/* Online Java - IDE, Code Editor, Compiler Online Java is a quick and easy tool that helps you to build, compile, test your programs online. */ //TO PREVENT LOTS REPEAT CODE, ENSURE NUMBER2 IS LONGER THAN NUMBER1 //IT MIGHT GET AWAY IF THERE ARE NO STREAKS OF 9's //BUT IF NUMBER1 is LONGER AND CONTAINS 9's, I have not coded this. //Since it will bring lots of duplicated code in a single method. //I have declared all variables class level and static to ensure //that the recursive method is kept tidy //this challenge also raises performance question //to ensure I can test the execution time, I will add stopwatch.. //I have not attempted this in my code executions so far //For moment, I am using long variable to ensure I can extend //the String to 2,147,483,648 characters (digits only!) public class Main { static int lengthFirstNumber; //number digits wide. This will get shorter during recursion. static int lengthSecondNumber; //number digits wide. This will get shorter during recursion. //these will work in conjunction with substring of the number to obtain last digit static String lastDigitFirstNumber; static String lastDigitSecondNumber; static int counter; static int temp; //extremely difficult to provide meaningful name //if there is NO frontal number (i.e) total less than 10, expect 0 in firstDigitTotalDigits //if total is between 10 - 18(9+9), expect firstDigitTotalDigits=1 //crucial part of adding two digits together static int firstDigitTotalDigits; static int lastDigitTotalDigits; static boolean successiveNinesNumber2 =false; static int total; //stores total of two digits //used to output the running grandtotal //similar to expectation of manually completing addition static String grandTotal=""; //all below variables are closely linked //in addition, if a number is wider with extra digits, those additional //MSD need to drop straight down into grandtotal static String remainingPortion; //if a number is more than 1 digit wider than number 2, the LSD of this portion //has to be excluded. (since in real life addition, this is subject to change if a number is carried forward). //if for instance, this is 9, then 9+1 =0, and carry 1. //in practice this could go on if all the digits in remainingPortionInteger is 9! static int remainingPortionInteger; //This is conversion of remainingPortionInteger into String static String remainingPortionString; //exactly this static int lastDigitFirstNumberRemainingPortion; //exactly this static int lastDigitSecondNumberRemainingPortion; public static void main(String[] args) { System.out.println("Welcome to Online IDE!! Happy Coding :)"); System.out.println("Addition program, ensure String has maximum characters: " + "2,147,483,648"); //Note this is a fixed version of previous code //but it required so much additional code to be added //it now functions formatted as below (without whitespace) //ALTHOUGH NO VALIDATION HAS BEEN PROVIDED, //DO NOT INCLUDE WHITESPACES IN STRING WHATSOEVER //THIS REPLICATES THE CHALLENGE //PASS = 711 //String number2 = "652"; //String number1 = " 59"; //PASS - 5210 //String number2 = "4666"; //String number1 = " 544"; //PASS - 100 //String number2="99"; //String number1= "1"; //PASS - 1764 //String number2="993"; //String number1="771"; //PASS - 160 //String number2 = "136"; //String number1 = "24"; //PASS - 82160 //String number2 = "81636"; //String number1 = "524"; //String number2 = "846369999999999"; //String number1 = "32454543"; //PASS - 400219 //String number2 = "399627"; //String number1 = "592"; //pass - 400019 //String number2 = "399947"; //String number1 = "72"; //PASS getting 4962 //String number2 = "3970"; //String number1 = "992"; //PASS 562 //String number1 = "370"; //String number2 = "192"; //PASS - getting 1000 //String number2 = "999"; //String number1 = "1"; //PASS 1998 //String number1 = "999"; //String number2 = "999"; //PASS 1098 //String number1 = "999"; //String number2 = "99"; //PASS ***** //String number1 = "99"; //String number2 = " 9"; //fail - getting 2000019 //String number2 = "1399947"; //String number1 = "72"; //1390219 - fail (getting 2000219) = > PASS*** //String number2 = "1389627"; //String number1 = "592"; //For an unknown reason, the addition method is unable to return the String, it just shows blank value //or just skips the return statement //Bizarre