algoadvance

Design a Neighbor Sum Service that supports the following operations on an integer array:

1. insert(idx: int, val: int)
   • Inserts val at index idx. If idx is out of bounds, do nothing.
2. update(idx: int, val: int)
   • Updates the integer at index idx to be val. If idx is out of bounds, do nothing.
3. sum_neighbors(idx: int) -> int
   • Returns the sum of the integer at index idx and its neighbors (i.e., idx-1, idx, and idx+1). If idx is out of bounds, return 0.

Example

nss = NeighborSumService([1, 2, 3, 4])
nss.insert(1, 5)  # Now, the array is [1, 5, 2, 3, 4]
nss.update(2, 8)  # Now, the array is [1, 5, 8, 3, 4]
print(nss.sum_neighbors(2))  # Returns 16 (5 + 8 + 3)
print(nss.sum_neighbors(4))  # Returns 7 (3 + 4)

Clarifying Questions

1. What should the service do if an insert operation is attempted at an index that is out of bounds?
   • Assume that nothing should happen if the index is out of bounds.
2. What should the service do if an update operation is attempted at an index that is out of bounds?
   • Assume that nothing should happen if the index is out of bounds.
3. What should the sum_neighbors function return if the given index has no valid neighbors?
   • It should sum the values of any existing valid neighbors only.

Strategy

• Use a list to store the elements.
• Operations like insertions and updates will be straightforward list operations.
• For sum_neighbors, check bounds before accessing the array to avoid out-of-bounds errors.

Code

class NeighborSumService:
    def __init__(self, initial_array):
        # Initialize with the given integer array
        self.array = initial_array
    
    def insert(self, idx, val):
        # Check index is within bounds
        if 0 <= idx <= len(self.array):
            self.array.insert(idx, val)
    
    def update(self, idx, val):
        # Check index is within bounds
        if 0 <= idx < len(self.array):
            self.array[idx] = val
    
    def sum_neighbors(self, idx):
        if 0 <= idx < len(self.array):
            neighbor_sum = 0
            # Include value at idx
            neighbor_sum += self.array[idx]
            # Include value at idx - 1 if valid
            if idx - 1 >= 0:
                neighbor_sum += self.array[idx - 1]
            # Include value at idx + 1 if valid
            if idx + 1 < len(self.array):
                neighbor_sum += self.array[idx + 1]
            return neighbor_sum
        # If idx is out of bounds, return 0
        return 0

Time Complexity

• insert(idx: int, val: int):
  • Average case: O(n), where n is the length of the array, due to the potential reallocation of elements.
  • Worst case: O(n) due to similar reallocation.
• update(idx: int, val: int):
  • O(1), as updating an element in a list by index is a constant-time operation.
• sum_neighbors(idx: int) -> int:
  • O(1), because checking and summing up to three elements involves a fixed number of operations.

Feel free to ask for any further clarifications or modifications!

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