Subtract the Product and Sum of Digits of an Integer

Refer to the problem statement.

Algorithm: Let p be the Integer.

• Step 1: Declare 3 variables i.e. sum = 0, prod = 1, temp.

• Step 2: If p > 0 then go to Step 3 else go to Step 7.

• Step 3: Extract the last digit of p i.e. temp = p%10.

• Step 4: Add the 'temp' variable to 'sum' i.e. sum = sum + temp.

• Step 5: Multiple the 'variable' to 'prod' i.e. prod = prod * temp.

• Step 6: Remove the last digit from n i.e. p = p/10. Goto Step 2.

• Step 7: Return (prod-sum).

Examples:

Let p = 234

Step 1: sum = 0, prod = 1, temp

• Iteration 1:

  Step 2: 234 > 0 then go to Step 3

  Step 3: temp = (234)%(10) = 4

  Step 4: sum = 0 + 4 = 4

  Step 5: prod = 1*4 = 4

  Step 6: p = (234)/(10) = 23

• Iteration 2:

  Step 2: 23 > 0 then go to Step 3

  Step 3: temp = (23)%(10) = 3

  Step 4: sum = 4 + 3 = 7

  Step 5: prod = 4*3 = 12

  Step 6: p = (23)/(10) = 2

• Iteration 3:

  Step 2: 2 > 0 then go to Step 3

  Step 3: temp = (2)%(10) = 2

  Step 4: sum = 7 + 2 = 9

  Step 5: prod = 12*2 = 24

  Step 6: p = (2)/(10) = 0

• Iteration 4:

  Step 2: 0 not greater than 0. goto Step 7

Step 7: Return (24-9) = 15

Time Complexity: O(n) where n is the number of digits in the integer p.

Code: Java