Finished timeSpacePractice#9
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auberonedu
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auberonedu
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Nice job! I think your final method doesn't quite meet the space complexity requirement though, see comments below.
| // Time Complexity: O(n) | ||
| // Space Complexity: O(n^2) | ||
| // Does the 'T' look confusing? Consider refreshing on generic methods | ||
| // We'll revisit generics as a class later | ||
| public static <T> Map<T, Integer> countFrequencies(T[] array) { |
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Take another look at space complexity for this one. How big is each entry going to be? How many entries are there going be total?
| // in O(n) time. n = nums.size() | ||
| return -1; | ||
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| HashMap<Integer, Integer> frequencies = new HashMap<>(); |
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Remember to use interface types where appropriate (Map).
| for (int i = 0; i < nums.length; i++) { | ||
| if (frequencies.containsKey(nums[i])) { | ||
| int count = frequencies.get(nums[i]); | ||
| count++; | ||
| frequencies.put(nums[i], count); | ||
| } else { | ||
| frequencies.put(nums[i], 1); | ||
| } | ||
| if (frequencies.get(nums[i]) >= highestCount) { | ||
| highestCount = frequencies.get(nums[i]); | ||
| mostCommon = nums[i]; | ||
| } |
| Map<Integer, Integer> mostCommonMap = new HashMap<>(); | ||
| // Keeps track of the most common number | ||
| int mostCommonCount = 0; | ||
| int mostCommonInt = nums[0]; | ||
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| // Loops over the input array to add each number to the HashMap | ||
| // Sets each numbers count to 1 | ||
| for (int num : nums) { | ||
| mostCommonMap.put(num, mostCommonMap.getOrDefault(num, 0)); | ||
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| // Variable that targets the count, that being the number that represents the | ||
| // amount of times a number appears in the array | ||
| int count = mostCommonMap.get(num); | ||
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| if (count > mostCommonCount) { | ||
| mostCommonCount = count; | ||
| mostCommonInt = num; | ||
| } | ||
| } |
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I like the idea, but this doesn't get us to our desired space complexity of O(1). Consider how the size of mostCommonMap will increase with our input.
| @Test | ||
| void mostCommonTimeEfficient() { | ||
| int[] nums = { 0, 1, 1, 2, 3, 4, 5, 6, 7, 8 }; | ||
| assertEquals(1, Practice.mostCommonTimeEfficient(nums)); | ||
| } | ||
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| @Test | ||
| void mostCommonSpaceEfficient() { | ||
| int[] nums = { 2, 2, 2, 2, 4, 4, 6, 6, 6, 8, 9, 10 }; | ||
| assertEquals(2, Practice.mostCommonSpaceEfficient(nums)); | ||
| } |
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I like these tests, but there needs to be more of them to give us more confidence that our code is working. What other cases could you consider testing here?
3 participants
Both methods have been completed and a testcase made for each of them.