Solution #31 - Daniel/Edited - 11.03.2025 #25
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2 problems to be added. |
danzang100
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JRS296
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Mar 17, 2025
danzang100
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JRS296
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Mar 21, 2025
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| Three possible methods: | ||
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| 1. Sorting: | ||
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| - Time Complexity: O(nlogn) | ||
| - Space Complexity: O(1) | ||
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| Simply sort the array and the middle element should always be the majority element. | ||
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| 2. Hashmap: | ||
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| - Time Complexity: O(n) | ||
| - Space Complexity: O(n) | ||
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| Map the frequencies of the elements to a hashmap and return the element with the highest frequency. | ||
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| 3. Moore's Voting Algorithm: | ||
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| - Time Complexity: O(n) | ||
| - Space Complexity: O(1) | ||
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| Initialise count and candidate element. If the element is the candidate element, increment count. Else if count is 0, change the candidate element to be the current element, else decrement the count. | ||
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| Works on the principle that upon complete iteration of the array, the majority element should always have the highest count and therefore end up being the candidate element all the time. |
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Kindly add the corresponding solutions as well, and also explain the time and space complexity + Try to give a brief on the various algorithms (+ Link articles) on your approach to this.
JRS296
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Mar 21, 2025
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| To solve this question, | ||
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| 1. We find the first decreasing element from the right calling it the pivot. | ||
| 2. We check if that element does not exist, which means that the array is sorted in non-ascending order, hence we just return a reversed array. | ||
| 3. If the element exists, we find the smallest element greater than the pivot element which is to the right of it. | ||
| 4. We swap the two elements. | ||
| 5. We reverse the suffix of the pivot element to obtain the answer. |
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Try to maintain consistency with the approach to explanations. Can be much better!
JRS296
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Mar 21, 2025
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