
Problem Statement
The task requires identifying the majority element within an array nums
, where the majority element is defined as the one that appears more than ⌊n / 2⌋
times in the array (n
being the size of the array). The condition given assures us that a majority element always exists. The definition of the array's majority element pivots on its frequent appearance. Essentially, this element appears in the array more frequently than all the other elements combined, occurring in more than half of the array's total entries.
Examples
Example 1
Input:
nums = [3,2,3]
Output:
3
Example 2
Input:
nums = [2,2,1,1,1,2,2]
Output:
2
Constraints
n == nums.length
1 <= n <= 5 * 104
-109 <= nums[i] <= 109
Approach and Intuition
To determine the majority element in the given examples, let's break down the steps that could be intuitively used, seen from the example inputs and outputs provided:
- Direct Observation: Simply by looking at the arrays for majority counts:
- For
[3, 2, 3]
, number3
appears 2 times in an array of length 3, which is more than⌊3/2⌋ = 1
. Hence,3
is the majority element. - In
[2, 2, 1, 1, 1, 2, 2]
,2
appears 4 times out of an array of length 7, which is more than⌊7/2⌋ = 3
. Thus,2
is the majority element.
- For
Frequency Calculation: Considering a straightforward method, we could maintain a frequency count for each element and return the one that has a count greater than
⌊n / 2⌋
.- Count each element’s frequency.
- Traverse through the count and return the element with the highest frequency that also satisfies the more than half condition.
Constraints Overview:
- Array size constraints ensure that naive methods (like counting frequencies for every element) are feasible for small
n
but might introduce performance issues whenn
reaches its upper limit (50,000). Thus, more efficient algorithms (like Boyer-Moore Voting Algorithm) might be considered in those cases, which operate in linear time and constant space. - Value constraints are wide (
-109
to109
), indicating a solution should accurately manage potential integer overflow or discrepancies in negative and positive majority scenarios.
- Array size constraints ensure that naive methods (like counting frequencies for every element) are feasible for small
From the overview, this problem can be tackled using various approaches such as hashing (using a map to count occurrences), sorting and picking the middle element (since the majority element would dominate the sorted median), or optimal divide-and-conquer methods that split the array and count majorities in halves. However, specific constraints direct the need for more efficient linear time solutions.
Solutions
- C++
- Java
- C
- JavaScript
- Python
class Solution {
public:
int findMajority(vector<int>& elements) {
int tally = 0;
int majorElement;
for (int& element : elements) {
if (tally == 0) {
majorElement = element;
}
tally += (element == majorElement) ? 1 : -1;
}
return majorElement;
}
};
The provided C++ code defines a solution class with a function findMajority
designed to find the majority element in a vector of integers. This function utilizes the Boyer-Moore voting algorithm, which efficiently determines the element that occurs more than n/2 times in the vector, where n is the size of the vector.
Here’s a high-level breakdown of how this algorithm functions in the code:
- Initialize
tally
to zero. This acts as a counter to track the difference between the number of timesmajorElement
has been encountered and times other elements have been encountered. - Initialize
majorElement
without a preset value. This variable will hold the candidate for the majority element. - Traverse through each
element
in theelements
vector using a for loop.- If
tally
is zero, updatemajorElement
to the currentelement
. - Increment or decrement
tally
depending if the currentelement
is the same asmajorElement
.
- If
- At the end of the loop,
majorElement
should hold the majority element since any majority would offset the count of all other elements.
Make sure to include the appropriate library if you are copying this code into a separate project:
#include <vector>
to support the vector container use in the function argument.
This code's strength lies in its linear time complexity O(n) and constant extra space O(1), making it highly efficient for processing large datasets.
class Solution {
public int findMajority(int[] elements) {
int tally = 0;
Integer major = null;
for (int element : elements) {
if (tally == 0) {
major = element;
}
tally += (element == major) ? 1 : -1;
}
return major;
}
}
The given Java solution addresses the problem of finding the majority element in an array. The method findMajority
employs the Boyer-Moore Voting Algorithm, which is efficient for this purpose.
Follow the process outlined below to understand how the solution works:
- Initialize two variables:
tally
to keep track of the count andmajor
to hold the potential majority element. - Iterate through each element in the array:
- If
tally
is zero, set the current element as the potential majority element (major
). - Increase the
tally
by 1 if the current element is the same asmajor
, otherwise decrease it by 1.
