Consider the following method.

public static int operation(int[][] mat)
{
  int currentVal = mat[0][0];
  int result = 0;

  for (int j = 0; j

Question

Consider the following method.

public static int operation(int[][] mat) 
{
  int currentVal = mat[0][0];
  int result = 0;

for (int j = 0; j < mat.length; j++) 
  {
    for (int k = 0; k < mat[j].length; k++) 
    {
      /* missing code */
    }
  }

return result;
}

Which of the following should replace /* missing code */ so that the method returns the index of the column which contains the smallest value in the two-dimensional array?

BlackTom AI · Accepted Answer

Compare each element to the current minimum and record the column index when you find a smaller value.

1) Relevant knowledge points and concepts
- Traversing a 2D array with nested loops: outer loop over rows (j), inner loop over columns (k) via mat[j][k].
- To find a minimum value, keep a running minimum (currentVal) and update it when a smaller element is found.
- The method must return the column index (k) of the smallest element, so when updating the minimum you must assign result = k (not j).

2) Step-by-step reasoning process
- Start with currentVal = mat[0][0] and result = 0 (column index of the current minimum).
- For every element mat[j][k], compare it to currentVal.
- If mat[j][k] is strictly less than currentVal, replace currentVal with mat[j][k] and set result = k.
- Continue until all elements are checked; result then holds the column index of the smallest element encountered.

3) Example with intermediate steps
Given mat = {{5, 2, 7}, {3, 9, 1}}:
- Initial: currentVal = 5, result = 0.
- j=0,k=0: mat[0][0]=5, 5<5? no.
- j=0,k=1: mat[0][1]=2, 2<5? yes → currentVal=2, result=1.
- j=0,k=2: mat[0][2]=7, 7<2? no.
- j=1,k=0: mat[1][0]=3, 3<2? no.
- j=1,k=1: mat[1][1]=9, 9<2? no.
- j=1,k=2: mat[1][2]=1, 1<2? yes → currentVal=1, result=2.
- End: return result = 2 (column index of smallest value).

4) Why the correct answer is correct
The provided snippet performs exactly the required update: when a strictly smaller value is found, it updates both the tracked minimum and the column index. Using result = k stores the column index (not the row), matching the problem requirement.

5) Why other plausible alternatives are incorrect
- Setting result = j would return the row index, not the column index.
- Assigning result = mat[j][k] would return the value, not an index.
- Using <= instead of < would change tie behavior (it would record a later column when equal minima occur), which is not requested and differs from the provided correct snippet.

6) Conclusion
The correct code to insert is:
if (mat[j][k] < currentVal)
{
  currentVal = mat[j][k];
  result = k;
}
This matches the provided answer and correctly returns the column index containing the smallest value.

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