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majority_element.py
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94 lines (72 loc) · 1.98 KB
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'''
Majority Element
Given an array of size n, find the majority element.
The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
Input: [3, 2, 3]
Output: 3
Input: [2, 2, 1, 1, 1, 2, 2]
Output: 2
=========================================
Sort the array and the result will the middle element.
Time Complexity: O(N LogN)
Space Complexity: O(1)
Use dictionary (hash map) and count the occurrences.
The result will be the one with more than ⌊ n/2 ⌋ occurrences.
Time Complexity: O(N)
Space Complexity: O(N)
Using Boyer-Moore voting algorithm. Choose a potential majority element and for each occurence add +1, but
if the current element isn't same substract -1.
When the counter is 0, the next element becomes the new potential majority element.
Time Complexity: O(N)
Space Complexity: O(1)
'''
##############
# Solution 1 #
##############
def majority_element_1(nums):
nums.sort()
return nums[len(nums) // 2]
##############
# Solution 2 #
##############
def majority_element_2(nums):
counter = {}
for num in nums:
if num in counter:
counter[num] += 1
else:
counter[num] = 1
half = len(nums) // 2
for num in counter:
if counter[num] > half:
return num
##############
# Solution 3 #
##############
def majority_element_3(nums):
majority = 0
count = 0
for num in nums:
if count == 0:
majority = num
if num == majority:
count += 1
else:
count -= 1
return majority
###########
# Testing #
###########
# Test 1
# Correct result => 3
arr = [3, 2, 3]
print(majority_element_1(arr))
print(majority_element_2(arr))
print(majority_element_3(arr))
# Test 2
# Correct result => 2
arr = [2, 2, 1, 1, 1, 2, 2]
print(majority_element_1(arr))
print(majority_element_2(arr))
print(majority_element_3(arr))