Problem Description

You have an integer array nums (0-indexed) and a positive integer k.

You can perform this operation as many times as you want:

• Select any contiguous subarray of exactly k elements
• Decrease every element in that subarray by 1

Your goal is to determine if it's possible to make all elements in the array equal to 0 through these operations.

Return true if you can reduce all array elements to 0, or false if it's impossible.

Key Points:

• A subarray must be contiguous (elements next to each other)
• Each operation affects exactly k consecutive elements
• You subtract 1 from all elements in the chosen subarray per operation
• You can perform the operation any number of times
• You can choose any valid subarray of size k for each operation

Example scenarios:

• If you have nums = [2, 2, 3, 1, 1, 0] and k = 3, you could:
  • Apply operation on indices 0-2: [1, 1, 2, 1, 1, 0]
  • Apply operation on indices 1-3: [1, 0, 1, 0, 1, 0]
  • Apply operation on indices 0-2: [0, -1, 0, 0, 1, 0]
  • This would result in a negative number, which means we've gone too far

The challenge is determining whether there exists a sequence of operations that reduces every element to exactly 0, without any element going negative at any point.

Intuition

Let's think about this problem step by step. When we look at the leftmost element nums[0], we have a crucial observation: this element can only be decreased by operations that start at index 0. Why? Because any operation starting at index 1 or later won't include nums[0].

So if nums[0] = 5, we must perform exactly 5 operations starting at index 0. There's no other way to make nums[0] equal to 0. This means we'll decrease elements nums[0] through nums[k-1] by 5.

After handling nums[0], we move to nums[1]. At this point, nums[1] might have already been reduced by some amount from the operations we did for nums[0]. Whatever value remains at nums[1], that's the exact number of operations we need to perform starting at index 1.

This pattern continues: for each position i, we must perform exactly nums[i] operations (after accounting for previous operations) starting at position i.

Now, how do we efficiently track the effect of all these operations? If we naively update all k elements for each operation, we'd have a time complexity issue. This is where the difference array technique comes in.

Instead of directly modifying the array elements, we can track the "changes" at specific positions:

• When we start an operation at position i that decreases k elements by some value x, we mark that the decrease starts at i (subtract x from our running sum)
• We mark that the decrease ends at position i+k (add x back to restore the running sum)

By maintaining a running sum s of these changes as we traverse the array, we can determine the actual value at each position after all previous operations.

The key insight is that we process the array from left to right, and at each position, we know exactly how many operations we need to perform. If at any point:

• The remaining value becomes negative (we've decreased too much)
• We need to perform an operation but don't have enough elements remaining (i + k > n)

Then it's impossible to make all elements 0, and we return false.

Learn more about Prefix Sum patterns.

Solution Approach

Let's implement the solution using a difference array combined with prefix sum to efficiently track the operations.

Step 1: Initialize the data structures

• Create a difference array d of size n + 1 to track where operations start and end
• Initialize a running sum s = 0 to track the cumulative effect of operations

Step 2: Process each element from left to right

For each position i with value nums[i]:

1. Update the running sum: Add d[i] to s. This accounts for any operations that start or end at position i.

2. Calculate the actual value: Add s to nums[i]. This gives us the current value after all previous operations have been applied.

   x = nums[i] + s

3. Check if already zero: If x = 0, we can skip to the next element.

4. Validate the operation:

   • If x < 0, it means previous operations have reduced this element below 0, which is invalid. Return false.
   • If i + k > n, we don't have enough elements to perform a subarray operation of size k. Return false.

5. Apply the operation:

   • We need to perform x operations starting at position i
   • Update the running sum: s -= x (this affects positions i through i+k-1)
   • Mark where the operation ends: d[i + k] += x (at position i+k, we restore the sum)

Why this works:

• The difference array d[i] stores the change that occurs at position i
• When we subtract x from s, it affects all subsequent positions until we add it back at d[i + k]
• This simulates decreasing elements in the range [i, i+k-1] by x

Example walkthrough:

If nums = [2, 2, 3, 1, 1, 0] and k = 3:

• Position 0: x = 2, perform 2 operations on [0, 2], set s = -2, d[3] = 2
• Position 1: s = -2, x = 2 + (-2) = 0, skip
• Position 2: s = -2, x = 3 + (-2) = 1, perform 1 operation on [2, 4], set s = -3, d[5] = 1
• Position 3: s = -3 + 2 = -1 (d[3] = 2), x = 1 + (-1) = 0, skip
• Position 4: s = -1, x = 1 + (-1) = 0, skip
• Position 5: s = -1 + 1 = 0 (d[5] = 1), x = 0 + 0 = 0, skip

All elements can be reduced to 0, so return true.

Time Complexity: O(n) - single pass through the array Space Complexity: O(n) - for the difference array

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: nums = [3, 1, 2] and k = 2

We'll use a difference array d and running sum s to track operations.

