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POAlgorithm: Max Counters
 

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    <p>I have the following problem:</p> <p>You are given N counters, initially set to 0, and you have two possible operations on them:</p> <ul> <li>increase(X) − counter X is increased by 1, </li> <li>max_counter − all counters are set to the maximum value of any counter.</li> </ul> <p>A non-empty zero-indexed array A of M integers is given. This array represents consecutive operations:</p> <ul> <li>if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),</li> <li>if A[K] = N + 1 then operation K is max_counter.</li> </ul> <p>For example, given integer N = 5 and array A such that:</p> <pre><code>A[0] = 3 A[1] = 4 A[2] = 4 A[3] = 6 A[4] = 1 A[5] = 4 A[6] = 4 </code></pre> <p>the values of the counters after each consecutive operation will be:</p> <pre><code>(0, 0, 1, 0, 0) (0, 0, 1, 1, 0) (0, 0, 1, 2, 0) (2, 2, 2, 2, 2) (3, 2, 2, 2, 2) (3, 2, 2, 3, 2) (3, 2, 2, 4, 2) </code></pre> <p>The goal is to calculate the value of every counter after all operations.</p> <p>I did the following solution but it runs at O(NK) where K = length of array A.</p> <pre><code>public int[] solution(int N, int[] A) { int[] result = new int[N]; int maximum = 0; for (int K = 0; K &lt; A.Length; K++) { if (A[K] &lt; 1 || A[K] &gt; N + 1) throw new InvalidOperationException(); if (A[K] &gt;= 1 &amp;&amp; A[K] &lt;= N) { result[A[K] - 1]++; if (result[A[K] - 1] &gt; maximum) { maximum = result[A[K] - 1]; } } else { // inefficiency here for (int i = 0; i &lt; result.Length; i++) result[i] = maximum; } } return result; } </code></pre> <p>Could anyone show me how this can be better done with O(N + K) where K is the length of array A? Sorry for may terrible coding, I am doing these exercises to improve my programming. Thanks!</p>
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    1. COI don't know too much about complexity analysis, but your code/algorithm looks like O(N) to me -- you iterate over the `A` array only once and if all counters are to be maxed sometimes, the for-loop you chose seems efficient enough.
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    2. COI tried this in codility and scored only 71. I know there is a much better solution than mine.
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      1. USRandal Cunanan
    3. COIts not O(n) since when setting everything to maximum is O(n) and assuming every call is to set Maximum, O(n) * O(n) = O(N^2). Btw code is wrong since maximun is updated on the run, and its not the global maximun.
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