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    <p>I'm not going to bother with a line-by-line code analysis since (a) you're changing it while people are answering your question and (b) it would literally take too long. But here's a summary (non-exhaustive) of klunkers:</p> <ol> <li>You are leaking memory (newArr) on each loop iteration starting with the second. </li> <li>You're out-of-bounds on your array access multiple times.</li> <li>You should not <em>need</em> to use a pointer array <em>at all</em> to solve this. A single array of dimension [N][2] where N is the number of unique values.</li> </ol> <p>One (of countless many) way you can solve this problem is presented below:</p> <pre><code>#include &lt;iostream&gt; #include &lt;algorithm&gt; int main() { // 0. Declare array and length int arr[]={6,7,3,1,3,2,4,4,7,5,1,1,5,6,6,4,5}; const size_t NR = sizeof(arr)/sizeof(arr[0]); // 1. sort the input array std::sort(arr, arr+NR); /* alternaive sort. for this input size bubble-sort is more than adequate, in case your limited to not being allowed to use the standard library sort */ /* for (size_t i=0;i&lt;NR;++i) for (size_t j=i+1;j&lt;NR;++j) if (arr[i] &gt; arr[j]) { arr[i] ^= arr[j]; arr[j] ^= arr[i]; arr[i] ^= arr[j]; } */ // 2. single scan to determine distinct values size_t unique = 1; for (size_t i=1;i&lt;NR;++i) if (arr[i] != arr[i-1]) unique++; // 3. Allocate a [unique][2] array int (*newArr)[2] = new int[unique][2]; // 4. Walk array once more, accumulating counts size_t j=0; newArr[j][0] = arr[0]; newArr[j][1] = 1; for (size_t i=1;i&lt;NR;++i) { if (arr[i] != arr[i-1]) { newArr[++j][0] = arr[i]; newArr[j][1] = 0; } ++newArr[j][1]; } // 5. Dump output for (size_t i=0;i&lt;unique;++i) cout &lt;&lt; newArr[i][0] &lt;&lt; " : " &lt;&lt; newArr[i][1] &lt;&lt; endl; delete [] newArr; return EXIT_SUCCESS; } </code></pre> <p><strong>Output</strong></p> <pre><code>1 : 3 2 : 1 3 : 2 4 : 3 5 : 3 6 : 3 7 : 2 </code></pre>
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