Majority Element Moore's Voting Algorithm in DSA C - Execution Trace

Concept Flow - Majority Element Moore's Voting Algorithm

Initialize candidate and count

↓

Iterate over array elements

↓

If count == 0

→Set candidate = current element

↓

If current element == candidate

→Increment count

↓

Else

→Decrement count

↓

Repeat until end of array

↓

Candidate is majority element

Start with no candidate and zero count. For each element, if count is zero, pick new candidate. Increase count if element matches candidate, else decrease. At end, candidate is majority.

Execution Sample

DSA C

int majorityElement(int* nums, int numsSize) {
    int count = 0, candidate = 0;
    for (int i = 0; i < numsSize; i++) {
        if (count == 0) candidate = nums[i];
        count += (nums[i] == candidate) ? 1 : -1;
    }
    return candidate;
}

Finds the majority element in the array using Moore's Voting Algorithm.

Execution Table

Step	Operation	Current Element	Candidate	Count	Visual State
1	Start	N/A	None	0	[]
2	Read element	2	None	0	[]
3	Count is 0, set candidate	2	2	0	[2]
4	Element == candidate, increment count	2	2	1	[2(count=1)]
5	Read element	2	2	1	[2(count=1)]
6	Element == candidate, increment count	2	2	2	[2(count=2)]
7	Read element	1	2	2	[2(count=2)]
8	Element != candidate, decrement count	1	2	1	[2(count=1)]
9	Read element	1	2	1	[2(count=1)]
10	Element != candidate, decrement count	1	2	0	[2(count=0)]
11	Read element	2	2	0	[2(count=0)]
12	Count is 0, set candidate	2	2	0	[2]
13	Element == candidate, increment count	2	2	1	[2(count=1)]
14	Read element	2	2	1	[2(count=1)]
15	Element == candidate, increment count	2	2	2	[2(count=2)]
16	End of array	N/A	2	2	[2(count=2)]

💡 Reached end of array; candidate 2 is majority element.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 6	After Step 8	After Step 10	After Step 12	After Step 13	Final
candidate	None	2	2	2	2	2	2	2	2
count	0	0	1	2	1	0	0	1	2

Key Moments - 3 Insights

Why do we reset the candidate when count reaches zero?

When count is zero (see step 3 and 12), it means previous candidate lost majority support. We pick the current element as new candidate to try again.

Why do we increment count when element equals candidate and decrement otherwise?

Incrementing count means current candidate is supported by this element (steps 4,6,13,15). Decrementing means opposition (steps 8,10). This balances votes to find majority.

How do we know the candidate at the end is the majority element?

Moore's Voting Algorithm guarantees that if a majority element exists, it will be the candidate after full iteration (step 16).

Visual Quiz - 3 Questions

Look at the execution table, what is the candidate after step 10?

A 1

B 2

C None

D 0

💡 Hint

Check the 'Candidate' column at step 10 in the execution_table.

At which step does the count first become zero after starting?

A Step 4

B Step 12

C Step 10

D Step 8

💡 Hint

Look at the 'Count' column in execution_table and find when it changes to zero.

If the array started with element 1 instead of 2, how would the candidate change at step 3?

A Candidate would be 1

B Candidate would be 2

C Candidate would be None

D Candidate would be 0

💡 Hint

At step 3, candidate is set to current element when count is zero.

Concept Snapshot

Moore's Voting Algorithm finds majority element in O(n) time.
Initialize candidate and count=0.
For each element: if count=0, set candidate to element.
If element == candidate, increment count else decrement.
At end, candidate is majority element if one exists.

Full Transcript

This visualization shows Moore's Voting Algorithm to find the majority element in an array. We start with no candidate and count zero. We scan each element. When count is zero, we pick the current element as candidate. If the element matches candidate, we increase count; otherwise, we decrease count. This balances votes. At the end, the candidate is the majority element. The execution table tracks each step, showing candidate and count changes. Key moments clarify why candidate resets and how count changes. The quiz tests understanding of candidate and count at specific steps. This method efficiently finds majority element in one pass.

Practice

1. What is the main purpose of Moore's Voting Algorithm in arrays?

easy

A. To sort the array in ascending order

B. To find the element that appears more than half the time

C. To find the smallest element in the array

D. To reverse the array elements

Solution

1. 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 time in the array.

2. Step 2: Compare other options

   Sorting, finding smallest, or reversing are unrelated to majority element detection.

