# 152. Maximum Product Subarray (1d-dp)

![](https://4272748102-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LekNH5IywF8mjBxFcnu%2F-Meo2HTzHXySBIkwfwUU%2F-MeqwvMaWVyb9ft7qW9e%2Fimage.png?alt=media\&token=6f4f4124-1302-46c2-9f7b-d36a508c3dfd)

![](https://4272748102-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LekNH5IywF8mjBxFcnu%2F-Meo2HTzHXySBIkwfwUU%2F-MeqwzVIic5NcYVu7-f_%2Fimage.png?alt=media\&token=81071d54-5d56-4044-bd4e-d841aa1bd99e)

#### 1. 其實有 0 或 奇數個neg 時應該要重新選擇起始點, 所以要跟 nums\[i] 比較

#### 2. 偶數個 neg 時可能會產生 max 值, 解法是紀錄 max 和 min value, 如果neg\*neg 時就會產生 max 值

#### idea - 

#### max( max product, min product, current value),  

#### min( max product, min product, current value), 

#### because of negative\*negative may generate max number!

最後取最大值

![](https://4272748102-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LekNH5IywF8mjBxFcnu%2F-Mg-A5RmsW8gxjnSLn1Z%2F-Mg-CsybNB46nuemkg1a%2Fimage.png?alt=media\&token=58c78d9c-37bf-49fa-9077-c817be0defed)

![](https://4272748102-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LekNH5IywF8mjBxFcnu%2F-Mg-A5RmsW8gxjnSLn1Z%2F-Mg-DlEP_gsb_7_CCLj-%2Fimage.png?alt=media\&token=37116c8e-78c9-400f-bf99-47b430262e10)

![](https://4272748102-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LekNH5IywF8mjBxFcnu%2F-Mg-A5RmsW8gxjnSLn1Z%2F-Mg-DxweRM4oVaaNvZyR%2Fimage.png?alt=media\&token=a66cd9cf-bef6-400e-938e-9426730bc0a5)

time: O(n)

space: O(1)

```java
class Solution {
    public int maxProduct(int[] nums) {
        int max = nums[0];
        int min = nums[0];
        int res = nums[0];
        for (int i = 1; i < nums.length; i++) {
            int temp = max;
            max = Math.max(Math.max(nums[i]*max, nums[i]*min), nums[i]);
            min = Math.min(Math.min(nums[i]*temp, nums[i]*min), nums[i]);
            res = Math.max(res, max);
        }
        return res;
    }
}
```

## original DP

```java
class Solution {
    public int maxProduct(int[] nums) {
        // [-2,0,-1], max = 0, 
        // 2 3 -3 4  max = 6
        // 2 -3 -3 4 max = 72 , so when even neg, will have max value
        // 2 -3 [-3 4 -3 -2 1 -4] max =  36* 8
        int dpMax[] = new int[nums.length];
        int dpMin[] = new int[nums.length];

        dpMax[0] = nums[0];
        dpMin[0] = nums[0];
        int res = nums[0];

        for (int i = 1; i < nums.length; i++) {
            dpMax[i] = Math.max(Math.max(nums[i]*dpMax[i-1], nums[i]*dpMin[i-1]), nums[i]) ;
            dpMin[i] = Math.min(Math.min(nums[i]*dpMax[i-1], nums[i]*dpMin[i-1]), nums[i]);
            res = Math.max(res, dpMax[i]);
        }
        return res;
    }
}
```

then use variables, or like first one way

```java
class Solution {
    public int maxProduct(int[] nums) {
        int max = nums[0];
        int min = nums[0];
        int res = nums[0];

        for (int i = 1; i < nums.length; i++) {
            int newMax = Math.max(Math.max(nums[i]*max, nums[i]*min), nums[i]) ;
            int newMin = Math.min(Math.min(nums[i]*max, nums[i]*min), nums[i]);
            max = newMax;
            min = newMin;
            res = Math.max(res, max);
        }
        return res;
    }
}
```


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