[geeksforgeeks] Count the number of occurrences in a sorted array

Count the number of occurrences in a sorted array

Given a sorted array arr[] and a number x, write a function that counts the occurrences of x in arr[]. Expected time complexity is O(Logn)

Examples:

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 2
  Output: 4 // x (or 2) occurs 4 times in arr[]

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 3
  Output: 1 

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 1
  Output: 2 

  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 4
  Output: -1 // 4 doesn't occur in arr[] 

Method 1 (Linear Search)
Linearly search for x, count the occurrences of x and return the count.

Time Complexity: O(n)

Method 2 (Use Binary Search)
1) Use Binary search to get index of the first occurrence of x in arr[]. Let the index of the first occurrence be i.
2) Use Binary search to get index of the last occurrence of x in arr[]. Let the index of the last occurrence be j.
3) Return (j – i + 1);

/* if x is present in arr[] then returns the count of occurrences of x,
otherwise returns -1. */
int count(int arr[], int x, int n)
{
int i; // index of first occurrence of x in arr[0..n-1]
int j; // index of last occurrence of x in arr[0..n-1]

/* get the index of first occurrence of x */
i = first(arr, , n-, x, n);

/* If x doesn't exist in arr[] then return -1 */
if(i == -)
    return i;

/* Else get the index of last occurrence of x. Note that we
    are only looking in the subarray after first occurrence */
j = last(arr, i, n-, x, n);     

/* return count */
return j-i+;
}

/* if x is present in arr[] then returns the index of FIRST occurrence
of x in arr[0..n-1], otherwise returns -1 */
int first(int arr[], int low, int high, int x, int n)
{
if(high >= low)
{
    int mid = (low + high)/; /*low + (high - low)/2;*/
    if( ( mid ==  || x > arr[mid-]) && arr[mid] == x)
    return mid;
    else if(x > arr[mid])
    return first(arr, (mid + ), high, x, n);
    else
    return first(arr, low, (mid -), x, n);
}
return -;
}

/* if x is present in arr[] then returns the index of LAST occurrence
of x in arr[0..n-1], otherwise returns -1 */
int last(int arr[], int low, int high, int x, int n)
{
if(high >= low)
{
    int mid = (low + high)/; /*low + (high - low)/2;*/
    if( ( mid == n- || x < arr[mid+]) && arr[mid] == x )
    return mid;
    else if(x < arr[mid])
    return last(arr, low, (mid -), x, n);
    else
    return last(arr, (mid + ), high, x, n);
}
return -;
}

/* driver program to test above functions */
int main()
{
int arr[] = {, , , , , , };
int x = ; // Element to be counted in arr[]
int n = sizeof(arr)/sizeof(arr[]);
int c = count(arr, x, n);
printf(" %d occurs %d times ", x, c);
getchar();
return ;
}

Time Complexity: O(Logn)
Programming Paradigm: Divide & Conquer