The following function computes the maximum value contained in an integer array p[] of size $n$ ($n \ge 1$).int max(int *p, int n) { int a=0, b=n-1; while (__________) { if (p[a] <= p[b]) { a = a+1; } else { b = b-1; } } return p[a]; }The missing loop condition is

The algorithm uses two pointers, $a$ initialized to $0$ and $b$ initialized to $n - 1$, to narrow down the range containing the maximum element.

• In each iteration, the elements $p[a]$ and $p[b]$ are compared:

  • If $p[a] \le p[b]$, then $p[a]$ cannot be the unique maximum, so index $a$ is incremented ($a = a + 1$).

  • Otherwise ($p[a] > p[b]$), $p[b]$ cannot be the maximum, so index $b$ is decremented ($b = b - 1$).

• The search space decreases by 1 in each step until only a single element remains, which happens when $a = b$.

• After the loop terminates with $a = b$, $p[a]$ is returned, which contains the maximum value in the array.

Therefore, the loop must continue as long as $a \neq b$, which is given by the condition b != a.

Thus, (D) is the correct option.