# 1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts

![](/files/-Mig0YWXyhATvJql5sKg)

![](/files/-Mig0aOK6BXetgGfCcTQ)

![](/files/-Mig0d8K6dGtYBxcrNZK)

![](/files/-Mig0gtePqi3jwUFBjaB)

just find **max height (height cut between height cut)** after each cut

and find **max width (width cut between width cut)** after each cut

ans = max height \* max width

time: O(nlogn)

space: O(1)

```java
class Solution {
    public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
        Arrays.sort(horizontalCuts);
        Arrays.sort(verticalCuts);
        int n = horizontalCuts.length;
        int m = verticalCuts.length;
        
        long maxH = Math.max(horizontalCuts[0], h - horizontalCuts[n-1]);
        for (int i = 1; i < n; i++) {
            maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i-1]);
        }
        
        long maxW = Math.max(verticalCuts[0], w - verticalCuts[m-1]);
        for (int i = 1; i < m; i++) {
            maxW = Math.max(maxW, verticalCuts[i] - verticalCuts[i-1]);
        }
        
        return (int)(maxH*maxW%1000000007); // notice here, (int) (product)
    }
}
```

```java
long maxH = horizontalCuts[0];
for (int i = 1; i < n; i++) {
    maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i-1]);
}
maxH = Math.max(maxH, h - horizontalCuts[n-1]);

可以寫成下面這樣, 合併再一起去比較了 (先
```

```java
        long maxH = Math.max(horizontalCuts[0], h - horizontalCuts[n-1]);

```

最後這個乘法, 要小心, 後面要一起乘玩, 再轉 int, 不然會壞掉

```java
return (int)(maxH*maxW%1000000007); // notice here, (int) (product)
```


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