Product of Array Except Self Explained: Tracing [1, 2, 0, 4]

Imagine you are handed the array $\text{nums}=[1,2,0,4]$. Your goal is to compute an output array where each element at index $i$ is the product of every number in $\text{nums}$ except $\text{nums}[i]$. For index 0, that means $2\times 0\times 4=0$. For index 3, it means $1\times 2\times 0=0$. The expected result for this entire array is $[0,0,8,0]$. If you try the naive nested-loop approach, your runtime balloons to $O(n^2)$, which instantly fails large inputs. If you instead try the intuitive shortcut—find the total product of the array and divide by $\text{nums}[i]$ for each position—you run headfirst into a catastrophic math wall when a zero appears in the input. In our array $[1,2,0,4]$, the total product is 0, meaning you would have to evaluate $0/0$, which is undefined, or you might accidentally divide by zero at index 2. Even worse, what happens if the array contains two zeros, such as $\text{nums}=[0,2,0,4]$? Every single position in the output must be 0, but division hides this structural reality behind floating-point errors and exceptions. We need a way to solve this purely through multiplication in linear time, without ever dividing.

When building an algorithm for the product of array except self explained approach, we must compute for every index $i$ the product of all elements to its left and all elements to its right, without ever including $num[i]$ itself and without using division. Instead of nested loops that take $O(n^2)$ time, we can maintain these two halves dynamically using two separate passes across our array. For our locked working example $\text{nums}=[1,2,0,4]$, the product at index 2 (which is 0) depends entirely on everything to its left ($1\times 2=2$) multiplied by everything to its right (4). In our first pass, we build a prefix products array that stores the running product of all elements strictly to the left of each index. In our second pass, we sweep backward from the right end of the array, maintaining a running suffix product that we multiply directly into our prefix results to form the final answer in-place.

We take our locked example $\text{nums}=[1,2,0,4]$ and execute the two-pass algorithm step by step. First, we compute the left prefix products where $\text{left}[i]$ holds the product of all elements to the left of index $i$. For the leftmost element at index 0, there are no elements to its left, so $\text{left}[0]=1$. Moving right, $\text{left}[1]=1\times 1=1$, $\text{left}[2]=1\times 2=2$, and $\text{left}[3]=2\times 0=0$, giving us a left prefix array of $[1,1,2,0]$. Next, we compute the right suffix products in a reverse pass. For the rightmost element at index 3, $\text{right}[3]=1$. Moving left, $\text{right}[2]=4\times 1=4$, $\text{right}[1]=0\times 4=0$, and $\text{right}[0]=2\times 0=0$, yielding a right suffix array of $[0,0,4,1]$. Finally, multiplying $\text{left}[i]\times \text{right}[i]$ at each position gives $[1\times 0,1\times 0,2\times 4,0\times 1]=[0,0,8,0]$.

def productExceptSelf(nums):
    n = len(nums)
    left = [1] * n
    right = [1] * n
    answer = [0] * n
    
    # Fill left prefix products
    for i in range(1, n):
        left[i] = left[i-1] * nums[i-1]
        
    # Fill right suffix products
    for i in range(n-2, -1, -1):
        right[i] = right[i+1] * nums[i+1]
        
    # Combine left and right
    for i in range(n):
        answer[i] = left[i] * right[i]
        
    return answer

Now that you have traced the single-zero case for $nums=[1,2,0,4]$, it is time to test your mental model on a trickier mutation. Consider what happens when the input array contains multiple zeros, such as $nums=[0,2,0,4]$. Work through the prefix and suffix logic step by step without running code. Remember how the left-to-right pass and right-to-left pass accumulate running products, and consider what happens when a running product encounters a zero.

We have successfully tracked our running prefix and suffix products across $[1,2,0,4]$ and reasoned through multiple zeros in $[0,2,0,4]$. The underlying pattern of the two-pass algorithm is not just a neat trick for array multiplication; it is a fundamental design template whenever every position in a collection needs global context excluding itself in $O(n)$ time. When faced with a new neighbor problem—such as finding running bounds or prefix/suffix combinations without division—you no longer need a nested loop or a division operator. You isolate the left accumulation state, isolate the right accumulation state, and fuse them in a clean backward sweep just as we did when assembling our final answer array.

def productExceptSelf(nums):
    # Apply the two-pass prefix/suffix template
    pass