human
Solve in Python:
Given an integer array arr and an integer k, modify the array by repeating it k times.
For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].
Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.
As the answer can be very large, return the answer modulo 10^9 + 7.
 
Example 1:
Input: arr = [1,2], k = 3
Output: 9

Example 2:
Input: arr = [1,-2,1], k = 5
Output: 2

Example 3:
Input: arr = [-1,-2], k = 7
Output: 0

 
Constraints:

1 <= arr.length <= 10^5
1 <= k <= 10^5
-10^4 <= arr[i] <= 10^4
 What is a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
    def kConcatenationMaxSum(self, arr: List[int], k: int) -> int:
        
        numPos, numNeg = 0,0
        
        for num in arr:
            if num >= 0:
                numPos += 1
            else:
                numNeg += 1
        
        if numPos == len(arr):
            return sum(arr)*k
        
        if numNeg == len(arr):
            return 0
        
        # Check for 2* arr
        arrTwo = arr*2
        for i in range(1, len(arrTwo)):
            if arrTwo[i - 1] > 0: 
                arrTwo[i] +=arrTwo[i - 1]
                
        # Check for 3*arr
        arrThree = arr*3
        for i in range(1, len(arrThree)):
            if arrThree[i - 1] > 0: 
                arrThree[i] +=arrThree[i - 1]
        
        arrDiff = max(arrThree) - max(arrTwo)
        
        
        return (max(arrTwo) + (k-2)*arrDiff) % (10**9 + 7)