Convert an array to reduced form | Set 1 (Simple and Hashing) | GeeksforGeeks

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  • Опубликовано: 14 янв 2025

Комментарии • 7

  • @gauravkvtamboli2660
    @gauravkvtamboli2660 4 года назад +1

    An alternate thing which can be done is, after sorting elements, instead of using Hashing, we can do a binary search of the element in sorted temp array
    1. After Sorting the temp array, iterate over the original array
    2. pick an element of the original array and search in a sorted array
    3. return index from binary search and update the same in the original array
    Hashing takes O(n) space and searching becomes O(1) and in Binary search, search time is O(log(n)) and space - O(1)

  • @gabrielzaragoza8220
    @gabrielzaragoza8220 5 лет назад +1

    On one hand your code has std:unordered_map which seems relatively modern. On the other hand you pass a C style array with the length, which seems older and error prone. Is there some reason you mix the two styles? Did you want to pass std::vector ?

  • @youd87
    @youd87 5 лет назад +2

    Why do we need a map? Cant we just sort the array and then assign them values from 0 to n-1?

  • @wecan2729
    @wecan2729 3 года назад

    void convert(int arr[], int n)
    {
    int t[n];
    for(int i=0;i

  • @gauravanand6937
    @gauravanand6937 7 лет назад +1

    What about the duplicates??

    • @GeeksforGeeksVideos
      @GeeksforGeeksVideos  7 лет назад +3

      It is mentioned in the problem statement that the array has distinct elements. :)