Majority Element Moore's Voting Algorithm in DSA C# - Time & Space Complexity
We want to understand how the time needed to find the majority element grows as the list gets bigger.
Specifically, how many steps does the Moore's Voting Algorithm take to find the majority element?
Analyze the time complexity of the following code snippet.
int FindMajorityElement(int[] nums) {
int count = 0, candidate = 0;
foreach (int num in nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
return candidate;
}
This code finds the majority element by tracking a candidate and adjusting a count as it scans the array once.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: One loop that goes through each element of the array exactly once.
- How many times: The loop runs n times, where n is the number of elements in the array.
As the input size grows, the number of steps grows directly with it.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 10 steps |
| 100 | About 100 steps |
| 1000 | About 1000 steps |
Pattern observation: The steps increase in a straight line as the input size increases.
Time Complexity: O(n)
This means the time to find the majority element grows directly with the size of the input list.
[X] Wrong: "Because there is a loop inside a conditional, the time complexity is more than linear."
[OK] Correct: The loop runs only once through the array, and the conditional inside does not add extra loops, so the time grows linearly.
Understanding this linear time solution shows you can find efficient answers without extra space, a skill valued in real coding challenges.
"What if we added a second pass to verify the candidate's count? How would the time complexity change?"
Practice
Solution
Step 1: Understand the problem Moore's Voting Algorithm solves
The algorithm is designed to find the majority element, which is the element that appears more than half the time in the array.Step 2: Compare with other options
Sorting, finding smallest element, or counting unique elements are not the goals of this algorithm.Final Answer:
To find the element that appears more than half the time -> Option AQuick Check:
Majority element = element with >50% frequency [OK]
- Thinking it sorts the array
- Confusing majority with most frequent element without >50%
- Assuming it finds smallest or unique counts
Solution
Step 1: Understand initialization requirements
Candidate should start with a default value (like 0) and count should start at 0 before processing the array.Step 2: Check each option
int candidate = 0; int count = 0; initializes candidate to 0 and count to 0, which is standard. int candidate; int count = 1; leaves candidate uninitialized. int candidate = -1; int count = 0; uses -1 which may be valid but less common. int candidate = 0; int count = 1; starts count at 1 which is incorrect before any element is processed.Final Answer:
int candidate = 0; int count = 0; -> Option CQuick Check:
Initialize candidate=0 and count=0 before loop [OK]
- Leaving candidate uninitialized
- Starting count at 1 before processing
- Using invalid default candidate values
int[] nums = {2, 2, 1, 1, 1, 2, 2};
int candidate = 0, count = 0;
foreach (int num in nums) {
if (count == 0) candidate = num;
count += (num == candidate) ? 1 : -1;
}
Console.WriteLine(candidate);Solution
Step 1: Trace the candidate and count updates
Start with candidate=0, count=0. Iterate nums:
- num=2: count=0, candidate=2, count=1
- num=2: same candidate, count=2
- num=1: different, count=1
- num=1: different, count=0
- num=1: count=0, candidate=1, count=1
- num=2: different, count=0
- num=2: count=0, candidate=2, count=1Step 2: Final candidate after loop
The final candidate is 2 after processing all elements.Final Answer:
2 -> Option BQuick Check:
Final candidate = 2 [OK]
- Not updating candidate when count is zero
- Confusing candidate with last element
- Ignoring count increments and decrements
int[] nums = {3, 3, 4, 2, 4, 4, 2, 4, 4};
int candidate = 0, count = 0;
foreach (int num in nums) {
if (count == 0) candidate = num;
if (num == candidate) count++;
else count++;
}
Console.WriteLine(candidate);Solution
Step 1: Analyze count updates in loop
Count is incremented in both if and else branches, so count never decreases. This breaks the algorithm logic.Step 2: Correct the else branch
In Moore's Voting Algorithm, count should decrement when current number is not the candidate.Final Answer:
Count is incremented in both if and else, should decrement in else -> Option AQuick Check:
Count must decrease when num != candidate [OK]
- Incrementing count in else instead of decrementing
- Not updating candidate when count is zero
- Assuming candidate initialization causes error
int[] nums = {1, 2, 3, 2, 2, 2, 5, 2, 2}, how can you verify after Moore's Voting Algorithm that the candidate is actually the majority element?Solution
Step 1: Understand Moore's Voting Algorithm limitation
The algorithm finds a candidate but does not guarantee it is majority if none exists.Step 2: Verify candidate by counting occurrences
Count how many times candidate appears and compare with half the array length to confirm majority.Final Answer:
Count the occurrences of candidate in the array and check if it is more than half the array length -> Option DQuick Check:
Verify candidate count > n/2 after algorithm [OK]
- Skipping verification step
- Assuming candidate is always majority
- Using sorting instead of counting
Ready to prove it?
Pass the test and this pattern moves unknown
