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Python3 Program for Check for Majority Element in a sorted array

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  • Last Updated : 13 Dec, 2021

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.

Python3




'''Python3 Program to check for majority element in a sorted array'''
  
def isMajority(arr, n, x):
    # get last index according to n (even or odd) */
    last_index = (n//2 + 1) if n % 2 == 0 else (n//2)
  
    # search for first occurrence of x in arr[]*/
    for i in range(last_index):
        # check if x is present and is present more than n / 2 times */
        if arr[i] == x and arr[i + n//2] == x:
            return 1
  
# Driver program to check above function */
arr = [1, 2, 3, 4, 4, 4, 4]
n = len(arr)
x = 4
if (isMajority(arr, n, x)):
    print ("% d appears more than % d times in arr[]"
                                            %(x, n//2))
else:
    print ("% d does not appear more than % d times in arr[]"
                                                    %(x, n//2))
  
  
# This code is contributed by shreyanshi_arun.

Output: 

4 appears more than 3 times in arr[]

Time Complexity: O(n)

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. 

Python3




'''Python3 Program to check for majority element in a sorted array'''
  
# This function returns true if the x is present more than n / 2
# times in arr[] of size n */
def isMajority(arr, n, x):
      
    # Find the index of first occurrence of x in arr[] */
    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 + n//2) <= (n -1)) and arr[i + n//2] == x:
        return True
    else:
        return False
  
# If x is present in arr[low...high] then returns the index of
# first occurrence of x, otherwise returns -1 */
def _binarySearch(arr, low, high, x):
    if high >= low:
        mid = (low + high)//2 # 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 or x > arr[mid-1]) and (arr[mid] == x):
            return mid
        elif x > arr[mid]:
            return _binarySearch(arr, (mid + 1), high, x)
        else:
            return _binarySearch(arr, low, (mid -1), x)
    return -1
  
  
# Driver program to check above functions */
arr = [1, 2, 3, 3, 3, 3, 10]
n = len(arr)
x = 3
if (isMajority(arr, n, x)):
    print ("% d appears more than % d times in arr[]"
                                            % (x, n//2))
else:
    print ("% d does not appear more than % d times in arr[]"
                                                    % (x, n//2))
  
# This code is contributed by shreyanshi_arun.

Output: 

3 appears more than 3 times in arr[]

Time Complexity: O(Logn) 
Algorithmic Paradigm: Divide and Conquer

METHOD 3: If it is already given that the array is sorted and there exists a majority element, checking if a particular element is as easy as checking if the middle element of the array is the number we are checking against.

Since a majority element occurs more than n/2 times in an array, it will always be the middle element. We can use this logic to check if the given number is the majority element.

Python3




def isMajorityElement(arr, 
                      n, key):
  
   if (arr[n // 2] == key):
        return True
      
   return False
  
# Driver code 
if __name__ == "__main__":
  
    arr = [1, 2, 3, 3,
           3, 3, 10]
    n = len(arr)
    x = 3
      
    if (isMajorityElement(arr, n, x)):
        print(x, " appears more than "
              n // 2 , " times in arr[]")
    else:
        print(x, " does not appear more than"
              n // 2, " times in arr[]")
  
# This code is contributed by Chitranayal
Output
3 appears more than 3 times in arr[]

Time complexity: O(1)
Auxiliary Space: O(1)

Please refer complete article on Check for Majority Element in a sorted array for more details!


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