Consider the following C-function in which a[n] and b[m] are two sorted integer arrays and c[n + m] be another integer array.

`void` `xyz(` `int` `a[], ` `int` `b [], ` `int` `c[])` `{` ` ` `int` `i, j, k;` ` ` `i = j = k = O;` ` ` `while` `((i<n) && (j<m))` ` ` `if` `(a[i] < b[j]) c[k++] = a[i++];` ` ` `else` `c[k++] = b[j++];` `}` |

Which of the following condition(s) hold(s) after the termination of the while loop? (GATE CS 2006)

(i) j < m, k = n+j-1, and a[n-1] < b[j] if i = n
(ii) i < n, k = m+i-1, and b[m-1] <= a[i] if j = m
**(A)** only (i)**(B)** only (ii)**(C)** either (i) or (ii) but not both**(D)** neither (i) nor (ii)**Answer:** **(D)****Explanation:** The function xyz() is similar to merge() of mergeSort().Both the conditions (i) and (ii) are false in the sections k = n+j-1 and k = m+i-1 respectively.

The while loop adds elements from a and b (whichever is smaller) to c and terminates when either of them exhausts. So, when the loop terminates either i=n or j=m.

Suppose i=n. This would mean all elements from array a are added to c=>k must be incremented by n. c would also contain j elements from array b. So, the number of elements in c would be n+j and hence k=n+j.

Similarly, when j=m, k=m+i.

Hence, the option (D) is correct. (Had k started from −1 and not 0 and we used ++k inside the loop, the answer would have been option (C))

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