Question
Given an array nums
of n integers where n > 1, return an array output such that output[i]
is equal to the product of all the elements of nums
except nums[i]
.
Example:
Input: [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
Note:
We can create two arrays left
andright
to store the multiplication of all left and right numbers of the corresponding position. Then, we just need to multiply them.
Solution:
class Solution {
public int[] productExceptSelf(int[] nums) {
int len = nums.length;
int[] left = new int[len];
int[] right = new int[len];
int[] ans = new int[len];
// left[i] contains the product of all the elements to the left
// Note: for the element at index '0', there are no elements to the left,
// so left[0] would be 1
left[0] = 1;
for (int i = 1; i < len; i++) {
left[i] = left[i - 1] * nums[i - 1];
}
// right[i] contains the product of all the elements to the right
// Note: for the element at index 'len - 1', there are no elements to the right,
// so right[len - 1] would be 1
right[len - 1] = 1;
for (int i = len - 1; i >= 0; i--) {
right[i] = right[i + 1] * nums[i + 1];
}
for (int i = 0; i < len; i++) {
ans[i] = left[i]*right[i];
}
return ans;
}
}
Solution of follow-up:
class Solution {
public int[] productExceptSelf(int[] nums) {
int len = nums.length;
int[] ans = int[len];
// first, we stored all the "left product" in the answer array
ans[0] = 1;
for (int i = 1; i < len; i++) {
ans[i] = ans[i - 1]*nums[i - 1];
}
int right = 1;
for (int i = len - 1; i >= 0; i--) {
ans[i] = ans[i]*right;
right = right*nums[i];
}
return ans;
}
}