Description#
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 space complexity? (The output array does not count as extra space for space complexity analysis.)
Solutions#
Solution 1: Two Passes#
We define two variables $left$ and $right$, which represent the product of all elements to the left and right of the current element respectively. Initially, $left=1$, $right=1$. Define an answer array $ans$ of length $n$.
We first traverse the array from left to right, for the $i$th element we update $ans[i]$ with $left$, then $left$ multiplied by $nums[i]$.
Then, we traverse the array from right to left, for the $i$th element, we update $ans[i]$ to $ans[i] \times right$, then $right$ multiplied by $nums[i]$.
After the traversal, the array ans is the answer.
The time complexity is $O(n)$, where $n$ is the length of the array nums. Ignore the space consumption of the answer array, the space complexity is $O(1)$.
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| class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [0] * n
left = right = 1
for i, x in enumerate(nums):
ans[i] = left
left *= x
for i in range(n - 1, -1, -1):
ans[i] *= right
right *= nums[i]
return ans
|
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| class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
for (int i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (int i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
}
|
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| class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n);
for (int i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (int i = n - 1, right = 1; ~i; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
};
|
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| func productExceptSelf(nums []int) []int {
n := len(nums)
ans := make([]int, n)
left, right := 1, 1
for i, x := range nums {
ans[i] = left
left *= x
}
for i := n - 1; i >= 0; i-- {
ans[i] *= right
right *= nums[i]
}
return ans
}
|
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| function productExceptSelf(nums: number[]): number[] {
const n = nums.length;
const ans: number[] = new Array(n);
for (let i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (let i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
|
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| impl Solution {
pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut ans = vec![1; n];
for i in 1..n {
ans[i] = ans[i - 1] * nums[i - 1];
}
let mut r = 1;
for i in (0..n).rev() {
ans[i] *= r;
r *= nums[i];
}
ans
}
}
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| /**
* @param {number[]} nums
* @return {number[]}
*/
var productExceptSelf = function (nums) {
const n = nums.length;
const ans = new Array(n);
for (let i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (let i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
};
|
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| public class Solution {
public int[] ProductExceptSelf(int[] nums) {
int n = nums.Length;
int[] ans = new int[n];
for (int i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (int i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
}
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| class Solution {
/**
* @param Integer[] $nums
* @return Integer[]
*/
function productExceptSelf($nums) {
$n = count($nums);
$ans = [];
for ($i = 0, $left = 1; $i < $n; ++$i) {
$ans[$i] = $left;
$left *= $nums[$i];
}
for ($i = $n - 1, $right = 1; $i >= 0; --$i) {
$ans[$i] *= $right;
$right *= $nums[$i];
}
return $ans;
}
}
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Solution 2#
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| function productExceptSelf(nums: number[]): number[] {
return nums.map((_, i) => nums.reduce((pre, val, j) => pre * (i === j ? 1 : val), 1));
}
|
