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May 2, 2021 10:07
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| public static int[] algoFunc(int[] numArray) { | |
| int totalMultiplication = 1; | |
| for(int num : numArray) { | |
| try{ | |
| totalMultiplication = totalMultiplication * num; | |
| }catch (ArithmeticException e){ | |
| totalMultiplication = 0; | |
| break; | |
| } | |
| } | |
| int arrayLength = numArray.length; | |
| int[] newArray = new int[arrayLength]; | |
| for(int i = 0 ; i < arrayLength ; i++) { | |
| if(numArray[i] == 0) { | |
| int multiplier = 1; | |
| //This comment encloses the short code I could come up with to solve the algorithm but I believe | |
| for(int j = 0 ; j < arrayLength ; j++) { | |
| if(j == i) continue; | |
| try{ | |
| multiplier = multiplier * numArray[j]; | |
| }catch (ArithmeticException e) { | |
| multiplier = 0; | |
| break; | |
| } | |
| } | |
| //it's madly inefficient when the array does not contain 0. | |
| //Hence my solution. | |
| newArray[i] = multiplier; | |
| continue; | |
| } | |
| try { | |
| newArray[i] = totalMultiplication /numArray[i]; | |
| }catch (ArithmeticException e){ | |
| newArray[i] = 0; | |
| } | |
| } | |
| return newArray; | |
| } |
Thank you @meekg33k for conducting Algrorithn Fridays.
I currently can't think of any better solution, I'm looking forward to the published solution.
It'll be really informative and interesting to see the best eay to solve this without a nested loop.
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Hello @Itz-kwaz, thank you for participating in Week 4 of Algorithm Fridays.
Your solution passes all the test cases in terms of correctness. Kudos to you!
However, in terms of time efficiency, your solution uses a nested loop which causes a time complexity of O(N). Do you think we can do better?
All the same, this is a good attempt! Let me know what you think.