![]() ![]() In the worst case, all numbers could be distinct, and then all of them would share the same place with exactly one occurrence. Suppose that we try to count occurrences of each number in the array in order to find which one occurs more times than others. ![]() With array mode, the difficulty comes from the fact that many numbers can share the winning position. There is even simpler and more efficient algorithm for finding majority element, which is covered in exercise Finding a Majority Element in an Array In case of proper (absolute) majority element, it is sufficient to find the median of the array and that element would either be the majority element, or the array would not contain the majority element at all. This is a different requirement compared to the Majority Element exerciseīut this loose requirement actually makes the problem harder to solve. It is sufficient for the winner to just appear more often than the second most frequent element. ![]() In this exercise we are dealing with simple majority, meaning that the winning element is not required to populate more than half of the array. ![]()
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