Reason About an Endpoint-Only Interval Score

Problem Statement

Consider an interval defined by two indices [start, end] in a zero-indexed array A of length N. Let the "score" of this interval be defined as the sum of elements from index start to end inclusive, minus the sum of elements in the strictly interior range (start, end).

1. Mathematically demonstrate why this score is equivalent to A[start] + A[end] if start < end, and explain the behavior when start == end.
2. Implement a function calculate_interval_score(A, start, end) that returns this value.
3. Given an array A, implement a function find_max_score_interval(A) that returns the maximum possible score for any valid interval 0 <= start <= end < N. If multiple intervals yield the same maximum score, return the interval that minimizes the start index, then minimizes the end index.

Examples

# Example 1
A = [1, 5, 2, 8, 3]
# For interval [1, 3]: A[1..3] = [5, 2, 8]. Interior = [2].
# Score = (5+2+8) - (2) = 13.
# Equivalent to A[1] + A[3] = 5 + 8 = 13.

# Example 2
A = [10, -2, 10]
# For interval [0, 2]: A[0..2] = [10, -2, 10]. Interior = [-2].
# Score = (10 - 2 + 10) - (-2) = 18.
# Equivalent to A[0] + A[2] = 10 + 10 = 20... Wait, ensure the logic handles negatives.

Constraints

• 1 <= N <= 10^5
• -10^6 <= A[i] <= 10^6

Edge Cases

• start == end: The interior is empty (sum = 0), so the score should be A[start].
• Single element arrays.