Skip to content

Latest commit

 

History

History

229. MajorityElement-ll

Folders and files

NameName
Last commit message
Last commit date

parent directory

..
README.md
README.md
 
 
Solution.cpp
Solution.cpp
 
 
Solution.java
Solution.java
 
 

README.md

The logic behind this code is to find the majority elements in an input array nums. A majority element is defined as an element that appears more than n/3 times in the array, where n is the length of the array. The code achieves this by iterating through the sorted array and keeping track of the count of each element.

Here's a step-by-step explanation of the logic:

1. Initial Setup:

  • An empty ArrayList called result is created to store the majority elements found.
  • If the length of the input array nums is less than 3, it's a special case because in small arrays, any element that appears once or twice is considered a majority element. In this case, the code simply adds unique elements to the result list (ensuring no duplicates) and returns it.

2. Sorting:

  • If the input array has 3 or more elements, it sorts the array in ascending order using Arrays.sort(nums). Sorting the array simplifies the process of counting the occurrences of each element because identical elements will be adjacent to each other after sorting.

3. Counting Elements:

  • The code then iterates through the sorted array using a loop.
  • It maintains a count variable initialized to 1 to keep track of how many times the current element appears.
  • If the current element is the same as the next element (i.e., nums[i] == nums[i+1]), it increments the count because it has encountered another occurrence of the same element.
  • If the count is greater than 1 (meaning there is a repeated element) and the current element is not the same as the next element, it means that the current element has ended its consecutive appearances. In this case, the count is reset to 1.

4. Checking for Majority Elements:

  • After counting the occurrences of each element in the sorted array, the code checks if the count of the current element is greater than or equal to gt + 1, where gt is calculated as nums.length / 3. If this condition is met, it means that the current element appears more than n/3 times in the original unsorted array.
  • When this condition is satisfied, the current element is added to the result list (if it's not already in the list to avoid duplicates), and the count is reset to 1.

5. Final Result:

  • Once the loop has iterated through the entire sorted array, the result list contains the majority elements that satisfy the condition of appearing more than n/3 times in the input array.