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    <p>My idea:</p> <ul> <li>Generate all squares between 0 and 2147483647</li> <li>For each X: <ul> <li>count = 0</li> <li>Have 2 iterators - one starting from the left and one starting from the right</li> <li>If <code>left + right = X</code>, increase the count, increase the left iterator and decrease the right one</li> <li>If <code>left + right &gt; X</code>, decrease the right one</li> <li>If <code>left + right &lt; X</code>, increase the left one</li> <li>Stop when <code>left</code> bypasses <code>right</code> </li> </ul></li> </ul> <p>As an optimization, we don't need to start all the way on the right, we can instead do a binary search for the starting position.</p> <p>Test code:</p> <pre><code>private static int binarySearch(long[] a, long key) { int low = 0; int high = a.length - 1; while (low &lt;= high) { int mid = (low + high) &gt;&gt;&gt; 1; long midVal = a[mid]; if (midVal &lt; key) low = mid + 1; else if (midVal &gt; key) high = mid - 1; else return mid; } return Math.min(low, a.length-1); } public static void main(String[] args) { long[] squares = new long[46341]; // ceil(sqrt(2147483647)) = 46341 long val = 0; int pos = 0; while (true) { long square = val*val; // sanity check, can also use pos &gt;= squares.length if (square &gt; 2147483647l) break; squares[pos++] = square; val++; } int X = 10; int left = 0; int right = binarySearch(squares, X); int count = 0; for (; left &lt;= right; ) { //Collections.b long l = squares[left] + squares[right]; if (l == X) { count++; left++; right--; } else if (l &gt; X) { right--; } else { left++; } } System.out.println(count); } </code></pre>
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    1. PODouble Squares (Facebook Hacker Cup 2011) in Java
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    1. USDukeling
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    1. COWhy not just generate all squares up to the given N? For example if it's 10, then you only need to generate 1^2, 2^2, 3^2 because 4^2 is greater than 10.
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      1. USShashank
    2. CO@ShashankGupta Because you're probably given many X's, so you only need to generate these squares once, and you can then reuse them for each X.
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      1. USDukeling
    3. COAn alternative to “Generate all squares between 0 and 2147483647” is “Read target values into array D; compute Dmax = {max value in D}; generate all squares between 0 and Dmax”. But the scope for time savings is minimal because it probably only takes a half millisecond to generate the 46340 squares below 2^31.
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      1. USJames Waldby - jwpat7
 

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