[i])
  
    return total_operations

Why it's wrong: This assumes the first element never needs modification, but what if nums = [5,1,2]? The first element might need to be decreased conceptually, but since we can only increase, we need to increase subsequent elements more.

Correct Approach: Initialize current_max to 0 (conceptually representing a value before the array) and process all elements uniformly.