Question

Suppose you have an array S indexed from 1 to n which contains n numbers, not in any particular order, and you wish to count how many times a given number x occurs in S. Consider the recursive algorithm below for this which finds the number of occurrences of x in the index range i...j in S. Of course, solving the problem would involve an initial call to the algorithm for the range 1.n: int CountOccur (int i,j) { int mid, count; if j

Answer 1

1.

int countOccur(int S[], int x, int n){
   int count = 0;
   for(int i = 0;i<n;i++){
       if(S[i] == x)
           count++;
   }
   return count;
  
}

In the above code, S is the array which contains all the digits, x is the one with which we want to match, n is the number of items inside the array S.

2. [2, 3, 6 , 2, 7, 8, 2, 9] x = 2

So here n = 8

So we need to check for each value of ranging from 0 to 7

i = 0, match, count = 1

i = 1, not match

i = 2, not match

i = 3, match, count = 2

i = 4, not match

i = 5 not match

i = 6 match, count = 3

i = 7 not match

return count as 3, 3 is the answer from the illustration.

3. For the following problem, time complexity would be O(n), where n is the number of elements in the array.

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