LeetCode Meditations: Maximum Product Subarray

The description for Maximum Product Subarray is:

Given an integer array nums, find a subarray that has the largest product, and return the product.

The test cases are generated so that the answer will fit in a 32-bit integer.

For example:

Input: nums = [2, 3, -2, 4]
Output: 6
Explanation: [2, 3] has the largest product 6.

Input: nums = [-2, 0, -1]
Output: 0
Explanation: The result cannot be 2, because [-2, -1] is not a subarray.

---

Now, using a brute force approach, we can solve it with a nested loop.

Since we eventually have to get the maximum product, let's first find out the maximum value in the array:

let max = Math.max(...nums);

Then, as we go through each number, we can continually multiply them with the other remaining numbers, building up a total. Once this total is more than max, we can update max to point to this new value:

for (let i = 0; i  max) {
      max = total;
    }
  }
}

At the end, we can just return max. So, the first attempt of our final solution looks like this:

function maxProduct(nums: number[]): number {
  let max = Math.max(...nums);
  for (let i = 0; i  max) {
        max = total;
      }
    }
  }

  return max;
}

Time and space complexity

The time complexity is $O(n^2)$O(n2) as we have a nested loop, doing a constant operation for each of the numbers for each one we iterate over.
The space complexity is $O(1)$O(1) because we don't need additional storage.

Again, this is only a brute force attempt. So, let's take one deep breath, and take a look at another solution.

---

The idea with this new solution is to keep two different values for the maximum and minimum as we go through each number in the array. The reason for that is handling negative values, as we'll see shortly.

First, let's start with initializing these values: we'll have a currentMax, a currentMin, and a result, all of which are initially pointing to the first value in the array:

let currentMax = nums[0];
let currentMin = nums[0];
let result = nums[0];

Now, starting with the second number, we'll loop through each value, updating the current maximum number and the current minimum number as well as the result (which will be the final maximum) as we go:

for (let i = 1; i 



<p>However, before that, let's see an example of what can happen if we just do that.</p>

<p>Let's say our array is [-2, 3, -4]. Initially, currentMax and currentMin are both -2. Now, to update currentMax, we have two options: it's either the current number or the current number multiplied by currentMax:<br>
</p>

<pre class="brush:php;toolbar:false">Math.max(3, 3 * -2)

Obviously, it's the first option, so our currentMax is now 3.
To update currentMin, we also have two options:

Math.min(3, 3 * -2)

It's again obvious, -6. For now, our values look like this:

currentMax // 3
currentMin // -6

On to the next number. We have two options for currentMax:

Math.max(-4, -4 * 3)

By itself, it has to be -4, but looking at our array, we see that this is not the case. Since multiplying two negative values results in a positive value, our currentMax should be 24 (-2 * 3 * -4).

Note

If we were to multiply it with currentMin, we reach this value: -4 * -6 = 24.

Also, let's look at our currentMin options:

Math.min(-4, -4 * -6)

This has to be -4 again, but something feels off.

The catch is that when we have negative numbers consecutively, our sign alternates, which affects the maximum result we need. That's why we're keeping track of the minimum value in the first case: to keep track of the sign.

Since the issue is just alternating signs, we can simply swap the maximum and minimum values when we're looking at a negative number before updating those values:

if (nums[i] 



<p>Also, note that we're taking the product of each previous subarray as we go, essentially solving a smaller portion of the problem.</p>

<p>And that's it, our final solution looks like this:<br>
</p>

<pre class="brush:php;toolbar:false">function maxProduct(nums: number[]): number {
  let currentMax = nums[0];
  let currentMin = nums[0];
  let result = nums[0];

  for (let i = 1; i 



<h4>
  
  
  Time and space complexity
</h4>

<p>The time complexity for this solution is 

  <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow>O(n) </semantics></math><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;">O(n)</span></span></span>
</span>
 because we go through each number once doing a constant operation. </p>

<div class="table-wrapper-paragraph"><table>
<thead>
<tr>
<th>Note</th>
</tr>
</thead>
<tbody>
<tr>
<td>
Math.max() and Math.min() are constant operations here, since we're comparing two values only. However, if we were to find max or min of a whole array, it would be 

  <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow>O(n) </semantics></math><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;">O(n)</span></span></span>
</span>
 as the time complexity of the operation would increase proportionately to the size of the array.</td>
</tr>
</tbody>
</table></div>

<p>The space complexity is 

  <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow>O(1) </semantics></math><span class="base"><span class="strut" style="height:1em;vertical-align:-0.25em;">O(1)</span></span></span>
</span>
 since we don't need any additional storage.</p>


<hr>

<p>The next problem on the list is called Word Break. Until then, happy coding.</p>


          

            
  

            
        

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