Majority Element

Clarify the problem:

• The problem requires finding the majority element in an array, which is an element that appears more than n/2 times, where n is the size of the array.
• We need to implement a function that takes an array of integers as input and returns the majority element.

Analyze the problem:

• Input: An array of integers.
• Output: The majority element.
• Constraints:
  • The array is non-empty.
  • The majority element always exists in the array.

Design an algorithm:

• We can solve this problem using the Boyer-Moore Voting Algorithm.
• The algorithm uses a variable to keep track of the potential majority element and a counter to maintain the count.
• We initialize the potential majority element as the first element of the array and the count as 1.
• We iterate over the remaining elements of the array.
• If the current element is equal to the potential majority element, we increment the count.
• If the current element is different from the potential majority element, we decrement the count.
• If the count becomes zero, we update the potential majority element to the current element and reset the count to 1.
• At the end of the iteration, the potential majority element will be the majority element.

Explain your approach:

• We will implement a function called majorityElement that takes an array of integers as input.
• We initialize two variables: majority to store the potential majority element and count to maintain the count.
• We set majority to the first element of the array and count to 1.
• We iterate over the array starting from the second element.
• If the current element is equal to majority, we increment count by 1.
• If the current element is different from majority, we decrement count by 1.
• If count becomes zero, we update majority to the current element and set count back to 1.
• Finally, we return majority as the majority element.

Write clean and readable code:

python

def majorityElement(nums):
    majority = nums[0]
    count = 1

    for i in range(1, len(nums)):
        if nums[i] == majority:
            count += 1
        else:
            count -= 1
            if count == 0:
                majority = nums[i]
                count = 1

    return majority

Test your code:

python

# Test case 1
# Input: [3, 2, 3]
# Majority element: 3
# Expected output: 3
assert majorityElement([3, 2, 3]) == 3

# Test case 2
# Input: [2, 2, 1, 1, 1, 2, 2]
# Majority element: 2
# Expected output: 2
assert majorityElement([2, 2, 1, 1, 1, 2, 2]) == 2

Optimize if necessary:

• The provided solution already uses the optimal Boyer-Moore Voting Algorithm for finding the majority element.
• There is no need for further optimization.

Handle error cases:

• The code assumes that the input array is non-empty.
• If the input array is empty, the behavior is undefined.

Discuss complexity analysis:

• Let n be the size of the input array.
• The time complexity of the solution is O(n) since we iterate over the array once.
• The space complexity is O(1) since we use a constant amount of extra space to store the majority and count variables.