Day 6 - Majority Element II.md

Difficulty	Source	Tags
Medium	160 Days of Problem Solving	Arrays Greedy Two Pointers

🚀 Day 6. Majority Element II 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given an array of integers arr[] where each number represents a vote for a candidate. Return the candidates that have votes greater than one-third of the total votes. If there's no majority vote, return an empty array.

Note: The answer should be returned in an increasing order.

🔍 Example Walkthrough:

Input:
arr[] = [2, 1, 5, 5, 5, 5, 6, 6, 6, 6, 6]
Output:
[5, 6]

Explanation:
5 and 6 occur more than n / 3 times in the array.

Input:
arr[] = [1, 2, 3, 4, 5]
Output:
[]

Explanation:
No candidate occurs more than n / 3 times.

Constraints:

• 1 <= arr.size() <= 10^6
• -10^9 <= arr[i] <= 10^9

🎯 My Approach:

1. Boyer-Moore Voting Algorithm:

   • The problem can be efficiently solved using the Boyer-Moore Voting Algorithm to find the top two candidates with the potential to exceed n / 3 votes.
   • First, identify the two potential majority elements.
   • Then, verify their counts to ensure they actually exceed the threshold.
   • This approach reduces the problem to a linear pass through the array.

2. Steps:

   • Traverse the array to find two majority candidates (num1 and num2) using count variables.
   • Traverse again to count occurrences of the candidates and validate the result.
   • Ensure the final output is sorted to meet the problem requirements.

🕒 Time and Auxiliary Space Complexity📝

• Expected Time Complexity: O(n), where n is the size of the array. The algorithm requires two linear scans of the array, making it efficient.
• Expected Auxiliary Space Complexity: O(1), as we use only a constant amount of additional space.

📝 Solution Code

Code (C)

int* findMajority(int arr[], int n, int* resultSize) {
    int num1 = -1, num2 = -1, c1 = 0, c2 = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == num1) {
            c1++;
        } else if (arr[i] == num2) {
            c2++;
        } else if (c1 == 0) {
            num1 = arr[i];
            c1 = 1;
        } else if (c2 == 0) {
            num2 = arr[i];
            c2 = 1;
        } else {
            c1--;
            c2--;
        }
    }
    c1 = c2 = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == num1) c1++;
        else if (arr[i] == num2) c2++;
    }
    int* result = (int*)malloc(2 * sizeof(int));
    *resultSize = 0;
    if (c1 > n / 3) result[(*resultSize)++] = num1;
    if (c2 > n / 3) result[(*resultSize)++] = num2;
    if (*resultSize == 2 && result[0] > result[1]) {
        int temp = result[1];
        result[1] = result[0];
        result[0] = temp;
    }

    return result;
}

Code (Cpp)

class Solution {
public:
    vector<int> findMajority(vector<int>& arr) {
        int n = arr.size();
        int num1 = INT_MIN, num2 = INT_MIN; 
        int c1 = 0, c2 = 0; 
        for (int x : arr) {
            if (x == num1) {
                c1++;
            } else if (x == num2) {
                c2++;
            } else if (c1 == 0) {
                num1 = x;
                c1 = 1;
            } else if (c2 == 0) {
                num2 = x;
                c2 = 1;
            } else {
                c1--;
                c2--;
            }
        }
        c1 = c2 = 0;
        for (int x : arr) {
            if (x == num1) c1++;
            else if (x == num2) c2++;
        }

        vector<int> res;
        if (c1 > n / 3) res.push_back(num1);
        if (c2 > n / 3) res.push_back(num2);

        sort(res.begin(), res.end()); 
        return res;
    }
};

Code (Java)

class Solution {
    public List<Integer> findMajority(int[] nums) {
        int num1 = Integer.MIN_VALUE, num2 = Integer.MIN_VALUE, c1 = 0, c2 = 0;
        int n = nums.length;
        for (int x : nums) {
            if (x == num1) {
                c1++;
            } else if (x == num2) {
                c2++;
            } else if (c1 == 0) {
                num1 = x;
                c1 = 1;
            } else if (c2 == 0) {
                num2 = x;
                c2 = 1;
            } else {
                c1--;
                c2--;
            }
        }
        c1 = c2 = 0;
        for (int x : nums) {
            if (x == num1) c1++;
            else if (x == num2) c2++;
        }
        List<Integer> res = new ArrayList<>();
        if (c1 > n / 3) res.add(num1);
        if (c2 > n / 3) res.add(num2);
        Collections.sort(res);
        return res;
    }
}

Code (Python)

class Solution:
    def findMajority(self, arr):
        num1, num2, c1, c2 = None, None, 0, 0
        n = len(arr)
        for x in arr:
            if x == num1:
                c1 += 1
            elif x == num2:
                c2 += 1
            elif c1 == 0:
                num1 = x
                c1 = 1
            elif c2 == 0:
                num2 = x
                c2 = 1
            else:
                c1 -= 1
                c2 -= 1
        c1, c2 = 0, 0
        for x in arr:
            if x == num1:
                c1 += 1
            elif x == num2:
                c2 += 1
        res = []
        if c1 > n // 3:
            res.append(num1)
        if c2 > n // 3:
            res.append(num2)
        res.sort()
        return res

🎯 Contribution and Support:

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