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Head of Career Skills Development & Coaching

*Based on past data of successful IK students

Given an array of numbers, return an array of the same size where i-th element is the product of all elements except the i-th one.

Calculate the products modulo 10^{9} + 7.

Using division is not allowed anywhere in the solution. Using more than constant auxiliary space is not allowed.

```
{
"nums": [1, 2, 3, 4, 5]
}
```

Output:

```
[120, 60, 40, 30, 24]
```

`[(2 * 3 * 4 * 5), (1 * 3 * 4 * 5), (1 * 2 * 4 * 5), (1 * 2 * 3 * 5), (1 * 2 * 3 * 4)]`

= `[120, 60, 40, 30, 24]`

Constraints:

- 2 <= length of the array <= 100000
- -10
^{9}<= a number in the array <= 10^{9}

If you are getting a wrong answer for some test cases, but think your logic is correct, check for the overflow.

Let `mod`

= 10^{9} + 7.

If `a`

= 10^{9} and `b`

= 10^{9}, then `int c = (a * b) % mod`

will overflow but this will work: `int c = (a * (long long) 1 * b) % mod`

. By multiplying with `(long long) 1`

, we make sure that the calculation is done in `long long`

, instead of `int`

.

We provided two solutions.

Let us refer to the length of input array `nums`

as `n`

.

A naive approach would be to find the `i`

-th element of the output array (i.e. `products[i]`

), iterate over the entire input array to get the product of all elements `nums[j]`

, such that `j != i`

.

O(n^{2}).

As we are iterating over the entire array to find `products[i]`

and as it can be `0 <= i <= (n - 1)`

. Each calculation of element of products array will take O(n) so total complexity will be O(n^{2}).

O(1).

As we are not storing anything extra and excluding space used to store output array products.

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(n).

So, total space complexity: O(n) + O(1) + O(n) = O(n).

```
/*
* Asymptotic complexity in terms of the size of the input array \`n\`:
* Time: O(n^2).
* Auxiliary space: O(1).
* Total space: O(n).
*/
static int mod = (int)Math.pow(10, 9) + 7;
static ArrayList<Integer> get_product_array(ArrayList<Integer> nums) {
// Size of output array is same as that of input array
ArrayList<Integer> products = new ArrayList<Integer>(Collections.nCopies(nums.size(), 1));
for (int currentIndex = 0; currentIndex < nums.size(); currentIndex++) {
for (int iterator = 0; iterator < nums.size(); iterator++) {
if (iterator != currentIndex) {
nums.set(iterator, (nums.get(iterator).intValue() > 0 ? (nums.get(iterator).intValue() % mod) : ((mod + nums.get(iterator)) % mod)));
products.set(currentIndex, (int)((products.get(currentIndex).intValue() * 1l * nums.get(iterator).intValue()) % mod));
}
}
}
return products;
}
```

Notice that for `products[i]`

, product of all input array elements other than `i`

-th element is nothing but (product of all elements `nums[j]`

, `0 <= j <= (i - 1)`

) * (product of all elements `nums[j]`

, `(i + 1) <= j <= (n - 1)`

) = `(nums[0] * nums[1] * ... * nums[i - 1]) * (nums[i + 1] * nums[i + 2] * ...* nums[n - 1])`

.

So, iterate over the input array twice to fill output array products, once for updating `products[i]`

with `(nums[0] * nums[1] *...* nums[i - 1])`

, and next one for updating `products[i]`

with `(nums[i + 1] * nums[i + 2] * … * nums[n - 1])`

.

O(n).

As we are iterating over the input array two times it will take O(n).

O(1).

We are not storing anything extra.

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(n).

