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  1. PO
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    <p>Here's a simple C version of the solution that takes advantage of the Ai + Ak must be even test:</p> <pre><code>#include &lt;stdio.h&gt; static int arr[] = {9, 4, 2, 3, 6, 10, 3, 3, 10}; int main () { int i, j, k; int sz = sizeof(arr)/sizeof(arr[0]); int count = 0; for (i = 0; i &lt; sz - 2; i++) { for (k = i + 2; k &lt; sz; k++) { int ik = arr[i] + arr[k]; int ikdb2 = ik / 2; if ((ikdb2 * 2) == ik) // if ik is even { for (j = i + 1; j &lt; k; j++) { if (arr[j] == ikdb2) { count += 1; printf("{%d, %d, %d}\n", arr[i], arr[j], arr[k]); } } } } } printf("Count is: %d\n", count); } </code></pre> <p>and the console dribble:</p> <pre><code>tmp e$ cc -o triples triples.c tmp e$ ./triples {9, 6, 3} {9, 6, 3} {2, 6, 10} {2, 6, 10} {3, 3, 3} Count is: 5 tmp e$ </code></pre> <p>This more complicated version keeps a list of Aj indexed by value to go from n-cubed to n-squared (kinda).</p> <pre><code>#include &lt;stdio.h&gt; #include &lt;stdint.h&gt; static uint32_t arr[] = {9, 4, 2, 3, 6, 10, 3, 3, 10}; #define MAX_VALUE 100000u #define MAX_ASIZE 30000u static uint16_t index[MAX_VALUE+1]; static uint16_t list[MAX_ASIZE+1]; static inline void remove_from_index (int subscript) { list[subscript] = 0u; // it is guaranteed to be the last element uint32_t value = arr[subscript]; if (value &lt;= MAX_VALUE &amp;&amp; subscript == index[value]) { index[value] = 0u; // list now empty } } static inline void add_to_index (int subscript) { uint32_t value = arr[subscript]; if (value &lt;= MAX_VALUE) { list[subscript] = index[value]; // cons index[value] = subscript; } } int main () { int i, k; int sz = sizeof(arr)/sizeof(arr[0]); int count = 0; for (i = 0; i &lt; sz - 2; i++) { for (k = i; k &lt; sz; k++) remove_from_index(k); for (k = i + 2; k &lt; sz; k++) { uint32_t ik = arr[i] + arr[k]; uint32_t ikdb2 = ik / 2; add_to_index(k-1); // A(k-1) is now a legal middle value if ((ikdb2 * 2) == ik) // if ik is even { uint16_t rover = index[ikdb2]; while (rover != 0u) { count += 1; printf("{%d, %d, %d}\n", arr[i], arr[rover], arr[k]); rover = list[rover]; } } } } printf("Count is: %d\n", count); } </code></pre> <p>and the dribble:</p> <pre><code>tmp e$ cc -o triples triples.c tmp e$ ./triples {9, 6, 3} {9, 6, 3} {2, 6, 10} {2, 6, 10} {3, 3, 3} Count is: 5 tmp e$ </code></pre>
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    1. POfastest algorithm count number of 3 length AP in array
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    1. USDoug Currie
    1. USDoug Currie
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      1. VTUpMod
    1. COI also thought of the same solution but the complexity of the solution is too high. Since n is 100,000 therefore n^2 solution does not suffice
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      1. PO
      1. USFluvid
 

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