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Algorithms & data structures

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4 matches
Algorithms & data structures easy

Find Elements Appearing More Than n/3 Times in Python

Return all elements that occur more than len(array)/3 times using a simple dictionary counter.

majority-element dictionary counting
Python
def majority_third(arr):
    """Return elements appearing more than len(arr)/3 times."""
    cutoff = len(arr) / 3
    counts = {}
    for x in arr:
        counts[x] = counts.get(x, 0) + 1
    return [x for x, c in counts.items() if c > cutoff]


if __name__ == "__main__":
    test1 = [3, 2, 3]
    test2 = [1, 1, 1, …
12 0 Open
Algorithms & data structures easy

Find First Duplicate Index in Python

Return the index of the first element that appears more than once in a list, using a dictionary for O(n) time.

duplicate dictionary arrays
Python
def find_first_duplicate(arr):
    seen = {}
    for index, value in enumerate(arr):
        if value in seen:
            return index
        seen[value] = index
    return -1

if __name__ == "__main__":
    test_array = [3, 5, 2, 8, 5, 1, 2]
    result = find_first_duplicate(test_array)
    print(f"Array: {test_arr…
13 0 Open
Algorithms & data structures easy

How to Count Occurrences of Each Value in Python

Count how many times each value appears in a list using Python's Counter from the collections module.

counter counting collections
Python
from collections import Counter

def count_occurrences(values):
    """Return a dictionary mapping each value to its count."""
    return dict(Counter(values))

if __name__ == "__main__":
    sample_data = ["apple", "banana", "apple", "cherry", "banana", "apple"]
    result = count_occurrences(sample_data)
    print(r…
10 0 Open
Algorithms & data structures easy

Rearrange array alternately max min in Python

Rearranges a sorted list so its elements alternate between the current maximum and current minimum using two pointers in O(n) time.

two-pointers array sorting
Python
def rearrange_alternately(arr):
    """
    Rearrange sorted array so elements alternate: max, min, next max, next min...
    Returns a new list in O(n) time using O(n) space.
    """
    n = len(arr)
    result = []
    left, right = 0, n - 1
    while left <= right:
        if left == right:
            result.appen…
14 0 Open

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