package array;

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

/**
 * Created by gouthamvidyapradhan on 03/01/2018.
 * Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum
 * equals to k.

 Example 1:
 Input:nums = [1,1,1], k = 2
 Output: 2
 Note:
 The length of the array is in range [1, 20,000].
 The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

 Solution: O(n) Maintain a hash-map of prefix sum and its count and check for range sum for each element.
 */
public class SubarraySumEqualsK {

    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        int[] A = {1, 2, 1, -2, 3, -1, -1};
        System.out.println(new SubarraySumEqualsK().subarraySum(A, 2));
    }

    public int subarraySum(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        for(int i : nums){
            sum += i;
            Integer count = map.get(sum);
            if(count == null){
                map.put(sum, 1);
            } else{
                map.put(sum, count + 1);
            }
        }
        sum = 0;
        int result = 0;
        for(int i : nums){
            int key = sum + k;
            if(map.containsKey(key)){
                int count = map.get(key);
                result += count;
            }
            sum += i;
            if(map.containsKey(sum)){
                int count = map.get(sum);
                if(count - 1 > 0){
                    map.put(sum, count - 1);
                } else{
                    map.remove(sum);
                }
            }
        }
        return result;
    }
}