16 lines
485 B
Crystal
16 lines
485 B
Crystal
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# the trick is that at any point
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# in the iteration if starting a new chain is
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# better than your current score with this element
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# added to it, then do so.
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# the interesting part is proving the math behind it
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def subarray_sum(arr)
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curr = max = 0
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first, last, curr_first = arr.size, 0, 0
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arr.each_with_index do |e, i|
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curr += e
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e > curr && (curr = e; curr_first = i)
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curr > max && (max = curr; first = curr_first; last = i)
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end
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return max, arr[first..last]
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end
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