Leetcode

Product of Array Except Self

Approach 1: $O(n)$ space

  • Time:O(n)
  • Space:O(n)

C++

class Solution {
 public:
  vector<int> productExceptSelf(vector<int>& nums) {
    const int n = nums.size();
    vector<int> ans(n);        // can also use nums as the ans array
    vector<int> prefix(n, 1);  // prefix product
    vector<int> suffix(n, 1);  // suffix product

    for (int i = 1; i < n; ++i)
      prefix[i] = prefix[i - 1] * nums[i - 1];

    for (int i = n - 2; i >= 0; --i)
      suffix[i] = suffix[i + 1] * nums[i + 1];

    for (int i = 0; i < n; ++i)
      ans[i] = prefix[i] * suffix[i];

    return ans;
  }
};

JAVA

class Solution {
  public int[] productExceptSelf(int[] nums) {
    final int n = nums.length;
    int[] ans = new int[n];    // can also use nums as the ans array
    int[] prefix = new int[n]; // prefix product
    int[] suffix = new int[n]; // suffix product

    prefix[0] = 1;
    for (int i = 1; i < n; ++i)
      prefix[i] = prefix[i - 1] * nums[i - 1];

    suffix[n - 1] = 1;
    for (int i = n - 2; i >= 0; --i)
      suffix[i] = suffix[i + 1] * nums[i + 1];

    for (int i = 0; i < n; ++i)
      ans[i] = prefix[i] * suffix[i];

    return ans;
  }
}

Python

class Solution:
  def productExceptSelf(self, nums: List[int]) -> List[int]:
    n = len(nums)
    prefix = [1] * n  # prefix product
    suffix = [1] * n  # suffix product

    for i in range(1, n):
      prefix[i] = prefix[i - 1] * nums[i - 1]

    for i in reversed(range(n - 1)):
      suffix[i] = suffix[i + 1] * nums[i + 1]

    return [prefix[i] * suffix[i] for i in range(n)]

Approach 2: $O(1)$ space

  • Time:O(n)
  • Space:O(1)

C++

class Solution {
 public:
  vector<int> productExceptSelf(vector<int>& nums) {
    const int n = nums.size();
    vector<int> ans(n, 1);

    // use ans as the prefix product array
    for (int i = 1; i < n; ++i)
      ans[i] = ans[i - 1] * nums[i - 1];

    int suffix = 1;  // suffix product
    for (int i = n - 1; i >= 0; --i) {
      ans[i] *= suffix;
      suffix *= nums[i];
    }

    return ans;
  }
};

JAVA

class Solution {
  public int[] productExceptSelf(int[] nums) {
    final int n = nums.length;
    int[] ans = new int[n];
    ans[0] = 1;

    // use ans as the prefix product array
    for (int i = 1; i < n; ++i)
      ans[i] = ans[i - 1] * nums[i - 1];

    int suffix = 1; // suffix product
    for (int i = n - 1; i >= 0; --i) {
      ans[i] *= suffix;
      suffix *= nums[i];
    }

    return ans;
  }
}

Python

class Solution:
  def productExceptSelf(self, nums: List[int]) -> List[int]:
    n = len(nums)
    ans = [1] * n

    # use ans as the prefix product array
    for i in range(1, n):
      ans[i] = ans[i - 1] * nums[i - 1]

    suffix = 1  # suffix product
    for i, num in reversed(list(enumerate(nums))):
      ans[i] *= suffix
      suffix *= num

    return ans