1426. Counting Elements-Two alternative solutions #206
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issue: #150 |
altay9
reviewed
May 15, 2021
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+4
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| public int countElements(int[] arr) { | ||
| boolean[] check = new boolean[1002]; | ||
| int count = 0; | ||
| for (int n : arr) { | ||
| check[n] = true; | ||
| } | ||
| for (int n : arr) { | ||
| if (check[n + 1]) { | ||
| count++; | ||
| } | ||
| } | ||
| return count; | ||
| } | ||
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altay9
reviewed
May 15, 2021
Comment on lines
+7
to
+20
| public int countElements(int[] arr) { | ||
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| Set<Integer> nums = new HashSet<>(); | ||
| int count = 0; | ||
| for (int n : arr) { | ||
| nums.add(n); | ||
| } | ||
| for (int n : arr) { | ||
| if (nums.contains(n + 1)) { | ||
| count++; | ||
| } | ||
| } | ||
| return count; | ||
| } |
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This is quite fine as well.
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How does the current solution with sorting perform?
altay9
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Thank you for these nice solutions Erdem.
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We can actually use the constraints given in the question to store the existence of numbers in a boolean array. The array is looped once again and we check if number+1 is true in the array. I personally do not really like this solution as it wouldn't be valid if the constraints were something unknown to us and not limited by LeetCode or the programming language itself (For example,
Integertype in Haskell represents an unbounded integer value).The other solution is a bit more general, and can be pretty fast as well depending on the implementation of the HashSet. We add all the numbers in the array to a set and after that, check if the sought value is in the set.