Majority Element Moore's Voting Algorithm in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the candidate variable for Moore's Voting Algorithm.

DSA Java

int candidate = [1];

Drag options to blanks, or click blank then click option'

A0

B-1

CInteger.MAX_VALUE

Dnums[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing candidate to 0 or -1 without considering array contents.

Using a constant like Integer.MAX_VALUE which is unrelated to array elements.

The candidate is initialized to the first element of the array to start tracking the majority element.

2fill in blank

medium

Complete the code to update the count when the current number matches the candidate.

DSA Java

if (nums[i] == candidate) {
    count [1];
}

Drag options to blanks, or click blank then click option'

A++

B--

C+=

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -- which decreases the count incorrectly.

Using = which assigns instead of increments.

When the current number matches the candidate, we increase the count by 1.

3fill in blank

hard

Fix the error in updating the candidate when count reaches zero.

DSA Java

if (count == 0) {
    candidate = [1];
    count = 1;
}

Drag options to blanks, or click blank then click option'

Ai

Bnums[i]

Cnums[0]

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using nums[0] which does not reflect the current element.

Assigning i or count which are not valid candidates.

When count is zero, we update the candidate to the current element nums[i].

4fill in blank

hard

Fill both blanks to complete the loop that implements Moore's Voting Algorithm.

DSA Java

for (int i = 1; i [1] nums.length; i++) {
    if (nums[i] == candidate) {
        count [2];
    } else {
        count--;
    }
}

Drag options to blanks, or click blank then click option'

A<

B<=

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of bounds error.

Using '--' to increment count decreases it incorrectly.

The loop runs while i is less than nums.length, and count is incremented when nums[i] equals candidate.

5fill in blank

hard

Fill all three blanks to complete the method that returns the majority element using Moore's Voting Algorithm.

DSA Java

public int majorityElement(int[] nums) {
    int candidate = [1];
    int count = 1;
    for (int i = 1; i [2] nums.length; i++) {
        if (nums[i] == candidate) {
            count[3];
        } else {
            count--;
            if (count == 0) {
                candidate = nums[i];
                count = 1;
            }
        }
    }
    return candidate;
}

Drag options to blanks, or click blank then click option'

Anums[0]

B<

C++

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of nums[0] for candidate initialization.

Using '<=' in loop condition causing errors.

Using '--' instead of '++' to increment count.

The candidate starts as nums[0], the loop runs while i < nums.length, and count is incremented with ++.

Practice

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

easy

A. To find the element that appears more than half the times

B. To sort the array in ascending order

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

   The algorithm is designed to find the majority element, which is the element that appears more than half the times in the array.

2. Step 2: Compare with other options

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

3. Final Answer:

   To find the element that appears more than half the times -> Option A

4. Quick Check:

   Majority element = element appearing > n/2 times [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 Java code snippets correctly initializes the count variable for Moore's Voting Algorithm?

easy

A. int count = null;

B. int count = -1;

C. int count = 0;

D. int count = 1;

Solution

1. Step 1: Check valid initialization for count

   Count starts at 1 because the first candidate is chosen and has one vote initially.

2. Step 2: Validate each option

   - -1 is invalid as count should not start negative.
   - 0 is incorrect because the first candidate is already counted once.
   - null is invalid for int type in Java.

3. Final Answer:

   int count = 1; -> Option D

4. Quick Check:

   Count starts at one for the first candidate [OK]

Hint: Count starts at one for the first candidate [OK]

Common Mistakes:

• Initializing count with 0 instead of 1
• Using null for primitive int
• Starting count negative

3. What will be the output of the following Java code using Moore's Voting Algorithm?

int[] arr = {2, 2, 1, 1, 1, 2, 2};
int candidate = arr[0];
int count = 1;
for (int i = 1; i < arr.length; i++) {
  if (arr[i] == candidate) count++;
  else count--;
  if (count == 0) {
    candidate = arr[i];
    count = 1;
  }
}
System.out.println(candidate);

medium

A. 2

B. 1

C. 0

D. Runtime Error

Solution

1. Step 1: Trace the candidate and count changes

   Start candidate=2, count=1
   i=1: arr[1]=2 equals candidate, count=2
   i=2: arr[2]=1 not equal, count=1
   i=3: arr[3]=1 not equal, count=0 -> reset candidate=1, count=1
   i=4: arr[4]=1 equals candidate, count=2
   i=5: arr[5]=2 not equal, count=1
   i=6: arr[6]=2 not equal, count=0 -> reset candidate=2, count=1

2. Step 2: Final candidate after loop

   Candidate is 2 after last iteration.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Final candidate = 2 [OK]

Hint: Track count changes to find final candidate [OK]

Common Mistakes:

• Not resetting candidate when count hits zero
• Confusing candidate value after last iteration
• Assuming output is the first element

4. Identify the error in this Moore's Voting Algorithm Java code snippet:

int candidate = 0;
int count = 0;
for (int num : arr) {
  if (count == 0) candidate = num;
  if (num == candidate) count++;
  else count = 0;
}
System.out.println(candidate);

medium

A. Printing candidate before loop ends

B. Candidate initialization outside loop

C. Resetting count to 0 instead of decrementing

D. Using enhanced for loop instead of index loop

Solution

1. Step 1: Analyze count update logic

   Count should decrement when current number is not candidate, but code resets count to 0.

2. Step 2: Effect of resetting count to 0

   This causes the algorithm to lose track of votes incorrectly, breaking the logic.

3. Final Answer:

   Resetting count to 0 instead of decrementing -> Option C

4. Quick Check:

   Count must decrement, not reset to zero [OK]

Hint: Decrement count when different element found [OK]

Common Mistakes:

• Resetting count to zero instead of decrementing
• Misunderstanding candidate update timing
• Confusing loop types as error

5. Given an array where no element appears more than half the time, what will Moore's Voting Algorithm return?
Example: int[] arr = {1, 2, 3, 4, 5, 6};

hard

A. The algorithm returns the element with highest frequency

B. The algorithm returns an incorrect candidate without verification

C. The algorithm throws an exception

D. The algorithm returns null

Solution

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

   The algorithm assumes a majority element exists (more than half the array).

2. Step 2: Behavior when no majority element exists

   It returns a candidate but this candidate may not actually be majority without a verification step.

3. Step 3: Verification necessity

   To confirm majority, a second pass counting candidate occurrences is needed.

4. Final Answer:

   The algorithm returns an incorrect candidate without verification -> Option B

5. Quick Check:

   No majority means candidate may be wrong [OK]

Hint: Verify candidate count if no guaranteed majority [OK]

Common Mistakes:

• Assuming algorithm throws error if no majority
• Thinking it returns null automatically
• Believing it always returns highest frequency element