Javascript Program to Check Majority Element in a sorted array

Last Updated : 15 Feb, 2023

Question: Write a function to find if a given integer x appears more than n/2 times in a sorted array of n integers.
Basically, we need to write a function say isMajority() that takes an array (arr[] ), array’s size (n) and a number to be searched (x) as parameters and returns true if x is a majority element (present more than n/2 times).

Examples:

Input: arr[] = {1, 2, 3, 3, 3, 3, 10}, x = 3
Output: True (x appears more than n/2 times in the given array)

Input: arr[] = {1, 1, 2, 4, 4, 4, 6, 6}, x = 4
Output: False (x doesn't appear more than n/2 times in the given array)

Input: arr[] = {1, 1, 1, 2, 2}, x = 1
Output: True (x appears more than n/2 times in the given array)

METHOD 1 (Using Linear Search)
Linearly search for the first occurrence of the element, once you find it (let at index i), check element at index i + n/2. If element is present at i+n/2 then return 1 else return 0.

Output:

4 appears more than 3 times in arr[]

Time Complexity: O(n)

Auxiliary Space: O(1)

As constant extra space is used.

METHOD 2 (Using Binary Search)
Use binary search methodology to find the first occurrence of the given number. The criteria for binary search is important here.

Javascript

<script>
    // Javascript Program to check for majority
    // element in a sorted array */
     
    // If x is present in arr[low...high]
    // then returns the index of first
    // occurrence of x, otherwise returns -1
    function _binarySearch(arr, low, high, x)
    {
        if (high >= low) {
            let mid = parseInt((low + high) / 2, 10);
            //low + (high - low)/2;
  
            // Check if arr[mid] is the first
            // occurrence of x.    arr[mid] is
            // first occurrence if x is one of
            // the following is true:
            // (i) mid == 0 and arr[mid] == x
            // (ii) arr[mid-1] < x and arr[mid] == x
              
            if ((mid == 0 || x > arr[mid - 1]) && (arr[mid] == x))
                return mid;
            else if (x > arr[mid])
                return _binarySearch(arr, (mid + 1), high, x);
            else
                return _binarySearch(arr, low, (mid - 1), x);
        }
  
        return -1;
    }
  
    // This function returns true if the x is
    // present more than n/2 times in arr[]
    // of size n
    function isMajority(arr, n, x)
    {
          
        // Find the index of first occurrence
        // of x in arr[]
        let i = _binarySearch(arr, 0, n - 1, x);
  
        // If element is not present at all,
        // return false
        if (i == -1)
            return false;
  
        // check if the element is present
        // more than n/2 times
        if (((i + parseInt(n / 2, 10)) <= (n - 1)) && arr[i + parseInt(n / 2, 10)] == x)
            return true;
        else
            return false;
    }
     
    let arr = [ 1, 2, 3, 3, 3, 3, 10 ];
    let n = arr.length;
    let x = 3;
    if (isMajority(arr, n, x) == true)
      document.write(x + " appears more than " + parseInt(n / 2, 10) + " times in arr[]");
    else
      document.write(x + " does not appear more than " + parseInt(n / 2, 10) + " times in arr[]");
     
</script>