Initial Setup:

• d = [0, 0, 0, 0] (size n+1 = 4)
• s = 0

Processing each position:

Position 0 (nums[0] = 3):

• Update running sum: s = s + d[0] = 0 + 0 = 0
• Calculate actual value: x = nums[0] + s = 3 + 0 = 3
• Since x = 3 > 0, we need to perform 3 operations starting at position 0
• Check if valid: i + k = 0 + 2 = 2 ≤ 3 ✓
• Apply operation:
  • s = s - x = 0 - 3 = -3 (this affects positions 0 and 1)
  • d[i + k] = d[2] = 3 (mark where the effect ends)
• State: d = [0, 0, 3, 0], s = -3

Position 1 (nums[1] = 1):

• Update running sum: s = s + d[1] = -3 + 0 = -3
• Calculate actual value: x = nums[1] + s = 1 + (-3) = -2
• Since x = -2 < 0, this means position 1 has been decreased below 0 by previous operations
• Return false - it's impossible to make all elements 0

Why it failed: When we performed 3 operations starting at position 0, we decreased both nums[0] and nums[1] by 3. This made nums[0] go from 3 to 0 (good!), but nums[1] went from 1 to -2 (bad!). We can't have negative values at any point.

Let's try another example that works: nums = [2, 1, 1] and k = 2

Initial Setup:

• d = [0, 0, 0, 0]
• s = 0

Position 0 (nums[0] = 2):

• Update running sum: s = 0 + d[0] = 0
• Calculate actual value: x = 2 + 0 = 2
• Apply 2 operations starting at position 0
• s = -2, d[2] = 2

Position 1 (nums[1] = 1):

• Update running sum: s = -2 + d[1] = -2
• Calculate actual value: x = 1 + (-2) = -1
• Since x = -1 < 0, return false

This also fails! The issue is that the operations on position 0 affect position 1 too much.

Final working example: nums = [1, 1, 2] and k = 2

Position 0 (nums[0] = 1):

• s = 0, x = 1 + 0 = 1
• Apply 1 operation starting at position 0
• s = -1, d[2] = 1

Position 1 (nums[1] = 1):

• s = -1 + 0 = -1
• x = 1 + (-1) = 0
• Since x = 0, skip (already zero after previous operations)

Position 2 (nums[2] = 2):

• s = -1 + 1 = 0 (d[2] = 1 restores the sum)
• x = 2 + 0 = 2
• Need to apply 2 operations, but i + k = 2 + 2 = 4 > 3
• Return false - not enough elements for the operation

This demonstrates how the algorithm correctly identifies when operations are impossible due to insufficient array length.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array exactly once with a single for loop that processes each element. Within each iteration, all operations (array access, arithmetic operations, and comparisons) are performed in constant time O(1).

The space complexity is O(n), where n is the length of the array nums. The algorithm creates an auxiliary array d of size n + 1 to store the difference values. Additionally, it uses a constant amount of extra space for variables s, i, and x, but these don't affect the overall space complexity which is dominated by the array d.

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

Common Pitfalls

1. Misunderstanding the Operation Direction

A common mistake is thinking we need to track which operations to perform rather than greedily applying operations from left to right. Some might try to find an optimal sequence of operations using backtracking or dynamic programming, which overcomplicates the problem.

Why this happens: The problem statement allows choosing "any valid subarray," which might suggest we need to consider all possible orderings.

Solution: Recognize that if a solution exists, there's always a valid sequence where we process elements from left to right. Once we reach position i, we must reduce it to exactly 0 before moving on, as we can't apply operations that start before position i once we've passed it.

2. Incorrect Handling of the Difference Array

Developers often make errors when updating the difference array, particularly:

• Forgetting to add the restoration value at position i + k
• Using difference_array[i + k - 1] instead of difference_array[i + k] (the operation affects indices [i, i+k-1], so the restoration happens at index i + k)

Example of incorrect code:

# WRONG: This would restore the value one position too early
difference_array[i + k - 1] += effective_value  

# CORRECT: Restore at position i + k (after the k-element window)
difference_array[i + k] += effective_value

3. Not Accounting for Cumulative Effects

A critical mistake is forgetting that operations accumulate. When processing position i, you must consider all previous operations that still affect this position.

Example of incorrect approach:

# WRONG: Only checking the original value
if nums[i] < 0:
    return False

# CORRECT: Check the effective value after all cumulative operations
effective_value = nums[i] + cumulative_decrement
if effective_value < 0:
    return False

4. Off-by-One Errors in Boundary Checking

When checking if we can perform an operation starting at position i, the condition should be i + k > n, not i + k >= n or i + k - 1 > n.

Why: An operation starting at position i affects positions [i, i+k-1]. The last valid starting position is when i + k - 1 = n - 1, which means i + k = n.

# WRONG: This would reject valid operations at the boundary
if i + k >= n:  # or if i + k - 1 >= n:
    return False

# CORRECT: Allow operations where the last affected element is at index n-1
if i + k > n:
    return False

5. Modifying the Original Array

Some solutions mistakenly modify nums[i] directly instead of tracking the effective value separately:

# WRONG: This modifies the input and loses track of the original values
nums[i] += cumulative_decrement
if nums[i] < 0:
    return False

# CORRECT: Calculate effective value without modifying the original array
effective_value = nums[i] + cumulative_decrement
if effective_value < 0:
    return False

This is problematic because:

• It modifies the input array (poor practice unless explicitly allowed)
• Future iterations might need the original values for validation
• It makes debugging harder as you lose the initial state