StackOverflow2013

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14545459
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2013-01-27T07:42:43.280
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2013-01-27T08:08:41.627
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2013-01-27T08:08:41.627
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text
Body
First note that even if you let the immediate results overflow, the final result will always be correct if it can be represented. This is because integral types of any size act like cyclic groups under addition in most languages including Java (not in C, where integer overflow is undefined behavior, and C#, which is able to throw you an overflow exception). If you still want to prevent overflow, here's how to perform it in-place and in linear time: <ul> <li>split the array in-place to its negative entries (in any order) and to its positive entries (in any order). Zero can end up anywhere. In other words, perform one quick-sort step with the pivot being zero.</li> <li>Let <code>ni</code> point to the start of the array (where negative entries are located).</li> <li>Let <code>pi</code> point to the end of the array.</li> <li>Let <code>sum</code> be zero.</li> <li>While <code>pi >= ni</code> <ul> <li>if <code>sum</code> is negative <ul> <li>add <code>arr[pi]</code> to the <code>sum</code>.</li> <li>if <code>arr[pi]</code> is negative (we've run out of positive addends) and <code>sum</code> is positive (an overflow has occured), the result overflows.</li> <li>decrement <code>pi</code></li> </ul></li> <li>else <ul> <li>add <code>arr[ni]</code> to the <code>sum</code>.</li> <li>if <code>arr[ni]</code> is positive and <code>sum</code> is negative, the result overflows.</li> <li>increment <code>ni</code>.</li> </ul></li> </ul></li> <li>Finally, check if <code>sum</code> has more than <code>K</code> bits. If it does, declare the result overflows.</li> </ul>
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1. POFind sum of integer array without overflow
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1. USJohn Dvorak
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1. USJohn Dvorak
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2. VO
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1. CO(+1) Great answer.
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2. COcan you give a code for that, I mean how do you see if it has more than k bits?.
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