Home
Algorithms Data Structures Interview Experiences
Amazon Google Facebook Microsoft Apple Netflix
Play 2048 How to

LeetCode Solutions

523. Continuous Subarray Sum

Time: $O(n)$

Space: $O(n)$

			


class Solution {
 public:
  bool checkSubarraySum(vector<int>& nums, int k) {
    unordered_map<int, int> prefixToIndex{{0, -1}};
    int prefix = 0;

    for (int i = 0; i < nums.size(); ++i) {
      prefix += nums[i];
      if (k != 0)
        prefix %= k;
      if (prefixToIndex.count(prefix)) {
        if (i - prefixToIndex[prefix] > 1)
          return true;
      } else {
        // Only add if absent, because the previous index is better
        prefixToIndex[prefix] = i;
      }
    }

    return false;
  }
};




			


class Solution {
  public boolean checkSubarraySum(int[] nums, int k) {
    int prefix = 0;
    Map<Integer, Integer> 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;
  }
}




			


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