is that it can see the variable value on screen before I try to return it. //So I have just used System.out.println(GRANDTOTAL) in recursive method. System.out.println(number1 + "+" + number2 + "= " + addition(number1, number2)); } public static String addition(String number1, String number2) { //it is still working constraints of the code //the Strings are note being converted to simplify addition, it is only the length of the Strings. lengthFirstNumber=(number1.length()); lengthSecondNumber=(number2.length()); /* //at this point it knowws that its on the most extreme left hand side of them... //looking at example 4666 // + 544 //it is just not a case of dropping the number 4 down.... into the grandTotal //this would be ok if (6+5 was not >=10) //there would be last addition undertaken //at this point its expected to perform 4 + //another example is just a case of dropping the remainder across //for instance performing 55 + 55 ------- 110 ------- //Another example is dealing with larger differences: 274666 + 544 //it would need to carry 1, add it to 4 //it would then have to truncate number1 again and drop 2 and 7 */ //if both lengths are 0, it means both had same lengths. if (lengthFirstNumber==0 && lengthSecondNumber==0) { //this is a slightly ambiguous name, but it identifies if //total from digit in number1 number2 is one or two digits long //firstDigitTotalDigits = (int)total/10; //if there is no frontal number (i.e) total less than 10, expect 0 in firstDigitTotalDigits //if total is between 10 - 18(9+9), expect firstDigitTotalDigits=1 //REFERS TO NO CARRY FORWARD, THIS SEEMS CORRECT //SINCE IT WOULD HAVE NOT BRANCHED OFF TO ANY CATCH STATEMENTS if (firstDigitTotalDigits==0) { System.out.println("GRANDTOTAL: " + grandTotal); System.out.println("FINAL TESTING 1"); return grandTotal; } // THIS APPEARS CORRECT SINCE IT WOULD HAVE NOT HIT ANY CATCH STATEMENT //SO IT IS IN MAIN CODE FLOW //IT WOULD HAVE ADDED lastDigitTotalDigits into grandTotal //just a case of adding extra 1 since no more digits left in number1 or number2 else { System.out.println("FINAL TESTING 2"); //it would append a 1 to the left of existing most significant digit grandTotal = "1" + grandTotal; System.out.println("GRANDTOTAL: " + grandTotal); return grandTotal; } } //unfortunately variable names are extremely confusing //need to remember it takes values from here onwards with one argument //if number1 was 9123 lengthFirstNumber=4 //there is a digit at index 3, so variable is reduced by 1 //There will be issue here if the Strings are not padded so that it fills the same //length System.out.println("**THIS IS ADDITION********"); System.out.println(number1); System.out.println(number2); //need something here since it was giving numberformat exception. //most likely since if no value in Strings (lastDigitFirstNumber //or lastDigitSecondNumber) BUT NOT BOTH. // it can not get Integer value henceforth. //note, it would have already dealt with circumstance of processing all //digits in both uniform numbers above in first loop in addition method. try { lastDigitFirstNumber=number1.substring(lengthFirstNumber-1); lastDigitSecondNumber=number2.substring(lengthSecondNumber-1); total = Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits; //checks if the total is greater than 10. This is the same index on number1, number2 and also //carry forward 1 if applicable if ((Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) + firstDigitTotalDigits)>=10) { System.out.println("\nOVER 10"); System.out.println("final test 3"); System.out.println("addition of: " + Integer.valueOf(lastDigitFirstNumber) + " + " + Integer.valueOf(lastDigitSecondNumber)); System.out.println("this is current total: " + total); //first digit carried forward, only feasible value is 1 firstDigitTotalDigits = (int)total/10; System.out.println("This is first digit, it will be carried forward: " + firstDigitTotalDigits); //getting last digit in total (10-18) lastDigitTotalDigits=total%10; //it has to be stored on the most left hand side (MSD) like real life grandTotal = Integer.toString(lastDigitTotalDigits) + grandTotal; System.out.println("running grand total: " + grandTotal); //recursive call. It truncates numbers by one addition(number1.substring(0, (lengthFirstNumber-1)), number2.substring(0, (lengthSecondNumber-1))); } else { //we expect total to be a single digit ///same principle above, added to MSD grandTotal = Integer.toString(total) + grandTotal; System.out.println("final test4"); System.out.println("\nless than 10"); System.out.println("addition of: " + Integer.valueOf(lastDigitFirstNumber) + " + " + Integer.valueOf(lastDigitSecondNumber)); System.out.println("this is current total: " + total); System.out.println("running grandtotal: " + grandTotal); //it needs to set the carry forward value to 0 firstDigitTotalDigits=0; //same style recursive call addition(number1.substring(0, (lengthFirstNumber-1)), number2.substring(0, (lengthSecondNumber-1))); } } //it would enter here if issue processing //lastDigitFirstNumber=number1.substring(lengthFirstNumber-1); //lastDigitSecondNumber=number2.substring(lengthSecondNumber-1); //Described above catch (NumberFormatException | StringIndexOutOfBoundsException s) { System.out.println("123This is last digit first number:" + lastDigitFirstNumber); System.out.println("123This is last digit second number:" + lastDigitSecondNumber); //now in here, it has to check if there is a carry over of 1 //if so it would add it to last digit of either number1 or number2 //for initial numbers such as, it is fortunately relatively more straight forward //4636 //+ 524 //but it was in a try and catch since if any of the values: //lastDigitFirstNumber="" or lastDigitSecondNumber="" //it would cause NumberFormatException since it can not format String => int //if the String is empty! We know however ONLY one of them is empty //if ((Integer.valueOf(lastDigitFirstNumber) + Integer.valueOf(lastDigitSecondNumber) //the alternate technique would be to add 0's on the front of the shortest number //to ensure both have uniform length, however this would affect readability when performing //System.out.println() // this is fine here System.out.println("wwww****************"); //for an unknown reason, I think it is related to do while loop prior to this //an extra zero is being inputted into grandTotal when dealing with streak of //9. Until I find a solution, I will just remove one zero if (successiveNinesNumber2) { System.out.println("final test FINAL!!!!!!"); System.out.println(lengthSecondNumber); remainingPortionString = number2.substring(0,lengthSecondNumber-counter+1); remainingPortionInteger=Integer.valueOf(remainingPortionString); System.out.println("REMAINING PORTION: " + remainingPortionString); System.out.println("counter: " + counter); //it would just simply drop remaining aspect of number2 downwards. //but it would add 1 onto the last digit in remaining portion. //it would get last digit on remaining portion //lastDigitSecondNumber = remainingPortionString.substring(0,1); lastDigitSecondNumberRemainingPortion = remainingPortionInteger%10; System.out.println(lastDigitSecondNumberRemainingPortion); System.out.println("******"); System.out.println(firstDigitTotalDigits); //more clarity than doing conversion of data type total = lastDigitSecondNumberRemainingPortion + firstDigitTotalDigits; //this is stripping off the 0, hence taking off 0 from front. grandTotal = grandTotal.substring(1); grandTotal = total + grandTotal; //now it would take off digit from the remainingPortionString //and drop rest numbers down. //remainingPortionInteger = Integer.valueOf(remainingPortionString); //remainingPortionInteger = remainingPortionInteger%10; //grandTotal = total + grandTotal; System.out.println("END FOR NUMBER 2 ARRANGEMENT, NEEED TO DO SAME NUMBER 1"); System.out.println(grandTotal); System.exit(0); } //still in catch, processing non uniform numbers.. System.out.println("number 1 at this point: " + number1); System.out.println("number 2 at this point: " + number2); //for instance in these numbers below, it would reach here and // lastDigitSecondNumber=3 and lastDigitFirstNumber=2 //it has not been able to progress to lastDigitSecondNumber=1 and //lastDigitFirstNumber since lastDigitFirstNumber has no value. //if I try to use these variables again, it would be a loop of errors. //A solution is to fill a 0 here, but once again I am refraining from this //to ensure screen readability mimics calculation in real life. //number2 = "136"; //number1 = "24"; //so now, it can no longer look at lastDigitFirstNumber or lastDigitSecondNumber //focus will be on number1 and