ITPEC FE Subject B October 2025 Question 2

The program finds the mode of an integer array using a brute-force frequency count:

• •Outer loop (i from 1 to n-1): picks each candidate element arr[i]
• •Counter reset: c ← 1 (the element counts itself)
• •Inner loop (j from i+1 to n): compares arr[i] with every subsequent element
• •Blank A = arr[i] = arr[j] — checks if arr[j] matches the current candidate; if yes, increments c
• •Blank B = m_c < c — after counting, checks if this candidate's frequency c exceeds the current best m_c; if yes, updates m_c ← c and m ← arr[i]

Why not others:
- Options a, b (B: m < arr[i] or m ≠ arr[i]): compare the mode value, not frequency — logically wrong
- Option d (B: m_c > c): would update the mode when frequency is lower, finding the least common element instead
- Options e–h (A: m = arr[j]): compare the current mode value m against arr[j] instead of comparing array elements to each other — breaks the counting logic

Key rule: When filling blanks in a frequency-counting algorithm, identify which variable tracks count (c, m_c) vs value (m, arr[i]). The inner loop condition compares values (arr[i] = arr[j]), while the update condition compares counts (m_c < c).

AI-generated — may contain errors