Question
Given an array of integers where each value 1 <= x <= len(array), write a function that finds all the duplicates in the array.
eg.
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dups([1, 2, 3]) = [] dups([1, 2, 2]) = [2] dups([3, 3, 3]) = [3] dups([2, 1, 2, 1]) = [1, 2] |
Once you think that you’ve solved the problem, click below to see the solution.
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Solution
How was that problem? You can check out the solution in the video below.
Here is the source code for the solution shown in the video (Github):
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// Return a list of duplicates in the array. To avoid using extra space, // we flag which elements we've seen before by negating the value at // indexed at that value in the array. public static List<Integer> findDuplicates(int[] arr) { // Use a set for results to avoid duplicates Set<Integer> resultSet = new HashSet<Integer>(); for (int i = 0; i < arr.length; i++) { // Translate the value into an index (1 <= x <= len(arr)) int index = Math.abs(arr[i]) - 1; // If the value at that index is negative, then we've already seen // that value so it's a duplicate. Otherwise, negate the value at // that index so we know we've seen it if (arr[index] < 0) { resultSet.add(Math.abs(arr[i])); } else { arr[index] = -arr[index]; } } // Return the array to it's original state for (int i = 0; i < arr.length; i++) { arr[i] = Math.abs(arr[i]); } return new ArrayList(resultSet); } |
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