number2 (string itself.) try { System.out.println("FAIL AGAIN"); System.out.println("number1: " + number1); System.out.println("number2: " + number2); //it shows number1 as blank //it shows number2 as 1 //NOTE BOTH CAN NOT BE BLANK! //String number2 = "136"; //String number1 = "24"; System.out.println("This is lastDigitSecondNumber: " + lastDigitSecondNumber); System.out.println("This is lastDigitFirstNumber: " + lastDigitFirstNumber); //String number2 = "136"; lastDigitSecondNumber=3 //String number1 = "24"; lastDigitFirstNumber=2 System.out.println("REMAINDER RIGHT NOW: " + firstDigitTotalDigits); //ie number2 is longer digits wide than number1 if (number2!=null) { System.out.println("final test 6"); //if this is the case, it has to get the last digit //and perform addition on this alone as per usual.... if(number2.length()>1) { System.out.println("final test 7 "); lastDigitSecondNumberRemainingPortion = (Integer.valueOf(number2)%10); System.out.println("lastDigitSecondNumberRemainingPortion: " + lastDigitSecondNumberRemainingPortion); //at this point, we still need to //have lots consideration. Before above it would carry across the //firstDigitTotalDigits (which would be 1 if total is 10-18) //if no extra logic is applied, it will perform: //String number2 = "9947"; //String number1 = "72"; //7+2 = 9 (grandtotal = 9) //4+7= 11 (store 1 and carry 1 across) (grandtotal = 19) //9+1 = 10 (it will not carry anything) (grandtotal = 1019) (INCORRECT) //drop the last digit number1 (grandtotal = 91019) (INCORRECT) //it is difficult to send it back to above loop for recursion... //so need to hardcode this logic in again //if total is less than 10, it can simply get total and drop numbers like above //total as usual, full visibility of firstDigitTotalDigits from legitimate calculation //on last digit at uniform location in number1 and number2 System.out.println("lastdigit remaining portion:" + lastDigitSecondNumberRemainingPortion); total = Integer.valueOf(lastDigitSecondNumberRemainingPortion) + firstDigitTotalDigits; //for example 9647 // + 72 //6+ 1 (carried forward) is less than 10. if (total<10) { System.out.println("WHAT IS TOTAL: " + total); System.out.println("final test 8"); //drops off last digit remaining portion remainingPortionInteger = (int) (Integer.valueOf(number2) / 10); remainingPortionString = Integer.toString(remainingPortionInteger); System.out.println("1This is remaining portion after conversion: " + remainingPortionString); remainingPortion = remainingPortionString.substring(0, (lengthSecondNumber-1)); System.out.println("This is remaining part: " + remainingPortion); //this is not applicable to both scenarios.... //this is applicable to if single carry forward does not cause MSD to exceed //9 //it can simply drop 9 (or infact the entire remainingPortion), (6+1), existing grandTotal grandTotal = remainingPortion + Integer.toString(total) + grandTotal; } //total is greater than or equal to 10 (9+1) //NOTE, the total can not exceed 10 anymore. //this else loop will get very involved later on. //since the carry forward could have a knock on effect //if the upcoming MSD are 9's! //for example 9947 // + 72 // else { System.out.println("final test 9"); System.out.println("1GREATER!!!!!!!!!"); firstDigitTotalDigits=(int) total/10; //expecting to get a 1 //it could equally be a 0. //it still has visibility from the addition on uniform length... System.out.println("This is the carry forward: " + firstDigitTotalDigits); lastDigitTotalDigits=total%10; //number between 0 and 8 //expecting a 1 here remainingPortionInteger = (int) (Integer.valueOf(number2) / 10); System.out.println("should be: " + remainingPortionInteger); lastDigitSecondNumberRemainingPortion =Integer.valueOf(Integer.valueOf(number2)%10); System.out.println("WHATSSSSS: " + lastDigitSecondNumberRemainingPortion); //it would enter in similar situation but not // "9627" "927" // + "92" + "92" //System.out.println("CONFIRM RIGHT TRACK TOTAL GREATER THAN 10: " + number2.substring((lengthSecondNumber-1))); //no cascading effect //it would know that even if carry 1 forward was applicable, it would stay single digit //NEED to be careful and understand that if remainder firstDigitTotalDigits was 0 (6+2=8) //The 9 in number.substring(lengthSecondNumber-1) would still be valid //since it would not be