**Question:** Write a C 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.

## C

/* C Program to check for majority element in a sorted array */ # include <stdio.h> # include <stdbool.h> bool isMajority(int arr[], int n, int x) { int i; /* get last index according to n (even or odd) */ int last_index = n%2? (n/2+1): (n/2); /* 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 1; } return 0; } /* Driver program to check above function */ int main() { int arr[] ={1, 2, 3, 4, 4, 4, 4}; int n = sizeof(arr)/sizeof(arr[0]); int x = 4; if (isMajority(arr, n, x)) printf("%d appears more than %d times in arr[]", x, n/2); else printf("%d does not appear more than %d times in arr[]", x, n/2); 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.

## C

/* Program to check for majority element in a sorted array */ # include <stdio.h> # include <stdbool.h> /* If x is present in arr[low...high] then returns the index of first occurrence of x, otherwise returns -1 */ int _binarySearch(int arr[], int low, int high, int x); /* This function returns true if the x is present more than n/2 times in arr[] of size n */ bool 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; } /* If x is present in arr[low...high] then returns the index of first occurrence of x, otherwise returns -1 */ 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; } /* Driver program to check above functions */ int main() { int arr[] = {1, 2, 3, 3, 3, 3, 10}; int n = sizeof(arr)/sizeof(arr[0]); int x = 3; if (isMajority(arr, n, x)) printf("%d appears more than %d times in arr[]", x, n/2); else printf("%d does not appear more than %d times in arr[]", x, n/2); return 0; }

## 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

Please write comments if you find any bug in the above program/algorithm or a better way to solve the same problem.