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This approach involves first calculating the remainder of the sum of the given array with respect to p. By maintaining a prefix sum and using the properties of modular arithmetic, we can determine the shortest subarray whose removal results in a remainder of zero.
Time Complexity: O(n), where n is the length of nums.
Space Complexity: O(n), for storing prefix sums in the hashmap.
1def min_subarray(nums, p):
2 total_mod = sum(nums) % p
3 if total_mod == 0:
4 return 0
5 prefix_sum = 0
6 prefix_map = {0: -1}
7 min_len = len(nums)
8 for i, num in enumerate(nums):
9 prefix_sum = (prefix_sum + num) % p
10 target = (prefix_sum - total_mod) % p
11 if target in prefix_map:
12 min_len = min(min_len, i - prefix_map[target])
13 prefix_map[prefix_sum] = i
14 return min_len if min_len != len(nums) else -1This implementation calculates the prefix sum and stores remainders in a hashmap. By checking if the target remainder exists in the hashmap, we find the minimal subarray that aligns with our requirement. We update the hashmap with current prefix sums iteratively.
This approach makes use of a sliding window or two-pointer technique to identify the minimal subarray which upon removal makes the sum of the remaining elements divisible by p.
Time Complexity: O(n), where n is the length of nums.
Space Complexity: O(n), for maintaining the hashmap.
1import java.util.*;
2
3The Java solution uses a HashMap to keep track of prefix sums as keys, linked to their indexes. The final loop computes prefix sums iteratively and checks for the desired remainder, computing the minimal subarray, if applicable.