affected. //for instance 9927 // + 62 //one inwards if (((Integer.valueOf(number2.substring((lengthSecondNumber-1)))+firstDigitTotalDigits)<10)) { System.out.println(grandTotal + " so far"); System.out.println("final test 10"); System.out.println("£££££££££££££££££"); //System.out.println(lastDigitSecondNumber); //we know there wont be cascading effect... //this would do 6 + carry forward), //remainingPortion(MSD) into grandtotal. On this instance it would be 9 //(9647 //+ 72) //remainingPortionString=number2.substring((lengthSecondNumber-1)); //remainingPortionInteger=String.valueOf(remainingPortionString); //total = remainingPortionInteger + firstDigitTotalDigits; //grandTotal = Integer.toString(total) + grandTotal; //System.out.println("G: " + grandTotal); //expecting this digit to be subject to remainder if applicable //NEED to use remaingPortion here... //***HERE*********** total = lastDigitSecondNumberRemainingPortion + firstDigitTotalDigits; //this will put the remainingpart(9) + total (9 + 0 carried forward) + existing grandTotal grandTotal = remainingPortionInteger + total + grandTotal; System.out.println("G: " + grandTotal); //it would now truncate single digit of number 2 //and place this in the grandtotal System.out.println("WHAT IS NUMBER2 RIGHT NOW:" + number2); } //there will be minimum 1 cascading effect //for instance 19947 // + 62 else { //at this point it has to examine all the remainingPortion (199) //backwards since if there are multiple 9 this will keep cascading... //a good starting point is being here and acknowleding there is a single cascading effect //at minimum System.out.println("final test 11"); System.out.println("123****** REMAINING ****: " + number2); System.out.println(number2.substring((lengthSecondNumber-1))); System.out.println(firstDigitTotalDigits); counter=0; //this can only be a single digit of 9 remaining //it would simply perform 9 + 1 = 10. //for example //947 //+ 62 if(lengthSecondNumber==1) { System.out.println("final test 12"); //total HAS to be 10 //we expect a record of a carry forward since it would be //9 appearing straight away after the lengths processed and differences at uniform location //number1 and number2 remainingPortionInteger=Integer.valueOf(number2); //there is a single digit hence number2 can be taken as a whole. total = remainingPortionInteger + firstDigitTotalDigits; //it can now be placed into grandtotal grandTotal = Integer.toString(total) + grandTotal; } //this is almost a recursive do while loop without calling //recursive method. //I am not sure if this is within restrictions of the challenge //but its not practical to call the addition recursive method again.. //it can be undertaken by setting up a flag perhaps consecutiveNines //and completing this code below in that part.. //but I think its best to keep flow going downwards than to move upwards.. //more than 1 digit remaining //and whilst they are all 9s else { System.out.println("final test 13"); //in practice it should still have the record of firstDigitTotalDigits //and we know it is not 0 otherwise //it would have executed this loop: //if (((Integer.valueOf(number2.substring((lengthSecondNumber-1)))+firstDigitTotalDigits)<10)) //CAREFUL, but ok to set it again. firstDigitTotalDigits=1; //also need to remember that still within //massive else statement of the extra digit in number2 equalling //10 (with 1 carry forward). //for instance the 39 part left of the number // 3974 // + 45 //This now includes entire block //note number2 has not got shorter at all remainingPortionInteger = (int) (Integer.valueOf(number2))/10; temp=Integer.valueOf(number2)%10; System.out.println(number2); System.out.println(remainingPortionInteger); //it takes a digit off the end. It is just to double verify it s a 9s //it should be the case lastDigitSecondNumberRemainingPortion = remainingPortionInteger % 10; System.out.println("WHAT is: " + lastDigitSecondNumberRemainingPortion); successiveNinesNumber2=true; do { System.out.println("reminder of : " + lastDigitSecondNumberRemainingPortion); System.out.println("final test 14"); if (lastDigitSecondNumberRemainingPortion==9) { System.out.println("\n\nfinal test 15 - definitely a 9"); //we know this can only be as a result of 9+1 total = 0; // (equivalent to 9+1 and obtaining last digit) //this begins to write