Question

Formatted question description: https://leetcode.ca/all/523.html

 523. Continuous Subarray Sum


 Given a list of non-negative numbers and a target integer k, write a function to check
 if the array has a continuous subarray of size at least 2 that sums up to a multiple of k,
 that is, sums up to n*k where n is also an integer.


 Example 1:

 Input: [23, 2, 4, 6, 7],  k=6
 Output: True
 Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.


 Example 2:

 Input: [23, 2, 6, 4, 7],  k=6
 Output: True
 Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.


 Constraints:
     The length of the array won't exceed 10,000.
     You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

Algorithm

bruce force

note if k is 0 then mod will not work

Code

Java

public class Continuous_Subarray_Sum {
    class Solution {
        public boolean checkSubarraySum(int[] nums, int k) {

            // input null check

            for (int i = 0; i < nums.length; ++i) {
                int sum = nums[i];
                for (int j = i + 1; j < nums.length; ++j) {
                    sum += nums[j];
                    if (sum == k) return true;
                    if (k != 0 && sum % k == 0) return true;
                }
            }

            return false;

        }
    }
}

// OJ: https://leetcode.com/problems/continuous-subarray-sum
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
  bool checkSubarraySum(vector<int>& nums, int k) {
    for (int i = 0; i < nums.size(); ++i) {
      int sum = nums[i];
      for (int j = i + 1; j < nums.size(); ++j) {
        sum += nums[j];
        if ((!k && !sum) || (k && sum % k == 0)) return true;
      }
    }
    return false;
  }
};

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        s = 0
        mp = {0: -1}
        for i, v in enumerate(nums):
            s += v
            r = s % k
            if r in mp and i - mp[r] >= 2:
                return True
            if r not in mp:
                mp[r] = i
        return False

############

class Solution(object):
  def checkSubarraySum(self, nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: bool
    """
    if k == 0:
      return "0,0" in ",".join([str(n) for n in nums])
    if len(nums) < 2:
      return False
    if len(nums) == 2:
      return sum(nums) % k == 0
    ppSum = 0
    subSum = nums[0] + nums[1]
    d = set([0])
    for i in range(2, len(nums)):
      ppSum = (ppSum + nums[i - 2]) % k
      d |= {ppSum}
      subSum = (subSum + nums[i]) % k
      if subSum % k in d:
        return True
    return False

Java

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        int length = nums.length;
        if (length < 2)
            return false;
        if (k == 0) {
            for (int i = 1; i < length; i++) {
                if (nums[i - 1] == 0 && nums[i] == 0)
                    return true;
            }
            return false;
        }
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        map.put(0, -1);
        int remainder = 0;
        for (int i = 0; i < length; i++) {
            remainder = (remainder + nums[i]) % k;
            if (map.containsKey(remainder)) {
                int prevIndex = map.get(remainder);
                if (i - prevIndex > 1)
                    return true;
            } else
                map.put(remainder, i);
        }
        return false;
    }
}

// OJ: https://leetcode.com/problems/continuous-subarray-sum
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
  bool checkSubarraySum(vector<int>& nums, int k) {
    for (int i = 0; i < nums.size(); ++i) {
      int sum = nums[i];
      for (int j = i + 1; j < nums.size(); ++j) {
        sum += nums[j];
        if ((!k && !sum) || (k && sum % k == 0)) return true;
      }
    }
    return false;
  }
};

class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        s = 0
        mp = {0: -1}
        for i, v in enumerate(nums):
            s += v
            r = s % k
            if r in mp and i - mp[r] >= 2:
                return True
            if r not in mp:
                mp[r] = i
        return False

############

class Solution(object):
  def checkSubarraySum(self, nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: bool
    """
    if k == 0:
      return "0,0" in ",".join([str(n) for n in nums])
    if len(nums) < 2:
      return False
    if len(nums) == 2:
      return sum(nums) % k == 0
    ppSum = 0
    subSum = nums[0] + nums[1]
    d = set([0])
    for i in range(2, len(nums)):
      ppSum = (ppSum + nums[i - 2]) % k
      d |= {ppSum}
      subSum = (subSum + nums[i]) % k
      if subSum % k in d:
        return True
    return False

523 - Continuous Subarray Sum

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