```python
def climb_budget(tolls: list) -> int:
    """Calculate the minimum cost to climb a scaffold given tolls for each rung.
    
    The climber starts below the first rung. Each move can step onto the next rung
    or skip exactly one rung. The climb ends once above the top rung.
    Returns 0 for an empty board or a board with a single rung (which can be skipped).
    Raises ValueError for invalid tolls (non-digit characters or leading zeros).
    """
    if not tolls:
        return 0
    
    # Validate all tolls first
    for i, toll_str in enumerate(tolls):
        if not isinstance(toll_str, str):
            raise ValueError(f"Toll at position {i} is not a string")
        if not toll_str:
            raise ValueError(f"Toll at position {i} is empty")
        if not toll_str.isdigit():
            raise ValueError(f"Toll at position {i} contains non-digit characters")
        if len(toll_str) > 1 and toll_str[0] == '0':
            raise ValueError(f"Toll at position {i} has a leading zero")
    
    n = len(tolls)
    # If there's only one rung, the climber can skip it (step from ground to above top)
    # Cost is 0.
    if n == 1:
        return 0
    
    # Convert tolls to integers
    costs = [int(t) for t in tolls]
    
    # Dynamic programming approach
    # dp[i] represents the minimum cost to reach rung i (0-indexed)
    # The climber starts "below" rung 0.
    # Moves: from current position, can step to next rung or skip one rung.
    # This means from position i, you can go to i+1 or i+2.
    # The climb ends once above the top rung (index n-1), so we need to reach index n or beyond.
    # Actually, let's redefine:
    # We want to find the minimum cost to get "above" the last rung.
    # Let's say positions are 0, 1, ..., n-1 (the rungs).
    # Start at position -1 (ground).
    # From position i, you can move to i+1 or i+2.
    # You pay the toll of the rung you land on.
    # You stop when you are at position >= n.
    
    # dp[i] = min cost to reach rung i
    # dp[0] = cost[0] (step from ground to rung 0)
    # dp[1] = cost[1] (step from ground to rung 1, skipping rung 0)
    # For i >= 2: dp[i] = cost[i] + min(dp[i-1], dp[i-2])
    
    # Final answer is min cost to reach position n (above last rung)
    # From n-1, you can step to n (cost 0, no toll)
    # From n-2, you can skip to n (cost 0, no toll)
    # So final answer is min(dp[n-1], dp[n-2]) if we consider reaching "above" as the end.
    # Wait, if we are at rung n-1, we can step to n (above top). Cost added is 0.
    # If we are at rung n-2, we can skip