- If
- By the end of the loop,
major
will contain the element that might be the majority in the array.
This method runs in linear time, making it highly efficient for large arrays. The use of the Boyer-Moore Voting Algorithm ensures that only a single pass through the data is necessary to find the majority element, if one exists.
int findMajority(int* array, int length) {
int tally = 0;
int leader = 0;
for (int index = 0; index < length; index++) {
if (tally == 0) {
leader = array[index];
}
tally += (array[index] == leader) ? 1 : -1;
}
return leader;
}
This summary provides a guide on a function written in C to identify the majority element in an array. The function findMajority
takes two parameters: a pointer to an integer array array
and an integer length
representing the size of the array.
The algorithm follows the Boyer-Moore voting principle. Here’s a breakdown of how it works:
- Initialize two integers,
tally
to track the net number of times a candidate appears andleader
to temporarily store the candidate for majority element. Start withtally
at 0 andleader
at the first element of the array. - Loop through each element of the array using a for loop.
- If
tally
is zero, update theleader
to the current element. - If the current array element is equal to
leader
, incrementtally
by 1. If not, decrementtally
by 1. - After the loop completes,
leader
holds the value of the candidate that may be the majority element.
The approach ensures an efficient scan through the array, using O(n) time complexity and O(1) space complexity, where n is the number of elements in the array. This method is effective for finding a potential majority element but does not verify if the leader
actually appears more than n/2
times. Additional checks might be necessary depending on the problem requirements.
var findMajority = function (elements) {
let tally = 0;
let major = null;
for (let element of elements) {
if (tally == 0) {
major = element;
}
tally += element == major ? 1 : -1;
}
return major;
};
The provided JavaScript solution determines the majority element within an array by utilizing the Boyer-Moore Voting Algorithm. This algorithm efficiently finds the element that appears more than half the time in the array, commonly known as the majority element.
In essence, the algorithm maintains a counter (tally
) to determine the potential majority element (major
). As the array elements
is iterated through:
- When
tally
is zero, set the current element as the new potential majority element. - Increase
tally
by one if the current element equals the potential majority element. If it doesn't, decreasetally
by one.
The final value of major
at the conclusion of this process is the majority element because any element not in the majority would have resulted in decreasing the tally, effectively not allowing it to carry over as the majority element.
This approach is very efficient with a time complexity of O(n), making it suitable for large datasets. The space complexity is O(1) as it only uses two additional variables regardless of the input size.
To implement this solution in your code:
- Define a function
findMajority
that takes an arrayelements
. - Initialize
tally
to 0 andmajor
to null. - Loop through each
element
ofelements
. - Adjust
tally
andmajor
based on the above conditions. - Return
major
which should be the majority element if one exists. If no majority element exists, the function will return the last setmajor
, which could be invalid, so further validation might be required depending on the problem constraints.
class Solution:
def findMajority(self, elements):
elem_count = 0
majority = None
for element in elements:
if elem_count == 0:
majority = element
elem_count += 1 if element == majority else -1
return majority
The provided Python code defines a class Solution
with a method findMajority
which is specifically designed to find the majority element in a given list, named elements
. The majority element is defined as an element that appears more than n/2
times in the list, where n
is the total number of elements in the list.
- In the algorithm, two variables
elem_count
andmajority
are initialized.majority
is used to keep track of the potential candidate for the majority element. - The program enters a loop iterating through each element in the provided list.
- If
elem_count
is zero (meaning no candidate is currently considered or previous candidates have been invalidated), it sets the currentelement
as the new candidate for a majority by assigning it tomajority
. - It then adjusts
elem_count
: incrementing it if the currentelement
matches the currentmajority
candidate, or decrementing it otherwise. - This adjustment essentially means that if the current element matches the suspected majority element, it gains a stronger position as a candidate; if it doesn’t, it weakens its position.
- After processing all elements via the loop, the
majority
variable, which now holds the candidate that has not been fully invalidated, is returned as the majority element of the list.
In terms of robustness, note that this algorithm effectively identifies a majority element if there is one, using Boyer-Moore voting algorithm logic, which is efficient and works in linear time complexity O(n) with constant space complexity O(1).
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