# 523. Continuous Subarray Sum

• Time: $O(n)$
• Space: $O(n)$
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 class Solution { public: bool checkSubarraySum(vector& nums, int k) { int prefix = 0; unordered_map prefixToIndex{{0, -1}}; for (int i = 0; i < nums.size(); ++i) { prefix += nums[i]; if (k != 0) prefix %= k; if (const auto it = prefixToIndex.find(prefix); it != prefixToIndex.cend()) { if (i - it->second > 1) return true; } else { // Only add if absent, because the previous index is better prefixToIndex[prefix] = i; } } return false; } }; 
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 class Solution { public boolean checkSubarraySum(int[] nums, int k) { int prefix = 0; Map prefixToIndex = new HashMap<>(); prefixToIndex.put(0, -1); for (int i = 0; i < nums.length; ++i) { prefix += nums[i]; if (k != 0) prefix %= k; if (prefixToIndex.containsKey(prefix)) { if (i - prefixToIndex.get(prefix) > 1) return true; } else { // Only add if absent, because the previous index is better prefixToIndex.put(prefix, i); } } return false; } } 
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 class Solution: def checkSubarraySum(self, nums: List[int], k: int) -> bool: prefix = 0 prefixToIndex = {0: -1} for i, num in enumerate(nums): prefix += num if k != 0: prefix %= k if prefix in prefixToIndex: if i - prefixToIndex[prefix] > 1: return True else: prefixToIndex[prefix] = i return False