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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.
1function minSubarray(nums, p) {
2 let totalMod This JavaScript implementation aligns with previous versions, utilizing an object to store prefix sums and applying similar logic to finalize the computation of the minimal subarray which satisfies the given condition.