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# Maximum Number of Non-Overlapping Subarrays With Sum Equals Target

Given an array `nums` and an integer `target`.

Return the maximum number of **non-empty** **non-overlapping** subarrays such that the sum of values in each subarray is equal to `target`.

**Example 1:**

```
Input: nums = [1,1,1,1,1], target = 2
Output: 2
Explanation: There are 2 non-overlapping subarrays [1,1,1,1,1] with sum equals to target(2).
```

**Example 2:**

```
Input: nums = [-1,3,5,1,4,2,-9], target = 6
Output: 2
Explanation: There are 3 subarrays with sum equal to 6.
([5,1], [4,2], [3,5,1,4,2,-9]) but only the first 2 are non-overlapping.
```

**Example 3:**

```
Input: nums = [-2,6,6,3,5,4,1,2,8], target = 10
Output: 3
```

**Example 4:**

```
Input: nums = [0,0,0], target = 0
Output: 3
```

**Constraints:**

* `1 <= nums.length <= 10^5`
* `-10^4 <= nums[i] <= 10^4`
* `0 <= target <= 10^6`

```java
class Solution {
    public int maxNonOverlapping(int[] nums, int target) {
        // Map of prefix sum -> Latest index for this prefix sum
        HashMap<Integer, Integer> map = new HashMap<>();
        // arr[i] -> Maximum Number of Non-Overlapping Subarrays With Sum Equals Target upto i
        int[] arr = new int[nums.length];
        map.put(0, -1);
        int prefix = 0;
        for (int i = 0; i < arr.length; i++) {
            prefix += nums[i];
            int ans = 0;
            if (map.containsKey(prefix - target)) {
                if (map.get(prefix - target) != -1)
                    ans = arr[map.get(prefix - target)];
                ans += 1;
            }
            map.put(prefix, i);
            arr[i] = Math.max(ans, i > 0 ? arr[i - 1] : Integer.MIN_VALUE);
        }
        return arr[arr.length - 1];
    }
}
```