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POReducing time complexity using HashMap, or is there a better method?
 

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  1. POReducing time complexity using HashMap, or is there a better method?
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    <p>Right, this is from an older exam which i'm using to prepare my own exam in january. We are given the following method:</p> <pre><code>public static void Oorspronkelijk() { String bs = "Dit is een boodschap aan de wereld"; int max = -1; char let = '*'; for (int i=0;i&lt;bs.length();i++) { int tel = 1; for (int j=i+1;j&lt;bs.length();j++) { if (bs.charAt(j) == bs.charAt(i)) tel++; } if (tel &gt; max) { max = tel; let = bs.charAt(i); } } System.out.println(max + " keer " + let); } </code></pre> <p>The questions are:</p> <ol> <li>what is the output? - Since the code is just an algorithm to determine the most occuring character, the output is "6 keer " (6 times space)</li> <li>What is the time complexity of this code? Fairly sure it's O(n²), unless someone thinks otherwise?</li> <li>Can you reduce the time complexity, and if so, how?</li> </ol> <p>Well, you can. I've received some help already and managed to get the following code:</p> <pre><code>public static void Nieuw() { String bs = "Dit is een boodschap aan de wereld"; HashMap&lt;Character, Integer&gt; letters = new HashMap&lt;Character, Integer&gt;(); char max = bs.charAt(0); for (int i=0;i&lt;bs.length();i++) { char let = bs.charAt(i); if(!letters.containsKey(let)) { letters.put(let,0); } int tel = letters.get(let)+1; letters.put(let,tel); if(letters.get(max)&lt;tel) { max = let; } } System.out.println(letters.get(max) + " keer " + max); } </code></pre> <p>However, I'm uncertain of the time complexity of this new code: Is it O(n) because you only use one for-loop, or does the fact we require the use of the HashMap's get methods make it O(n log n) ?</p> <p>And if someone knows an even better way of reducing the time complexity, please do tell! :)</p>
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    1. COI'm not good in determining complexity and wounder - why O(n^2)? The problem spot for me is: In the 2nd loop u iterate not N times, but (N - M, and M increasing by 1 on each iteration).
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    2. CO@Vanger that means the second loop will be performed (1+2+3+...+N) times which is [(N+1)*N]/2 = (N^2+N)/2 = O(N^2)
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    3. CO@yurib Thx :) totally forgot series.
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