At the time of sorting an array of size N by normal selection sort, the number of comparisons made in first iteration is :

The correct answer is: option 2: N−1

Concept:

In the Selection Sort algorithm, during each iteration, we find the smallest (or largest) element from the unsorted portion of the array and place it at the correct position.

In the first iteration:

• You compare the first element with all the other N−1 elements to find the minimum.
• So, the number of comparisons made in the first iteration is exactly N−1.

General selection sort comparison count:

• Total comparisons = (N−1) + (N−2) + ... + 1 = N(N−1)/2

Explanation of options:

• Option 1 – N: ❌ Too many, only N−1 comparisons are needed.
• Option 2 – N−1: ✅ Correct. This is the actual number of comparisons in the first iteration.
• Option 3 – N×(N−1)/2: ❌ This is the total number of comparisons in the entire sorting process, not just the first iteration.
• Option 4 – NlogN: ❌ This is the time complexity of more efficient algorithms like Merge Sort or Heap Sort, not Selection Sort.

Hence, the correct answer is: option 2: N−1