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    <blockquote> <p>You are given an array of n+2 elements. All elements of the array are in range 1 to n. And all elements occur once except two numbers which occur twice</p> </blockquote> <p>Lets modify this slightly, and go with just <code>n</code>, not <code>n+2</code>, and the first part of the problem statement, it becomes</p> <blockquote> <p>You are given an array of n elements. All elements of the array are in range 1 to n</p> </blockquote> <p>So now you know you have an array, the numbers in the array start at 1 and go up by one for every item in the array. So if you have 10 items, the array will contain the numbers 1 to 10. 5 items, you have 1 to 5 and so forth.</p> <p>It follows that the numbers stored in the array can be used to index the array. i.e. you can always say <code>A[A[i]]</code> where i &lt;= size of A. e.g. <code>A={5,3,4,1,2}; print A[A[2]]</code></p> <p>Now, lets add in one duplicate number. The algorithm takes the value of each number in the array, and visits that index. We know if we visit the same index twice, we know we have found a duplicate.<br> How do we know if we visit the same index twice?<br> Yup, we change the sign of the number in each index we visit, if the sign has already changed, we know we've already been here, ergo, this index (not the value stored at the index) is a duplicate number.</p> <p>You could achieve the same result by keeping a second array of booleans, initialised to false. That algroithm becomes</p> <pre><code>A={123412} B={false, false, false, false} for(i = 1; i &lt;= 4; i++) { if(B[A[i]]) // Duplicate else B[A[i]] = true; } </code></pre> <p>However in the MS question you're changing the sign of the element in A instead of setting a boolean value in B.</p> <p>Hope this helps, </p>
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    1. POFind the two repeating elements in a given array
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    1. CO+1. For me, it takes your explanation to understand the other explanations.
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    2. COThanks Binary Worrier for your lucid explanation.
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    3. CO@yogi15490: You're more than welcome sir (If you're happy that this answers your question, you could mark it as the Accepted Answer, thanks).
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