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def maxMaintainedRepos(resolved: list[int], required: list[int], k: int) -> int:
    """
    Calculates the maximum number of repositories that can be marked as maintained
    by optimally distributing k issue resolutions.
    """
    n = len(resolved)
    
    # 1. Calculate the 'need' for each repository
    # needed[i] = max(0, required[i] - resolved[i])
    needs = []
    
    # Repositories already maintained
    maintained_count = 0 
    
    for i in range(n):
        need = required[i] - resolved[i]
        
        if need <= 0:
            # Repository is already maintained (resolved[i] >= required[i])
            maintained_count += 1
        else:
            # Repository needs 'need' more fixes
            needs.append(need)
            
    # 2. Filter and Sort the positive needs in ascending order
    # We want to fix the cheapest repositories first
    needs.sort()
    
    # 3. Spend Greedily
    for need in needs:
        if k >= need:
            # We have enough fixes to maintain this repository
            k -= need
            maintained_count += 1
        else:
            # Not enough fixes for this one (or any remaining, since they are sorted)
            break
            
    return maintained_count