So, total space complexity: O(n) + O(1) + O(n) = O(n).

```
/*
* Asymptotic complexity in terms of the size of the input array \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
static int mod = (int)Math.pow(10, 9) + 7;
static ArrayList<Integer> get_product_array(ArrayList<Integer> nums) {
// Size of output array is same as that of input array
ArrayList<Integer> products = new ArrayList<>(Collections.nCopies(nums.size(), 0));
// For finding value of products[i], product of all nums elements
// other than ith element is nothing but
// (product of all nums[j], 0 <= j <= (i - 1)) * (product of all nums[j], (i + 1) <= j <= (nums.size() - 1))
// i.e. (nums[0] * nums[1] * ...* nums[i - 1]) * (nums[i + 1] * nums[i + 2] * ... * nums[nums.size() - 1])
int leftProduct = 1;
// Filling products, such that products[i] contains
// product of all elements nums[j], 0 <= j <= (i - 1)
for (int currentIndex = 0; currentIndex < nums.size(); currentIndex++) {
// Here, leftProduct contains product of all elements
// nums[j], 0 <= j <= (currentIndex - 1)
products.set(currentIndex, leftProduct);
// After this updation of leftProduct, leftProduct contains product of all
// elements nums[j], 0 <= j <= currentIndex
nums.set(currentIndex, (nums.get(currentIndex) > 0 ? nums.get(currentIndex) : (mod + nums.get(currentIndex)) % mod));
leftProduct = (int)((leftProduct * 1l * nums.get(currentIndex)) % mod);
}
int rightProduct = 1;
// Updating products, such that products[i] contains new value
// ((products[i]) * (product of all elements nums[j], 0 <= j <= (i - 1)))
for (int currentIndex = nums.size() - 1; currentIndex >= 0; currentIndex--) {
// Here, rightProduct contains product of all elements
// nums[j], (currentIndex + 1) <= j <= (nums.size() - 1)
products.set(currentIndex, ((int)((products.get(currentIndex) * 1l * rightProduct) % mod)));
// after this updation of rightProduct, rightProduct contains product of all
// elements nums[j], currentIndex <= j <= (nums.size() - 1)
rightProduct = (int)((rightProduct * 1l * nums.get(currentIndex)) % mod);
}
return products;
}
```

We hope that these solutions to array product problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

Note: Input and Output will already be taken care of.

Given an array of numbers, return an array of the same size where i-th element is the product of all elements except the i-th one.

Calculate the products modulo 10^{9} + 7.

Using division is not allowed anywhere in the solution. Using more than constant auxiliary space is not allowed.

```
{
"nums": [1, 2, 3, 4, 5]
}
```

Output:

```
[120, 60, 40, 30, 24]
```

`[(2 * 3 * 4 * 5), (1 * 3 * 4 * 5), (1 * 2 * 4 * 5), (1 * 2 * 3 * 5), (1 * 2 * 3 * 4)]`

= `[120, 60, 40, 30, 24]`

Constraints:

- 2 <= length of the array <= 100000
- -10
^{9}<= a number in the array <= 10^{9}

If you are getting a wrong answer for some test cases, but think your logic is correct, check for the overflow.

Let `mod`

= 10^{9} + 7.

If `a`

= 10^{9} and `b`

= 10^{9}, then `int c = (a * b) % mod`

will overflow but this will work: `int c = (a * (long long) 1 * b) % mod`

. By multiplying with `(long long) 1`

, we make sure that the calculation is done in `long long`

, instead of `int`

.

We provided two solutions.

Let us refer to the length of input array `nums`

as `n`

.

A naive approach would be to find the `i`

-th element of the output array (i.e. `products[i]`

), iterate over the entire input array to get the product of all elements `nums[j]`

, such that `j != i`

.

O(n^{2}).

As we are iterating over the entire array to find `products[i]`

and as it can be `0 <= i <= (n - 1)`

. Each calculation of element of products array will take O(n) so total complexity will be O(n^{2}).

O(1).

As we are not storing anything extra and excluding space used to store output array products.

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(n).

