Problem Description

You have an array nums and need to determine the minimum number of operations to make it "positive". An array is considered positive when every subarray containing more than two elements has a positive sum.

The operation you can perform is:

• Replace any single element in nums with any integer value (between -10^18 and 10^18)

The goal is to find the minimum number of operations needed to ensure that all subarrays with length greater than 2 have positive sums.

For example, if you have a subarray of length 3 or more, its sum must be greater than 0 for the array to be considered positive. Subarrays of length 1 or 2 don't need to satisfy this condition.

The solution uses a greedy approach with a sliding window technique. It tracks:

• l: The left boundary (last position where an operation was performed)
• s: Running sum of the current segment
• pre_mx: Maximum prefix sum that can be achieved by potentially replacing elements
• ans: Count of operations needed

The algorithm iterates through the array and decides when to perform an operation. When a subarray of length > 2 would have a non-positive sum, it performs an operation (increments ans) and resets the tracking variables to start a new segment.

The key insight is that we can greedily decide to "cut" the array at certain positions by performing operations, ensuring each segment between cuts maintains the positive sum property for all its subarrays of length > 2.

Intuition

The key observation is that when we replace an element with a very large positive number (like 10^18), we can effectively "break" the array into independent segments. Any subarray containing this large element will automatically have a positive sum, so we only need to worry about subarrays within each segment.

Think of it this way: if we place a huge positive number at position i, then:

• Any subarray crossing position i will be positive (due to the huge value)
• We only need to check subarrays entirely to the left of i or entirely to the right of i

This means we can divide the problem into finding optimal "cut points" where we place these large values, creating segments that independently satisfy the condition.

For each segment between cuts, we need to ensure all its subarrays of length > 2 have positive sums. The critical insight is that if we're processing elements left to right and encounter a situation where adding the current element would create a subarray of length > 2 with non-positive sum, we should cut immediately.

Why is this greedy approach optimal? Because once we detect a violation (a subarray that would have non-positive sum), we must use at least one operation to fix it. The most efficient way is to cut right at the current position, which fixes all problematic subarrays ending at this position while giving us a fresh start for the next segment.

The variable pre_mx tracks the maximum sum we could achieve by potentially modifying earlier elements in the current segment. When s <= pre_mx for a subarray of length > 2, it means even with the best possible modification within the segment, we can't make all subarrays positive - so we must cut and start a new segment.

This greedy strategy of cutting as soon as we detect an unavoidable violation ensures we use the minimum number of operations.

Learn more about Greedy and Prefix Sum patterns.

Solution Approach

The solution implements a greedy algorithm with a sliding window technique to track segments and determine when to perform operations.

Variables Used:

• l: Marks the last position where we performed an operation (initially -1, meaning no operation yet)
• ans: Counts the number of operations performed
• pre_mx: Tracks the maximum prefix sum we could achieve in the current segment
• s: Running sum of elements in the current segment
• r: Current position in the array (right pointer)

Algorithm Steps:

1. Initialize variables: Start with l = -1 (no previous cut), ans = 0 (no operations), pre_mx = 0, and s = 0.

2. Iterate through the array: For each position r with value x:

   • Add current element to running sum: s += x

3. Check for violation: When we have a subarray of length > 2 (r - l > 2):

   • If s <= pre_mx, it means the current sum is not greater than what we could achieve by modifying previous elements
   • This indicates we cannot make all subarrays positive within this segment
   • Perform an operation:
     • Increment ans += 1
     • Set l = r (mark current position as cut point)
     • Reset pre_mx = s = 0 for the new segment

4. Update maximum prefix: When we have at least 2 elements in the segment (r - l >= 2):

   • Update pre_mx = max(pre_mx, s - x - nums[r-1])
   • This represents the maximum sum we could get by potentially replacing the last two elements

Why This Works:

The algorithm essentially decides: "Can I continue the current segment, or must I cut here?"

• When s <= pre_mx for a subarray of length > 2, it means even if we optimally modified elements within the segment, we couldn't make all subarrays positive
• By cutting at position r (replacing nums[r] with a huge positive value), we ensure all subarrays crossing this position become positive
• The greedy choice to cut as soon as a violation is detected ensures we use the minimum number of operations

Time Complexity: O(n) - single pass through the array
Space Complexity: O(1) - only using a constant amount of extra variables

Example Walkthrough

Let's walk through the algorithm with nums = [1, -2, 3, -4, 2].

