Moore's Voting Algorithm
Finding the Majority Element: Brute Force vs. Moore's Voting Algorithm
The Majority Element problem asks us to find the element that appears more than half the time in a given array. Let's explore two approaches: brute force and Moore's Voting Algorithm, highlighting their trade-offs.
1. Brute Force (O(n^2) Time, O(1) Space)
The brute-force method checks the count of each element by iterating through the array for every element.
Java
public int majorityElementBruteForce(int[] nums) {
int n = nums.length;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (nums[i] == nums[j]) {
count++;
}
}
if (count > n / 2) {
return nums[i];
}
}
return -1; // No majority element
}
Advantage: Simple to understand.
Disadvantage: Inefficient for large datasets.
2. Moore's Voting Algorithm (O(n) Time, O(1) Space)
Moore's Voting Algorithm is a clever approach using a candidate and a counter.
Java
public int majorityElementMoore(int[] nums) {
int n = nums.length;
int candidate = nums[0];
int count = 1;
for (int i = 1; i < n; i++) {
if (nums[i] == candidate) {
count++;
} else {
count--;
if (count == 0) {
candidate = nums[i];
count = 1;
}
}
}
// Verification (Important!)
int actualCount = 0;
for (int num : nums) {
if (num == candidate) {
actualCount++;
}
}
if (actualCount > n / 2) {
return candidate;
} else {
return -1; // No majority element
}
}
Advantage: Highly efficient, linear time complexity.
Disadvantage: Slightly more complex to understand.
Which Approach to Choose?
Moore's Voting Algorithm is the clear winner for its efficiency. The brute-force method should only be considered for very small datasets. Understanding Moore's Voting Algorithm is crucial for any programmer dealing with array-based problems.