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.
Java
/* Program to check for majority element in a sorted array */import java.io.*; class Majority { static boolean isMajority(int arr[], int n, int x) { int i, last_index = 0; /* get last index according to n (even or odd) */ last_index = (n%2==0)? n/2: n/2+1; /* search for first occurrence of x in arr[]*/ for (i = 0; i < last_index; i++) { /* check if x is present and is present more than n/2 times */ if (arr[i] == x && arr[i+n/2] == x) return true; } return false; } /* Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 2, 3, 4, 4, 4, 4}; int n = arr.length; int x = 4; if (isMajority(arr, n, x)==true) System.out.println(x+" appears more than "+ n/2+" times in arr[]"); else System.out.println(x+" does not appear more than "+ n/2+" times in arr[]"); }}/*This article is contributed by Devesh Agrawal*/ |
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.
Java
/* Java Program to check for majority element in a sorted array */import java.io.*; class Majority { /* If x is present in arr[low...high] then returns the index of first occurrence of x, otherwise returns -1 */ static int _binarySearch(int arr[], int low, int high, int x) { if (high >= low) { int 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 || 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 */ static boolean isMajority(int arr[], int n, int x) { /* Find the index of first occurrence of x in arr[] */ int 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)) && arr[i + n/2] == x) return true; else return false; } /*Driver function to check for above functions*/ public static void main (String[] args) { int arr[] = {1, 2, 3, 3, 3, 3, 10}; int n = arr.length; int x = 3; if (isMajority(arr, n, x)==true) System.out.println(x + " appears more than "+ n/2 + " times in arr[]"); else System.out.println(x + " does not appear more than " + n/2 + " times in arr[]"); }}/*This code is contributed by Devesh Agrawal*/ |
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.
Java
import java.util.*; class GFG{ static boolean isMajorityElement(int arr[], int n, int key){ if (arr[n / 2] == key) return true; else return false;} // Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 3, 3, 3, 3, 10 }; int n = arr.length; int x = 3; if (isMajorityElement(arr, n, x)) System.out.printf("%d appears more than %d " + "times in arr[]", x, n / 2); else System.out.printf("%d does not appear more " + "than %d times in " + "arr[]", x, n / 2);}} // This code is contributed by aashish1995 |
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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