a 0 for every 9 that it finds in remainingPortionInteger grandTotal = Integer.toString(total) + grandTotal; //this will get shorter by a digit each time remainingPortionInteger = remainingPortionInteger % 10; //at same time expecting a 9 to be filled on left hand side and carry over //CAREFUL WHAT IS THE PURPOSE OF THIS? //this is so that once it breaks out of the loop, //it has a total of MSD before the 9. //We know the previous MSD would have been 9+1 = 0 //so it includes the lastDigitSecondNumberRemainingPortion //CAREFUL //it might be ok infact total = lastDigitSecondNumberRemainingPortion + firstDigitTotalDigits; //it will now get new last digit //the purpose of this is so that once it finishes the do while loop //of storing 9s, it can process anything else left //remainingPortionInteger //CAREFUL, does this need to be %10 ? lastDigitSecondNumberRemainingPortion = remainingPortionInteger%10; counter++; System.out.println("\nnew grand total: " + grandTotal); System.out.println("ddd: " + number2.charAt(lengthSecondNumber-counter-1)); System.out.println("COUNTER: " + counter); System.out.println("bf: " + lastDigitSecondNumberRemainingPortion); //didnt want to use this in recursive challenge since it appears to be //indexing. //}while(counter1) //this would be the condition number2==1) //so it would need to undertake manipulation on number1 //and drop it into the grandTotal else { System.out.println("DOUBLE CHECK REMAINDER: " + firstDigitTotalDigits); System.out.println("final test 17 - dropping if excess digit is one number wide"); //the value has to be made blank since it has no content and default it is null //this will impact presentation of grandTotal System.out.println(lastDigitFirstNumber); total = Integer.valueOf(lastDigitFirstNumber) + firstDigitTotalDigits; System.out.println("uuTOTAL NOW: " + total); //CAREFUL, it was commented out //remainingPortion=""; remainingPortionInteger= Integer.valueOf(number2); System.out.println("REMAINING PORTION: " + remainingPortionInteger); //this will just simply take the number and any carry forward from previous MSD total = remainingPortionInteger + firstDigitTotalDigits; System.out.println("TOTALLLL: " + total); grandTotal = total + grandTotal; System.out.println("GRANDDDD TOTAL: " + grandTotal); return grandTotal; } //at this point we have dealt with scenarios such as these since it has logic //to carry forward 1 // 4636 // + 324 // 4636 // + 524 //but if the numbers were: 84636 // + 524 //it would fail since it has not had opportunity to process the 8 in number1 //so simply, once it has performed total with 4 and no value in number2 //it needs to take all the remaining numbers in the substring //and append it to the grand total //in effect it will shorten length number again by 1.. //note this time the shortening is normal. } // this is the end of number2!=null } //end try //exactly same as above, but flipped to number1 //repeat code but it is required rather than passing value into another //method (is this allowed if recursive?). It will create more variables //and hit on performance. //NOTE THIS IS STILL WITHIN //catch (NumberFormatException | StringIndexOutOfBoundsException s) //It has decided at this point that number2=null //so we know that number1 is longer. //I AM REFRAINING FROM EXTENDING THIS CODE OUTWARDS.. //it can be seen its a massive duplication in code //just for the sake of making number1 more digits wide than number2 //this code will become redundant... //no test cases are taken through this area... //The challenge does show that number2 can be shorter than number1 and vice versa. //ALL that is required is copying exactly code above from number2!=null and pasting it here //and changing all variables to reflect number1 catch (NumberFormatException n) { } //this part needs addressing. System.out.println("FFFFFtotal: " + grandTotal); System.out.println("remaining portion: " + remainingPortion); System.out.println("total: " + total); //EXTRA LINE ADDED - CAREFUL //grandTotal = Integer.toString(total) + grandTotal; System.out.println("final test 17"); //LAST ADDITION - CAREFUL grandTotal=total + grandTotal; System.out.println("grand23333: " +grandTotal); return grandTotal; } //end of catch (NumberFormatException | StringIndexOutOfBoundsException s) return "test"; //} // end of if number%10!=0 } }