3. Final Answer:

   To find the element that appears more than half the time -> Option B

4. Quick Check:

   Majority element = element with > n/2 occurrences [OK]

Hint: Majority element appears more than half times [OK]

Common Mistakes:

• Confusing majority element with most frequent element
• Thinking it sorts the array
• Assuming it finds smallest or largest element

2. Which of the following is the correct initialization step in Moore's Voting Algorithm in C?

easy

A. int count = 0, candidate = -1;

B. int count = 1, candidate = 0;

C. int candidate = 0, count = 0;

D. int candidate = -1, count = 1;

Solution

1. Step 1: Identify correct variable initialization

   Moore's Voting Algorithm starts with count = 0 and candidate = -1 (or any invalid index) to indicate no candidate chosen yet.

2. Step 2: Check other options

   Other options start count or candidate incorrectly, which can cause logic errors.

3. Final Answer:

   int count = 0, candidate = -1; -> Option A

4. Quick Check:

   Start with no candidate and zero count [OK]

Hint: Initialize count=0 and candidate invalid (-1) [OK]

Common Mistakes:

• Starting count at 1 instead of 0
• Assigning candidate to 0 prematurely
• Swapping order of variables

3. What is the output of the following C code snippet using Moore's Voting Algorithm?

int arr[] = {2, 2, 1, 1, 1, 2, 2};
int n = 7;
int count = 0, candidate = -1;
for (int i = 0; i < n; i++) {
  if (count == 0) {
    candidate = arr[i];
  }
  if (arr[i] == candidate) {
    count++;
  } else {
    count--;
  }
}
printf("%d", candidate);

medium

A. 1

B. 0

C. 2

D. No majority element

Solution

1. Step 1: Trace the candidate and count updates

   Start count=0, candidate=-1
   i=0: count=0, candidate=2, count=1
   i=1: arr[i]=2 == candidate=2, count=2
   i=2: arr[i]=1 != candidate=2, count=1
   i=3: arr[i]=1 != candidate=2, count=0
   i=4: count=0, candidate=1, count=1
   i=5: arr[i]=2 != candidate=1, count=0
   i=6: count=0, candidate=2, count=1

2. Step 2: Final candidate after loop

   Candidate is 2 after loop ends.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Majority element found = 2 [OK]

Hint: Track count and candidate changes stepwise [OK]

Common Mistakes:

• Not resetting candidate when count hits zero
• Confusing candidate with last element
• Ignoring count decrement on mismatch

4. Identify the bug in this Moore's Voting Algorithm implementation snippet:

int count = 0, candidate = -1;
for (int i = 0; i < n; i++) {
  if (count == 0) {
    candidate = arr[i];
  }
  if (arr[i] != candidate) {
    count++;
  } else {
    count--;
  }
}

medium

A. Candidate initialized incorrectly

B. Candidate is not updated when count is zero

C. Loop does not run for all elements

D. Count increment and decrement conditions are reversed

Solution

1. Step 1: Check count update logic

   Correct logic: increment count if element equals candidate, decrement otherwise.
   Given code increments count when arr[i] != candidate, which is wrong.

2. Step 2: Confirm candidate update and loop correctness

   Candidate updates correctly when count == 0.
   Loop runs correctly for all elements.
   Candidate initialization is valid.

3. Final Answer:

   Count increment and decrement conditions are reversed -> Option D

4. Quick Check:

   Increment on match, decrement on mismatch [OK]

Hint: Increment count on match, decrement on mismatch [OK]

Common Mistakes:

• Swapping increment and decrement conditions
• Not updating candidate when count is zero
• Incorrect loop bounds

5. Given an array int arr[] = {3, 3, 4, 2, 4, 4, 2, 4, 4}, how can you verify if the candidate found by Moore's Voting Algorithm is actually a majority element?

hard

A. Count occurrences of candidate and check if it is more than n/2

B. Check if candidate is at the first position of the array

C. Verify if candidate is the maximum element in the array

D. No verification needed; candidate is always majority

Solution

1. Step 1: Understand Moore's Voting Algorithm limitation

   The algorithm finds a candidate but does not guarantee it is majority if none exists.

2. Step 2: Verify candidate by counting occurrences

   Count how many times candidate appears in array.
   If count > n/2, candidate is majority; else no majority element.

3. Final Answer:

   Count occurrences of candidate and check if it is more than n/2 -> Option A

4. Quick Check:

   Verify candidate count > n/2 to confirm majority [OK]

Hint: Always verify candidate count > n/2 after algorithm [OK]

Common Mistakes:

• Assuming candidate is always majority without check
• Checking candidate position instead of count
• Confusing majority with max element