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 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.

//https://discuss.leetcode.com/topic/80793/java-o-n-time-o-k-space

//We iterate through the input array exactly once, keeping track of the running sum mod k of the elements in the process. If we find that a running sum value at index j has been previously seen before in some earlier index i in the array, then we know that the sub-array (i,j] contains a desired sum.

public class Solution {

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

Map<Integer, Integer> map = new HashMap<Integer, Integer>(){{put(0,-1);}};;

int runningSum = 0;

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

runningSum += nums[i];

if (k != 0) runningSum %= k;

Integer prev = map.get(runningSum);

if (prev != null) {

if (i - prev > 1) return true;

}

else map.put(runningSum, i);

}

return false;

}

}