GATE 2008

The Best way to find out whether an integer appears more than n/2 times in a sorted array(Ascending Order) of n integers, would be binary search approach.
1. The First occurrence of an element can be found out in O(log(n)) time using divide and conquer technique, lets say it is i.
2. The Last occurrence of an element can be found out in O(log(n)) time using divide and conquer technique,lets say it is j.
3. Now number of occurrence of that element(count) is (j-i+1).
Overall time complexity = log n +log n +1 = O(logn)
Program:
/* 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;
}
Note: The code finds out the first occurrence of the element in the array and checks if the element at (n/2+1)^th position is same(as the array is sorted). Takes O(logn) time.