GATE CS 2016 Set 1 — Question 44

MCQ+2 / -0.67EasyArray Operations (1D, 2D)Arrays & StringsProgramming & Data Structures

Programming & Data Structures → Arrays & Strings → Array Operations (1D, 2D)

Last updated 5 September 2026

Question

The following function computes the maximum value contained in an integer array p[] of size n (n >= 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

Correct answer (D) b != a Solution The function uses a two-pointer approach to find the maximum element in the array. 1. Initialization : a points to the start (0) and b points to the end (n-1) of the array. 2. Logic : In each iteration, we compare p[a] and p[b] . If p[a] <= p[b], then p[a] cannot be the unique maximum (since p[b] is greater or equal), so we increment a to discard it. If p[a] > p[b], then p[b] cannot be the maximum, so we decrement b to discard it. 3. Termination : The loop should continue as long as there are at least two candidates to compare, i.e., a and b are distinct indices. When a == b, only one element remains, which must be the maximum. Therefore, the loop condition should be b != a (or a < b). Checking the options: (A) a != n : Incorrect, a stops when it meets b . (B) b != 0 : Incorrect. (C) b > (a + 1) : This would stop when b and a are adjacent, leaving two elements unchecked against each other. (D) b != a : Correct. The loop runs until a and b converge to the single maximum element. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (D) b != a Solution The function uses a two-pointer approach to find the maximum element in the array. 1. Initialization : a points to the start (0) and b points to the end (n-1) of the array. 2. Logic : In each iteration, we compare p[a] and p[b] . If p[a] <= p[b], then p[a] cannot be the unique maximum (since p[b] is greater or equal), so we increment a to discard it. If p[a] > p[b], then p[b] cannot be the maximum, so we decrement b to discard it. 3. Termination : The loop should continue as long as there are at least two candidates to compare, i.e., a and b are distinct indices. When a == b, only one element remains, which must be the maximum. Therefore, the loop condition should be b != a (or a < b). Checking the options: (A) a != n : Incorrect, a stops when it meets b . (B) b != 0 : Incorrect. (C) b > (a + 1) : This would stop when b and a are adjacent, leaving two elements unchecked against each other. (D) b != a : Correct. The loop runs until a and b converge to the single maximum element.

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