C Program to Check for Majority Element in a sorted array (original) (raw)

Last Updated : 31 May, 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.

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; }

`

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 `

/* 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; }

`

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.

C] [tabby title=C `

#include <stdio.h> #include <stdbool.h>

bool isMajorityElement(int arr[], int n, int key) { if (arr[n / 2] == key) return true; else return false; }

int main() { int arr[] = { 1, 2, 3, 3, 3, 3, 10 }; int n = sizeof(arr) / sizeof(arr[0]); int x = 3; if (isMajorityElement(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; }

`

Output

3 appears more than 3 times in arr[]

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

Algorithm

  1. Initialize a variable candidate to the first element of the input array nums, and a variable count to 1.
  2. Loop through the elements of the input array nums starting from the second element.
  3. If the current element is equal to the candidate, increment count by 1.
  4. If the current element is not equal to the candidate, decrement count by 1. If count becomes 0, set the candidate to the current element and set count to 1.
  5. After the loop, loop through the elements of the input array nums again, counting how many times the candidate appears.
  6. If the count of the candidate is greater than n/2 where n is the size of the input array nums, return the candidate as the majority element.
  7. Otherwise, return -1 to indicate that there is no majority element. C `

#include <stdio.h>

int findMajorityElement(int* nums, int n) { int candidate = nums[0], count = 1;

for (int i = 1; i < n; i++) {
    if (nums[i] == candidate) {
        count++;
    } else {
        count--;
        if (count == 0) {
            candidate = nums[i];
            count = 1;
        }
    }
}

count = 0;
for (int i = 0; i < n; i++) {
    if (nums[i] == candidate) {
        count++;
    }
}

if (count > n/2) {
    return candidate;
} else {
    return -1;  // no majority element
}

}

int main() { int nums[] = {1, 2, 3, 4, 4, 4, 4}; int n = sizeof(nums)/sizeof(nums[0]); int majority = findMajorityElement(nums, n);

if (majority != -1) {
    printf("The majority element is %d\n", majority);
} else {
    printf("There is no majority element\n");
}

return 0;

}

`

Output

The majority element is 4

The time complexity is O(n), and the auxiliary space is O(1)

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