**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

You are given a 2D integer array `stockPrices`

where `stockPrices[i] = [day`

indicates the price of the stock on day _{i}, price_{i}]`day`

is _{i}`price`

. A _{i}**line chart** is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below:

Return *the minimum number of lines needed to represent the line chart*.

**Example 1:**

```
Input: stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
Output: 3
Explanation:
The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price.
The following 3 lines can be drawn to represent the line chart:
- Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4).
- Line 2 (in blue) from (4,4) to (5,4).
- Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1).
It can be shown that it is not possible to represent the line chart using less than 3 lines.
```

**Example 2:**

```
Input: stockPrices = [[3,4],[1,2],[7,8],[2,3]]
Output: 1
Explanation:
As shown in the diagram above, the line chart can be represented with a single line.
```

**Constraints:**

`1 <= stockPrices.length <= 10`

^{5}`stockPrices[i].length == 2`

`1 <= day`

_{i}, price_{i}<= 10^{9}- All
`day`

are_{i}**distinct**.

```
class Solution {
public:
int minimumLines(vector<vector<int>>& stk) {
sort(stk.begin(),stk.end());
int n = stk.size();
if(n==1)
return 0;
int cnt = 1;
for(int i=2;i<n;i++){
long x1 = stk[i][0],x2 = stk[i-1][0],x3 = stk[i-2][0];
long y1 = stk[i][1],y2 = stk[i-1][1],y3 = stk[i-2][1];
long diff1 = (y3-y2) * (x2-x1);
long diff2 = (y2-y1) * (x3-x2);
if(diff1 != diff2)
cnt++;
}
return cnt;
}
};
```

```
public int minimumLines(int[][] A) {
int n = A.length, res = n - 1;
Arrays.sort(A, (a, b) -> Integer.compare(a[0], b[0]));
for (int i = 1; i < n - 1; ++i)
if (1L * (A[i][0] - A[i - 1][0]) * (A[i + 1][1] - A[i][1]) == 1L * (A[i + 1][0] - A[i][0]) * (A[i][1] - A[i - 1][1]))
res--;
return res;
}
```

```
def minimumLines(self, A: List[List[int]]) -> int:
n = len(A)
res = n - 1
A.sort()
for i in range(1, n - 1):
a, b, c = A[i-1], A[i], A[i+1]
if (b[0] - a[0]) * (c[1] - b[1]) == (c[0] - b[0]) * (b[1] - a[1]):
res -= 1
return res
```

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