manual trace the following code int numbers [5][3] = { {1, 2, 5} , {2, 4, 3}, {1

Question

manual trace the following code

int numbers [5][3] = { {1, 2, 5} , {2, 4, 3}, {1, 9, 8} , {2, 6, 5}, {7, 4, 2}};

int counts [9] = {0};

int digit = 1; //keeps track of numbers from 1 through 9

while (digit <= 9)

{

for (int row = 0; row < 5; row = row + 1)

for (int col = 0; col < 3; col = col + 1)

//count the number of times the digit appears in the numbers array

if (numbers[row] [col} == digit)

counts[digit - 1] = counts[digit - 1] + 1;

//end if

//end for

//end for

digit = digit + 1; //look for next digit

}//end while

//display counts

for (int x = 0; x < 9; x = x + 1)

cout << "The number " << x + 1 << " appears "

<< counts [x] << " time(s) ." << endl;

// end for

Explanation / Answer

digit=1

while ->true

     checking each row and each column for equivalent of digit i.e. 1

   as 1 has only 2 time in numbers array increase 2 to counts[1-1] i.e. counts[0]=2

    digit=digit+1 => digit=2

while-> true

    checking each row and each column for equivalent of digit i.e. 2

   as 2 has only 4 times in numbers array increase 4 to counts[2-1] i.e. counts[1]=4

    digit=digit+1 => digit=3

while-> true

    checking each row and each column for equivalent of digit i.e. 3

   as 3 has only 1 times in numbers array increase 1 to counts[3-1] i.e. counts[2]=1

    digit=digit+1 => digit=4

while-> true

    checking each row and each column for equivalent of digit i.e. 4

   as 4 has only 2 times in numbers array increase 2 to counts[4-1] i.e. counts[3]=2

    digit=digit+1 => digit=5

while-> true

    checking each row and each column for equivalent of digit i.e. 5

   as 5 has only 2 times in numbers array increase 2 to counts[5-1] i.e. counts[4]=2

    digit=digit+1 => digit=6

while-> true

    checking each row and each column for equivalent of digit i.e. 6

   as 6 has only 1 times in numbers array increase 1 to counts[6-1] i.e. counts[5]=1

    digit=digit+1 => digit=7

while-> true

    checking each row and each column for equivalent of digit i.e. 7

   as 7 has only 1 times in numbers array increase 1 to counts[7-1] i.e. counts[6]=1

    digit=digit+1 => digit=8

while-> true

    checking each row and each column for equivalent of digit i.e. 8

   as 8 has only 1 times in numbers array increase 1 to counts[8-1] i.e. counts[7]=1

    digit=digit+1 => digit=9

while-> true

    checking each row and each column for equivalent of digit i.e. 9

   as 9 has only 1 times in numbers array increase 1 to counts[9-1] i.e. counts[8]=1

    digit=digit+1 => digit=10

while->false.

Display counts array i.e. [2,4,1,2,2,1,1,1,1]

i.e

The number 1 appears 2 times.

The number 2 appears 4 times.

The number 3 appears 1 times.

The number 4 appears 2 times.

The number 5 appears 2 times.

The number 6 appears 1 times.

The number 7 appears 1 times.

The number 8 appears 1 times.

The number 9 appears 1 times.

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