Majority Element Moore's Voting Algorithm in DSA Go - Time & Space Complexity
We want to understand how the time needed to find the majority element grows as the list gets bigger.
How many steps does the algorithm take when the input size changes?
Analyze the time complexity of the following code snippet.
func majorityElement(nums []int) int {
count := 0
candidate := 0
for _, num := range nums {
if count == 0 {
candidate = num
}
if num == candidate {
count++
} else {
count--
}
}
return candidate
}
This code finds the majority element in a list using Moore's Voting Algorithm by scanning the list once.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Single loop through the list of numbers.
- How many times: Exactly once for each element in the list.
As the list gets bigger, the number of steps grows directly with the number of elements.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 10 steps |
| 100 | About 100 steps |
| 1000 | About 1000 steps |
Pattern observation: Doubling the input size roughly doubles the work done.
Time Complexity: O(n)
This means the time to find the majority element grows in a straight line with the size of the input list.
[X] Wrong: "The algorithm checks pairs or does nested loops, so it must be slower than linear."
[OK] Correct: The algorithm only scans the list once, updating counters without extra loops, so it runs in linear time.
Understanding this linear time solution shows you can find efficient answers by clever counting, a skill valued in many coding challenges.
"What if we needed to verify the candidate by counting its occurrences in a second pass? How would the time complexity change?"
Practice
Solution
Step 1: Understand the problem Moore's Voting Algorithm solves
It is designed to find the majority element, which is the element that appears more than half the times in the array.Step 2: Compare with other options
Sorting, finding smallest element, or counting unique elements are not the algorithm's purpose.Final Answer:
To find the element that appears more than half the times -> Option AQuick Check:
Majority element = element appearing > n/2 times [OK]
- Confusing majority element with most frequent element
- Thinking it sorts the array
- Assuming it finds smallest or unique elements
Solution
Step 1: Identify correct variable types and initialization
Moore's Voting Algorithm uses an integer count starting at 0 and an integer candidate initialized to 0 or any valid integer.Step 2: Check each option for type correctness
count := 0; candidate := 0 uses count and candidate as integers with proper initialization. var count int = 1; var candidate int = -1 initializes count to 1 which is not standard start. count := 0; candidate := '' uses empty string for candidate which is wrong type. var count string = '0'; var candidate int = 0 uses string type for count which is incorrect.Final Answer:
count := 0; candidate := 0 -> Option AQuick Check:
Integer count and candidate initialized to 0 [OK]
- Using wrong data types for count or candidate
- Starting count at 1 instead of 0
- Initializing candidate with empty string
arr := []int{2, 2, 1, 1, 1, 2, 2}
count := 0
candidate := 0
for _, num := range arr {
if count == 0 {
candidate = num
}
if num == candidate {
count++
} else {
count--
}
}
fmt.Println(candidate)Solution
Step 1: Trace the loop to find candidate
Start with count=0, candidate=0.
Iteration 1: count=0, candidate=2, count=1
Iteration 2: num=2 == candidate=2, count=2
Iteration 3: num=1 != candidate=2, count=1
Iteration 4: num=1 != candidate=2, count=0
Iteration 5: count=0, candidate=1, count=1
Iteration 6: num=2 != candidate=1, count=0
Iteration 7: count=0, candidate=2, count=1Step 2: Final candidate after loop
The candidate after the loop ends is 2.Final Answer:
2 -> Option CQuick Check:
Majority element found = 2 [OK]
- Not updating candidate when count is zero
- Confusing candidate with last element
- Miscounting increments and decrements
arr := []int{3, 3, 4, 2, 4, 4, 2, 4, 4}
count := 0
candidate := 0
for i := 0; i < len(arr); i++ {
if count == 0 {
candidate = arr[i]
}
if arr[i] == candidate {
count--
} else {
count++
}
}
fmt.Println(candidate)Solution
Step 1: Analyze count update logic
The algorithm should increase count when current element equals candidate, and decrease count otherwise.Step 2: Identify the bug in code
Here, count is decreased when arr[i] == candidate and increased otherwise, which is reversed.Final Answer:
Count increments and decrements are reversed -> Option DQuick Check:
Correct count update: increment if equal, decrement if not [OK]
- Swapping increment and decrement operations
- Not updating candidate when count is zero
- Incorrect loop bounds causing runtime errors
Solution
Step 1: Understand Moore's Voting Algorithm limitation
The algorithm finds a candidate but does not guarantee it is a majority if none exists.Step 2: Adaptation to confirm majority
After finding candidate, count its occurrences in the array. If count > n/2, candidate is majority; else no majority element exists.Final Answer:
Run the algorithm to find a candidate, then verify its count exceeds half the array length -> Option BQuick Check:
Verify candidate count > n/2 to confirm majority [OK]
- Assuming candidate is always majority without verification
- Using sorting instead of Moore's Voting Algorithm
- Picking max frequency without checking majority condition
Ready to prove it?
Pass the test and this pattern moves unknown
