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+# 1191. K-Concatenation Maximum Sum
+
+"""
+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 1e9+7
+"""
+
+
+class Solution(object):
+ def max_sum(self, arr):
+ curr_max, max_end = int("-inf"), 0
+ for i in arr:
+ max_end += i
+ if max_end < 0:
+ max_end = 0
+ curr_max = max(curr_max, max_end)
+ return curr_max
+
+ def kConcatenationMaxSum(self, arr, k):
+ """
+ :type arr: List[int]
+ :type k: int
+ :rtype: int
+ """
+ total = sum(arr)
+ kadane = self.max_sum(arr)
+ if k < 2:
+ return kadane(10**9 + 7)
+ if total > 0:
+ return (kadane + (k - 1) * total) % (10**9 + 7)
+ double = self.max_sum(arr * 2)
+ return double % (10**9 + 7)
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.kConcatenationMaxSum(arr=[1, 2], k=3))
+ print(obj.kConcatenationMaxSum(arr=[1, -2, 1], k=5))
+ print(obj.kConcatenationMaxSum(arr=[1, -2, 1], k=5))