CMI2013-A-09

Question

The below question is based on the following program.

procedure mystery (A : array [1..100] of int)
    int i,j,position,tmp;
    begin
        for j := 1 to 100 do
            position := j;
            for i := j to 100 do
                if (A[i] > A[position]) then
                    position := i;
                endfor
            tmp := A[j];
            A[j] := A[position];
            A[position] := tmp;
        endfor
end

When the procedure terminates, the array A has been:

• A. Reversed
• B. Left unaltered
• C. Sorted in descending order
• D. Sorted in ascending order

Comments

@kenzou, It would be better to shift the "endfor" stmt to align with "for".  May get confused as break in the pseudocode.

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This code is finding the position of max element in the unsorted part of the array and interchanging it with the first element in the unsorted part.

Answer 1

Answer is $B$. Sorted in descending order ( selection sorting algorithm is used ).

Answer 2

void selectionSort(int arr[], int n)

{

    int i, j, min_idx;

 

    // One by one move boundary of unsorted subarray

    for (i = 0; i < n-1; i++)

    {

        // Find the minimum element in unsorted array

        min_idx = i;

        for (j = i+1; j < n; j++)

          if (arr[j] < arr[min_idx])

            min_idx = j;

 

        // Swap the found minimum element with the first element

        swap(&arr[min_idx], &arr[i]);

    }

}

 

 

 

Selection sort arranges elements in descending order.

Answer : B

Answer 3