WAP to Find Occurance of a Number N in Sorted Array in O(logn)

Given a sorted array and a number n.How can u find the number of occurance of n in the array . should be o(logn)


Modified Binary Search Algorithm

mid=low+high/2;
if(a[mid]>num)
search in right side & set low=mid+1 & return ;
else if(a[mid]search in left side of mid set high=mid-1 & return ;
else //its important instead of just of printing the num or incrementing the counter //i tried if you will do like this then it will be O(n) not O(logn) , si i will add 1 to recursively call for left side + recursively call for right side so every time this line executes we are incrementing the counter &
return 1+left_binsearch()+right_binsearch thus it will be in O(logn)


you people can dry run for this 1 2 2 3 3 3 4 4 4 low=0 high=8 mid=4 & run it



#include


int CountFreq (int *A, int value, int low, int high)
{
int mid;
if (high < low)
return 0;
mid = low + (high - low);
if (A[mid] > value)
return CountFreq (A, value, low, mid - 1);
else if (A[mid] < value)
return CountFreq (A, value, mid + 1, high);
else
return 1 + CountFreq (A, value, low, mid - 1) + CountFreq (A, value, mid + 1, high);
}

int main() {

int A[] = { 1,2,2,2,2,3, 3, 3,4,4,4, 4};
int value = 2;



printf("%d\n", CountFreq(A, value, 0, sizeof(A)/sizeof(int)-1));

return 0;

}

TC O(n) consider ar[]={3,3,3,3,3,3}, & x=3 T(n)=2*T(n/2)+c leads to O(n)
Sc O(1)
Run Here https://ideone.com/clone/mTGes


Method 2 Modified 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);

why it works because we know that array i sorted so if find the first occurrence at position i & last occurrence at position j then we are sure that all the elements between i & j are the same value what they at ith & jth position it works because array is sorted & we have done ....Cheers

#include


/* 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)/2; /*low + (high - low)/2;*/
if( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x)
return mid;
else if(x > arr[mid])
return first(arr, (mid + 1), high, x, n);
else
return first(arr, low, (mid -1), x, n);
}
return -1;
}

/* 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)/2; /*low + (high - low)/2;*/
if( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x )
return mid;
else if(x < arr[mid])
return last(arr, low, (mid -1), x, n);
else
return last(arr, (mid + 1), high, x, n);
}
return -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, 0, n-1, x, n);

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

/* Else get the index of last occurrence of x */
j = last(arr, 0, n-1, x, n);

/* return count */
return j-i+1;
}



/* driver program to test above functions */
int main()
{
int arr[] = {1, 2, 2, 2, 2, 3, 3};
int x = 2; // Element to be counted in arr[]
int n = sizeof(arr)/sizeof(arr[0]);
int c = count(arr, x, n);
printf(" %d occurs %d times ", x, c);
getchar();
return 0;
}

TC O(logn)
Sc O(1)
Run here https://ideone.com/J0enE