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+# 1590. Make Sum Divisible by P
+
+"""
+given an array of positive integers 'nums', remove the smallest subarray
+(possibly empty) such that the sum of the remaining elements is divisible by
+'p'. it is not allowed to remove the whole array. return the length of the
+smallest subarray that you need to remove, or -1 if it is impossible. a
+subarray is defined as a contiguous block of elements in the array.
+"""
+
+
+class Solution(object):
+ def minSubarray(self, nums, p):
+ """
+ :type nums: List[int]
+ :type p: int
+ :rtype: int
+ """
+ req = sum(nums) % p
+ dp, curr = {0: -1}, 0
+ ans = n = len(nums)
+ for i, j in enumerate(nums):
+ curr = (curr + j) % p
+ dp[curr] = i
+ if (curr - req) % p in dp:
+ ans = min(ans, i - dp[(curr - req) % p])
+ return ans if ans < n else -1
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minSubarray(nums=[3, 1, 4, 2], p=6))
+ print(obj.minSubarray(nums=[6, 3, 5, 2], p=9))
+ print(obj.minSubarray(nums=[1, 2, 3], p=3))