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LeetCode Solutions

238. Product of Array Except Self

Time: $O(n)$

Space: $O(n)$

			


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;
  }
};




			


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;
  }
}




			


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)]