238. Product of Array Except Self

Description of Problem

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:

Input: nums = [1,2,3,4]
Output: [24,12,8,6]

Example 2:

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]

Constraints:

2 <= nums.length <= 105
-30 <= nums[i] <= 30
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in O(1) extra Auxiliary Space? (The output array does not count as extra space for Auxiliary Space analysis.)

Without consideration of division, we can express the answer by the following equations \[ f(x_i) = X_{0..i} \cdot X_{(i+1)..n} \text{ where } X_{i..j} = \prod_{k=i}^{j-1} x_k \text{ and } x_k \text{ is element in the array with index k}\]

By observation, \(X_{0..i}\) is the product of prefixes of \(x_i\) and \(X_{(i+1)..n}\) is the product of suffixes of \(x_i\).

For Prefixes: \[ \begin{matrix} \left [ x_0, x_1, ... ,x_{n-2} ,x_{n-1} \right ] & \longrightarrow & \left [ 1, x_0, ... ,x_{n-3} ,x_{n-2} \right ] \\ & \longrightarrow & \left [ 1, x_0, ... ,\prod_{k=0}^{n-3} x_k, \prod_{k=0}^{n-2} x_k \right ] \\ & \longrightarrow & \left [ 1, X_{0..1}, ... ,X_{0..n-2}, X_{0..n-1} \right ] \end{matrix} \]

For Suffixes: \[ \begin{matrix} \left [ x_0, x_1, ... ,x_{n-2} ,x_{n-1} \right ] & \longrightarrow & \left [ x_1, x_2, ... ,x_{n-1} ,1 \right ] \\ & \longrightarrow & \left [ \prod_{k=1}^{n-1} x_k, \prod_{k=2}^{n-1} x_k, ... ,\prod_{k=n-1}^{n-1} x_k, 1 \right ] \\ & \longrightarrow & \left [ X_{1..n}, X_{2..n}, ... , X_{n-1..n}, 1 \right ] \end{matrix} \]

And combine the prefixes and suffixes into single result.

Code (Rust)

impl Solution {
    pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let mut result = vec![1; n];

        let (mut left_state, mut right_state) = (1, 1);
        for i in 0..n {
            result[i] *= left_state;
            left_state *= nums[i];

            result[n - i - 1] *= right_state;
            right_state *= nums[n - i - 1];
        }

        return result;
    }
}

Complexity

n is number of elements in the array

Time complexity:

\(\Theta(n)\)

Auxiliary Space:

\(O(1)\)

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