sol.md

Solution

The provided solution is implemented in C++ and uses two passes through the nums array to calculate the answer vector without using division.

1. Initialization:

   • Initialize an empty vector answer.
   • Initialize two variables m and m1 to 1, which will be used to track the product of non-zero elements and all elements, respectively.
   • Initialize a variable count to keep track of the number of zeros in the array.

2. First Pass:

   • Iterate through the nums array.
   • For each element nums[i], add it to the answer vector.
   • Update m and m1 by multiplying them with the current element nums[i].
   • If the current element is 0, increment the count variable.

3. Second Pass:

   • Iterate through the nums array again.
   • For each element at index i, calculate the value for answer[i] based on the following conditions:
     • If count is greater than 1 (meaning there are more than one zero in the array), set answer[i] to 0 because the product of all elements would be 0.
     • If nums[i] is 0, set answer[i] to m1. This is because the product of all elements except the zero element is equal to m1.
     • Otherwise, set answer[i] to the result of dividing m by nums[i, which effectively gives the product of all elements except the current one.

4. Return:

   • Return the answer vector, which contains the desired result.

The provided solution correctly handles different cases and satisfies the O(n) time complexity constraint, but it uses O(n) extra space to store the answer vector. Achieving O(1) extra space complexity would require a different approach, such as using prefix and suffix products with careful computation.