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/**

* 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 the 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.

*

* Note:

* 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.

*

*/

public class ContinuousSubarraySum523 {

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

if (k == 0) return kIsZero(nums);

return kIsNotZero(nums, k);

}

public boolean kIsNotZero(int[] nums, int k) {

Map<Integer, Integer> map = new HashMap<>();

map.put(0, -1);

int sum = 0;

for (int i=0; i<nums.length; i++) {

sum += nums[i];

int mod = sum % k;

if (map.containsKey(mod)) {

if (i - map.get(mod) >= 2) return true;

} else {

map.put(mod, i);

}

}

return false;

}

public boolean kIsZero(int[] nums) {

int count = 0;

for (int i=0; i<nums.length; i++) {

if (nums[i] == 0) {

count++;

if (count >= 2) return true;

} else {

count = 0;

}

}

return false;

}

/**

* https://leetcode.com/problems/continuous-subarray-sum/solution/

*/

public boolean checkSubarraySum2(int[] nums, int k) {

int sum = 0;

HashMap < Integer, Integer > map = new HashMap < > ();

map.put(0, -1);

for (int i = 0; i < nums.length; i++) {

sum += nums[i];

if (k != 0)

sum = sum % k;

if (map.containsKey(sum)) {

if (i - map.get(sum) > 1)

return true;

} else

map.put(sum, i);

}

return false;

}

}