Solution Approach

The solution uses the boundary-finding binary search template combined with a difference array to efficiently find the minimum number of queries needed.

Binary Search Template Structure: We search for the minimum k where feasible(k) returns True. The search space is [0, m] where m is the total number of queries.

left, right = 0, m
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1  # Look for smaller k
    else:
        left = mid + 1

return first_true_index if first_true_index != -1 else -1

The Feasible Function: feasible(k) determines if the first k queries provide enough reduction capacity to transform nums into a Zero Array.

1. Initialize Difference Array: Create an array d of length n + 1 filled with zeros.

2. Process First k Queries: For each query [l, r, val] in the first k queries:

   • Add val to d[l] (marking the start of the range)
   • Subtract val from d[r + 1] (marking the end of the range)

3. Verify Feasibility:

   • Initialize a running sum s = 0
   • Iterate through each index i from 0 to n-1
   • Add d[i] to the running sum s
   • If nums[i] > s, return False (insufficient capacity)

4. Return True if all indices have sufficient capacity.

Why This Template Works: The feasible function creates a monotonic pattern: False, False, ..., True, True, True. Once we have enough queries, adding more queries only provides more reduction capacity. We use the template to find the first True (minimum sufficient k).

Time Complexity: O((m + n) × log m) where binary search runs O(log m) iterations, each check takes O(m + n).

Space Complexity: O(n) for the difference array.

Example Walkthrough

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

Given:

• nums = [3, 2, 1]
• queries = [[0, 1, 2], [1, 2, 1], [0, 2, 1]]

We need to find the minimum number of queries k such that we can transform nums into a Zero Array.

Step 1: Binary Search Setup We'll binary search on the range [0, 3] (since we have 3 queries total).

Step 2: Check if k=2 queries are sufficient

Let's check if the first 2 queries [[0, 1, 2], [1, 2, 1]] provide enough reduction capacity.

Building the Difference Array:

• Initialize: d = [0, 0, 0, 0] (length n+1 = 4)
• Query 1 [0, 1, 2]:
  • d[0] += 2 → d = [2, 0, 0, 0]
  • d[2] -= 2 → d = [2, 0, -2, 0]
• Query 2 [1, 2, 1]:
  • d[1] += 1 → d = [2, 1, -2, 0]
  • d[3] -= 1 → d = [2, 1, -2, -1]

Computing Reduction Capacity via Prefix Sum:

• Index 0: s = 0 + d[0] = 0 + 2 = 2
  • Check: nums[0] = 3 > 2? Yes! Cannot reduce 3 to 0 with only 2 units.
  • Return False

Since k=2 is insufficient, binary search will try a larger value.

Step 3: Check if k=3 queries are sufficient

Using all 3 queries [[0, 1, 2], [1, 2, 1], [0, 2, 1]]:

Building the Difference Array:

• Initialize: d = [0, 0, 0, 0]
• Query 1 [0, 1, 2]: d = [2, 0, -2, 0]
• Query 2 [1, 2, 1]: d = [2, 1, -2, -1]
• Query 3 [0, 2, 1]:
  • d[0] += 1 → d = [3, 1, -2, -1]
  • d[3] -= 1 → d = [3, 1, -2, -2]

Computing Reduction Capacity:

• Index 0: s = 0 + 3 = 3
  • Check: nums[0] = 3 ≤ 3? Yes! Can reduce to 0.
• Index 1: s = 3 + 1 = 4
  • Check: nums[1] = 2 ≤ 4? Yes! Can reduce to 0.
• Index 2: s = 4 + (-2) = 2
  • Check: nums[2] = 1 ≤ 2? Yes! Can reduce to 0.
• Return True

Step 4: Binary Search Conclusion The binary search finds that:

• k=2 is insufficient
• k=3 is sufficient

Therefore, the minimum k is 3.

Key Insight: The difference array elegantly tracks how reduction capacity accumulates across overlapping ranges. When we compute the prefix sum, we get the exact total reduction available at each position, making it easy to verify if we have enough "budget" to reduce each element to zero.

Time and Space Complexity

Time Complexity: O((n + m) × log m)

The algorithm uses binary search on the range [0, m] where m is the length of queries. Binary search performs O(log m) iterations.

For each binary search iteration, the check function is called, which:

• Iterates through k queries (at most m queries) to build the difference array: O(m)
• Iterates through nums array to verify the condition using the difference array: O(n)

Therefore, each check call takes O(n + m) time, and with O(log m) binary search iterations, the total time complexity is O((n + m) × log m).

Space Complexity: O(n)

The space complexity is determined by:

• The difference array d of size n + 1: O(n)
• Other variables like s, l, k use constant space: O(1)

The total space complexity is O(n), where n is the length of the nums array.

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

Common Pitfalls

1. Using Wrong Binary Search Template Variant

A critical mistake is using the while left < right with right = mid pattern instead of the correct template.

Incorrect approach:

left, right = 0, total_queries + 1
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid
    else:
        left = mid + 1
return left if left <= total_queries else -1

Correct approach (template-compliant):

left, right = 0, total_queries
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index

The template uses while left <= right with right = mid - 1 and tracks the answer in first_true_index.

2. Incorrect Difference Array Boundary Handling

Forgetting to allocate n + 1 elements for the difference array, or incorrectly handling the right + 1 index.

Common Mistake:

# Wrong: This would cause IndexError when right == n-1
diff_array = [0] * len(nums)  # Should be len(nums) + 1
diff_array[right + 1] -= value  # Would fail for last element

Correct Approach: Always allocate one extra element to safely handle the range end marker at right + 1.

3. Not Handling the -1 Case Properly

When no valid k exists (even using all queries is insufficient), the template naturally returns -1 because first_true_index is never updated from its initial value of -1.