slightly more understandable variation of the solution (in python)

arr = [0] * (n + 1)

for query in queries:
    # where a represents the lower bound of the range,
    # and b represents the upper bound.
    a, b, k = query
    arr[a] += k
    try:
        arr[b+1] -= k
    except IndexError:
        pass

max_val = running_sum = 0

for delta in arr:
    # delta can be either positive or negative,
    # but in any case it is added to the running sum.
    # however max_val always contains the greatest sum so far
    running_sum += delta
    if max_val < running_sum:
        max_val = running_sum

return max_val