ContinuousSubarraySum.java

package dynamic_programming;

import java.util.HashMap;
import java.util.Map;

/**
 * Created by gouthamvidyapradhan on 10/12/2017.
 * 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.

 Solution: O(n) sum the elements and maintain a hashmap of key value pair of (sum % k) -> index. If the key is
 already found in the hashmap and the difference in the current_index and hashmap index is > 1 then return true.
 */
public class ContinuousSubarraySum {

    /**
     * Main method
     * @param args
     */
    public static void main(String[] args) throws Exception{
        int[] A = {1, 3, 6, 12, 7};
        System.out.println(new ContinuousSubarraySum().checkSubarraySum(A, 6));
    }

    public boolean checkSubarraySum(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        map.put(0, -1);
        for(int i = 0; i < nums.length; i ++){
            sum += nums[i];
            int mod = (k == 0) ? sum : sum % k; //this is to handle case where k is 0
            if(map.containsKey(mod)){
                if(i - map.get(mod) > 1){
                    return true;
                }
            } else{
                map.put(mod, i);
            }
        }
        return false;
    }
}