Question

Recall the pseudocode from lectures for identifying a good VLSI chip. The idea can be reused to finding the majority item in a list, where the majority is defined by the element in a list of size n which occurs at least n/2 + 1 times. Write a linear-time program to find the majority element in an array. What is the invariant of your loop? What is the variant of your loop? Explain why your design has linear time performance. Write JUnit tests to check your program.

Answer 1

First of all, we will discuss What is the invariant of the loop: -A loop invariant is a condition [among program variables] that is necessarily true immediately before and immediately after each iteration of a loop. By itself, a loop invariant doesn't do much. However, given an appropriate invariant, it can be used to help prove the correctness of an algorithm.

A majority element in an array A[] of size n is an element that appears more than n/2 times (and hence there is at most one such element).
Examples :

Input : {3, 3, 4, 2, 4, 4, 2, 4, 4}
Output : 4 

Input : {3, 3, 4, 2, 4, 4, 2, 4}
Output : No Majority Element

The basic solution is to have two loops and keep track of the maximum count for all different elements. If maximum count becomes greater than n/2 then break the loops and return the element having maximum count. If the maximum count doesn’t become more than n/2 then majority element doesn’t exist.
Below is the implementation of the above approach :

void find majority(int arr[], int n) {     int maxCount = 0;     int index = -1; // sentinels     for(int i = 0; i < n; i++)     {         int count = 0;         for(int j = 0; j < n; j++)         {             if(arr[i] == arr[j])             count++;         }                    // update maxCount if count of         // current element is greater         if(count > maxCount)         {             maxCount = count;             index = i;         }     }            // if maxCount is greater than n/2     // return the corresponding element     if (maxCount > n/2)        cout << arr[index] << endl;            else         cout << "No Majority Element" << endl; }    // Driver code int main() {     int arr[] = {1, 1, 2, 1, 3, 5, 1};     int n = sizeof(arr) / sizeof(arr[0]);            // Function calling     findMajority(arr, n);        return 0; }

Time Complexity : O(n*n).