So, total space complexity: O(n) + O(1) + O(n) = O(n).

```
/*
* Asymptotic complexity in terms of the size of the input array \`n\`:
* Time: O(n^2).
* Auxiliary space: O(1).
* Total space: O(n).
*/
static int mod = (int)Math.pow(10, 9) + 7;
static ArrayList<Integer> get_product_array(ArrayList<Integer> nums) {
// Size of output array is same as that of input array
ArrayList<Integer> products = new ArrayList<Integer>(Collections.nCopies(nums.size(), 1));
for (int currentIndex = 0; currentIndex < nums.size(); currentIndex++) {
for (int iterator = 0; iterator < nums.size(); iterator++) {
if (iterator != currentIndex) {
nums.set(iterator, (nums.get(iterator).intValue() > 0 ? (nums.get(iterator).intValue() % mod) : ((mod + nums.get(iterator)) % mod)));
products.set(currentIndex, (int)((products.get(currentIndex).intValue() * 1l * nums.get(iterator).intValue()) % mod));
}
}
}
return products;
}
```

Notice that for `products[i]`

, product of all input array elements other than `i`

-th element is nothing but (product of all elements `nums[j]`

, `0 <= j <= (i - 1)`

) * (product of all elements `nums[j]`

, `(i + 1) <= j <= (n - 1)`

) = `(nums[0] * nums[1] * ... * nums[i - 1]) * (nums[i + 1] * nums[i + 2] * ...* nums[n - 1])`

.

So, iterate over the input array twice to fill output array products, once for updating `products[i]`

with `(nums[0] * nums[1] *...* nums[i - 1])`

, and next one for updating `products[i]`

with `(nums[i + 1] * nums[i + 2] * … * nums[n - 1])`

.

O(n).

As we are iterating over the input array two times it will take O(n).

O(1).

We are not storing anything extra.

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(n).

So, total space complexity: O(n) + O(1) + O(n) = O(n).

```
/*
* Asymptotic complexity in terms of the size of the input array \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
static int mod = (int)Math.pow(10, 9) + 7;
static ArrayList<Integer> get_product_array(ArrayList<Integer> nums) {
// Size of output array is same as that of input array
ArrayList<Integer> products = new ArrayList<>(Collections.nCopies(nums.size(), 0));
// For finding value of products[i], product of all nums elements
// other than ith element is nothing but
// (product of all nums[j], 0 <= j <= (i - 1)) * (product of all nums[j], (i + 1) <= j <= (nums.size() - 1))
// i.e. (nums[0] * nums[1] * ...* nums[i - 1]) * (nums[i + 1] * nums[i + 2] * ... * nums[nums.size() - 1])
int leftProduct = 1;
// Filling products, such that products[i] contains
// product of all elements nums[j], 0 <= j <= (i - 1)
for (int currentIndex = 0; currentIndex < nums.size(); currentIndex++) {
// Here, leftProduct contains product of all elements
// nums[j], 0 <= j <= (currentIndex - 1)
products.set(currentIndex, leftProduct);
// After this updation of leftProduct, leftProduct contains product of all
// elements nums[j], 0 <= j <= currentIndex
nums.set(currentIndex, (nums.get(currentIndex) > 0 ? nums.get(currentIndex) : (mod + nums.get(currentIndex)) % mod));
leftProduct = (int)((leftProduct * 1l * nums.get(currentIndex)) % mod);
}
int rightProduct = 1;
// Updating products, such that products[i] contains new value
// ((products[i]) * (product of all elements nums[j], 0 <= j <= (i - 1)))
for (int currentIndex = nums.size() - 1; currentIndex >= 0; currentIndex--) {
// Here, rightProduct contains product of all elements
// nums[j], (currentIndex + 1) <= j <= (nums.size() - 1)
products.set(currentIndex, ((int)((products.get(currentIndex) * 1l * rightProduct) % mod)));
// after this updation of rightProduct, rightProduct contains product of all
// elements nums[j], currentIndex <= j <= (nums.size() - 1)
rightProduct = (int)((rightProduct * 1l * nums.get(currentIndex)) % mod);
}
return products;
}
```

We hope that these solutions to array product problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart's FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

**Attend our free webinar to amp up your career and get the salary you deserve.**

Hosted By

Ryan Valles

Founder, Interview Kickstart