Check for Majority Element in a sorted array | GeeksforGeeks


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.
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.
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;
}
METHOD 1 (Using Linear Search)
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;
}

Find popular item in sorted array of natural numbers.An item is popular is if its repeated n/4 times or more.
For Ex:
Input: 123444445678
Popular item is 4.
Liner scan is O(n), but solution needs to be in O(logN) complexity and O(1) space complexity.
This is actually a very interesting problem. Since the popular item is defined as the element is repeated more than 1 / 4 times, and since it is a sorted array, so it can only occurs on 0, n / 4, n /2 and 3n/4 index. And the rest is just do binary search and get the range.

 public static void popular(int[] n){
  if(n == null || n.length == 0)
   return;
  int len = n.length;
  int[] check = {0, len / 4, len / 2, 3 * len / 4};
  for(int i = 0; i < 4; i++){
   if(i > 0 && n[check[i]] == n[check[i - 1]])
    continue;
   int l = check[i];
   int start = binarySearchStart(n, n[l]);
   int end = binarySearchEnd(n, n[l]);
   //need to be larger than the ceil in case len / 4.0 is not an integer
   if(end - start + 1 >= Math.ceil(len / 4.0))
    System.out.println(n[l]);
  }
 }
 private static int binarySearchEnd(int[] n, int target){
  int len = n.length;
  int start = 0;
  int end = len - 1;
  while(start + 1 < end){
   int mid = (start + end) / 2;
   if(n[mid] <= target)
    start = mid;
   else
    end = mid;
  }
  if(n[end] == target)
   return end;
  else return start;
 }
 private static int binarySearchStart(int[] n, int target){
  int len = n.length;
  int start = 0;
  int end = len - 1;
  while(start + 1 < end){
   int mid = (start + end) / 2;
   if(n[mid] >= target)
    end = mid;
   else
    start = mid;
  }
  if(n[start] == target)
   return start;
  else
   return end;
 }
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