Products excluding each element of the array
$begingroup$
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6].
Is there a better solution for this problem?
Note: Division is not allowed.
public class DailyCodingProblem2 {
public static void main(String args) {
int arr = { 1, 2, 3, 4, 5 };
int ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
arr = new int { 3, 2, 1 };
ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
}
private static int solution(int arr, int n) {
int left = new int[n];
int right = new int[n];
int res = new int[n];
left[0] = 1;
right[n - 1] = 1;
for (int i = 1; i < n; i++) {
left[i] = arr[i - 1] * left[i - 1];
}
for (int i = n - 2; i >= 0; i--) {
right[i] = arr[i + 1] * right[i + 1];
}
for (int i = 0; i < n; i++) {
res[i] = left[i] * right[i];
}
return res;
}
}
java algorithm array
edited 12 mins ago
Solomon Ucko
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
$endgroup$
add a comment |
$begingroup$
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6].
Is there a better solution for this problem?
Note: Division is not allowed.
public class DailyCodingProblem2 {
public static void main(String args) {
int arr = { 1, 2, 3, 4, 5 };
int ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
arr = new int { 3, 2, 1 };
ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
}
private static int solution(int arr, int n) {
int left = new int[n];
int right = new int[n];
int res = new int[n];
left[0] = 1;
right[n - 1] = 1;
for (int i = 1; i < n; i++) {
left[i] = arr[i - 1] * left[i - 1];
}
for (int i = n - 2; i >= 0; i--) {
right[i] = arr[i + 1] * right[i + 1];
}
for (int i = 0; i < n; i++) {
res[i] = left[i] * right[i];
}
return res;
}
}
java algorithm array
edited 12 mins ago
Solomon Ucko
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
$endgroup$
$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago
add a comment |
$begingroup$
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6].
Is there a better solution for this problem?
Note: Division is not allowed.
public class DailyCodingProblem2 {
public static void main(String args) {
int arr = { 1, 2, 3, 4, 5 };
int ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
arr = new int { 3, 2, 1 };
ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
}
private static int solution(int arr, int n) {
int left = new int[n];
int right = new int[n];
int res = new int[n];
left[0] = 1;
right[n - 1] = 1;
for (int i = 1; i < n; i++) {
left[i] = arr[i - 1] * left[i - 1];
}
for (int i = n - 2; i >= 0; i--) {
right[i] = arr[i + 1] * right[i + 1];
}
for (int i = 0; i < n; i++) {
res[i] = left[i] * right[i];
}
return res;
}
}
java algorithm array
edited 12 mins ago
Solomon Ucko
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
$endgroup$
Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i.
For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6].
Is there a better solution for this problem?
Note: Division is not allowed.
public class DailyCodingProblem2 {
public static void main(String args) {
int arr = { 1, 2, 3, 4, 5 };
int ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
arr = new int { 3, 2, 1 };
ans = solution(arr, arr.length);
System.out.println(Arrays.toString(ans));
}
private static int solution(int arr, int n) {
int left = new int[n];
int right = new int[n];
int res = new int[n];
left[0] = 1;
right[n - 1] = 1;
for (int i = 1; i < n; i++) {
left[i] = arr[i - 1] * left[i - 1];
}
for (int i = n - 2; i >= 0; i--) {
right[i] = arr[i + 1] * right[i + 1];
}
for (int i = 0; i < n; i++) {
res[i] = left[i] * right[i];
}
return res;
}
}
java algorithm array
java algorithm array
edited 12 mins ago
Solomon Ucko
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
edited 12 mins ago
Solomon Ucko
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
edited 12 mins ago
Solomon Ucko
1,112415
edited 12 mins ago
Solomon Ucko
1,112415
edited 12 mins ago
Solomon Ucko
1,112415
1,112415
asked 2 days ago
Maclean PintoMaclean Pinto
1225
asked 2 days ago
Maclean PintoMaclean Pinto
1225
asked 2 days ago
Maclean PintoMaclean Pinto
1225
1225
$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago
add a comment |
$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago
$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Time complexity-wise, I think not. You're required to 'visit' all numbers, and your solution is O(n), so that can't be improved.
Code clarity could be improved, as it's not very obvious what the intention is. Some comments would help that.
I think shifting indices by 1 might make it clearer (left[i+1] = arr[i] * left[i]), but then maybe not because it'd mess up the last loop.
Have you explored different algorithms? I wonder if straightforward memoization makes a very clear solution for this.
answered 2 days ago
domendomen
1762
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Time complexity-wise, I think not. You're required to 'visit' all numbers, and your solution is O(n), so that can't be improved.
Code clarity could be improved, as it's not very obvious what the intention is. Some comments would help that.
I think shifting indices by 1 might make it clearer (left[i+1] = arr[i] * left[i]), but then maybe not because it'd mess up the last loop.
Have you explored different algorithms? I wonder if straightforward memoization makes a very clear solution for this.
answered 2 days ago
domendomen
1762
$endgroup$
add a comment |
$begingroup$
Time complexity-wise, I think not. You're required to 'visit' all numbers, and your solution is O(n), so that can't be improved.
Code clarity could be improved, as it's not very obvious what the intention is. Some comments would help that.
I think shifting indices by 1 might make it clearer (left[i+1] = arr[i] * left[i]), but then maybe not because it'd mess up the last loop.
Have you explored different algorithms? I wonder if straightforward memoization makes a very clear solution for this.
answered 2 days ago
domendomen
1762
$endgroup$
add a comment |
$begingroup$
Time complexity-wise, I think not. You're required to 'visit' all numbers, and your solution is O(n), so that can't be improved.
Code clarity could be improved, as it's not very obvious what the intention is. Some comments would help that.
I think shifting indices by 1 might make it clearer (left[i+1] = arr[i] * left[i]), but then maybe not because it'd mess up the last loop.
Have you explored different algorithms? I wonder if straightforward memoization makes a very clear solution for this.
answered 2 days ago
domendomen
1762
$endgroup$
Time complexity-wise, I think not. You're required to 'visit' all numbers, and your solution is O(n), so that can't be improved.
Code clarity could be improved, as it's not very obvious what the intention is. Some comments would help that.
I think shifting indices by 1 might make it clearer (left[i+1] = arr[i] * left[i]), but then maybe not because it'd mess up the last loop.
Have you explored different algorithms? I wonder if straightforward memoization makes a very clear solution for this.
answered 2 days ago
domendomen
1762
answered 2 days ago
domendomen
1762
answered 2 days ago
domendomen
1762
answered 2 days ago
domendomen
1762
1762
add a comment |
add a comment |
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$begingroup$
Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
$endgroup$
– Mast
2 days ago
$begingroup$
"Note: Division is not allowed." Why? Is it because of a restriction for the assignment?
$endgroup$
– Solomon Ucko
2 days ago
$begingroup$
@SolomonUcko Yes it is a restriction for the assignment
$endgroup$
– Maclean Pinto
12 mins ago