Initial State:

• l = -1, ans = 0, pre_mx = 0, s = 0

Step 1: r = 0, x = 1

• Add to sum: s = 0 + 1 = 1
• Check violation: r - l = 0 - (-1) = 1 ≤ 2 (no check needed)
• Update pre_mx: r - l = 1 < 2 (no update)

Step 2: r = 1, x = -2

• Add to sum: s = 1 + (-2) = -1
• Check violation: r - l = 1 - (-1) = 2 ≤ 2 (no check needed)
• Update pre_mx: r - l = 2 ≥ 2, so pre_mx = max(0, -1 - (-2) - 1) = max(0, -1 + 2 - 1) = 0

Step 3: r = 2, x = 3

• Add to sum: s = -1 + 3 = 2
• Check violation: r - l = 2 - (-1) = 3 > 2
  • Check if s ≤ pre_mx: 2 ≤ 0? No, so continue
• Update pre_mx: pre_mx = max(0, 2 - 3 - (-2)) = max(0, 1) = 1

Step 4: r = 3, x = -4

• Add to sum: s = 2 + (-4) = -2
• Check violation: r - l = 3 - (-1) = 4 > 2
  • Check if s ≤ pre_mx: -2 ≤ 1? Yes! Violation detected
  • Perform operation: ans = 1, l = 3, reset pre_mx = 0, s = 0
• Update pre_mx: r - l = 3 - 3 = 0 < 2 (no update)

Step 5: r = 4, x = 2

• Add to sum: s = 0 + 2 = 2
• Check violation: r - l = 4 - 3 = 1 ≤ 2 (no check needed)
• Update pre_mx: r - l = 1 < 2 (no update)

Result: ans = 1

The algorithm detected that continuing the segment would create the subarray [1, -2, 3, -4] with sum -2, which violates the positive sum requirement. By performing one operation at position 3 (replacing -4 with a large positive number), we split the array into two segments: [1, -2, 3] and [2], both of which satisfy the condition.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a single pass through the input array nums using a for loop that iterates through each element exactly once. Within each iteration, all operations are constant time:

• Array access operations (nums[r-1]) take O(1)
• Arithmetic operations (addition, subtraction) take O(1)
• Comparison operations and max function take O(1)
• Variable assignments take O(1)

Since we iterate through n elements where n is the length of the input array, and each iteration performs only constant time operations, the overall time complexity is O(n).

Space Complexity: O(1)

The algorithm uses only a fixed number of variables regardless of the input size:

• Integer variables: l, ans, pre_mx, s, r, x

All these variables occupy constant space. The algorithm doesn't create any additional data structures that scale with the input size. The input array is not modified and no auxiliary arrays or other collections are created. Therefore, the space complexity is O(1) or constant space.

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

Common Pitfalls

1. Incorrect Boundary Handling After Operation

Pitfall: When performing an operation (cutting the segment), developers often incorrectly handle the boundary conditions. A common mistake is setting left = right - 1 instead of left = right, or forgetting to reset the tracking variables properly.

# INCORRECT implementation
if right - left > 2 and current_sum <= previous_max:
    operations_count += 1
    left = right - 1  # Wrong! This creates an off-by-one error
    # Forgot to reset previous_max and current_sum

Why it's wrong: Setting left = right - 1 would mean the new segment starts including the element at position right, but we've conceptually "replaced" this element to create the cut. The tracking variables also need to be reset to start fresh for the new segment.

Correct approach:

if right - left > 2 and current_sum <= previous_max:
    operations_count += 1
    left = right  # Correct: new segment starts after position right
    previous_max = 0  # Reset maximum prefix
    current_sum = 0  # Reset running sum

2. Misunderstanding the previous_max Calculation

Pitfall: Incorrectly calculating what previous_max represents, leading to wrong comparisons.

# INCORRECT calculation
elif right - left >= 2:
    previous_max = max(previous_max, current_sum)  # Wrong!

Why it's wrong: previous_max should represent the maximum sum we could achieve by potentially replacing the last two elements in the current window. It's not just the maximum sum seen so far.

Correct approach:

elif right - left >= 2:
    # Sum of middle elements (excluding first and last)
    middle_sum = current_sum - value - nums[right - 1]
    previous_max = max(previous_max, middle_sum)

3. Off-by-One Errors in Length Checks

Pitfall: Confusing when to check > 2 vs >= 2 for segment lengths.

# INCORRECT conditions
if right - left >= 2 and current_sum <= previous_max:  # Wrong comparison!
    # ...
elif right - left > 2:  # Wrong condition for update!
    # ...

Why it's wrong:

• We need to check violation when segment length is greater than 2 (subarrays of length > 2 must be positive)
• We update previous_max when segment length is at least 2 (we need at least 2 elements to have "middle" elements)

Correct approach:

if right - left > 2 and current_sum <= previous_max:
    # Check violation for segments longer than 2
  
elif right - left >= 2:
    # Update previous_max when we have at least 2 elements

4. Not Handling Edge Cases Properly

Pitfall: The algorithm might fail on edge cases like very small arrays or arrays with all negative/positive values.

Solution: Ensure the algorithm handles:

• Arrays with length ≤ 2 (should return 0 operations)
• Arrays where no operations are needed
• Arrays where multiple consecutive operations might be needed

The current implementation handles these correctly by:

• Starting with left = -1 to properly handle the first elements
• Only checking violations when segment length > 2
• Resetting properly after each operation to handle consecutive cuts