## Question

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

**Example 1**

```
Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences :
1, 2, 3
3, 4, 5
```

**Example 2**

```
Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences :
1, 2, 3, 4, 5
3, 4, 5
```

**Example 3**

```
Input: [1,2,3,4,4,5]
Output: False
```

Note:

The length of the input is in range of [1, 10000]

## Thinking

At first, I have no idea about it. I thought I should use DP to solve it. But I

couldn’t find the base case.

Check the discuss, I think this is the good idea:

- Save all numbers in a map and count them
- In another loop, check each number and make sure it can be added to exist subsquences or create a new one.
- If the number cannot create a new subsequences or be added to an exist subsquence . return false.

## Solution

### Java

```
public class Solution {
public boolean isPossible(int[] nums) {
Map<Integer, Integer> freqMap = new HashMap<Integer, Integer>();
Map<Integer, Integer> nextMap = new HashMap<Integer, Integer>();
for (int num : nums) {
freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
}
for (int num : nums) {
if (freqMap.get(num) == 0) {
continue; // All this number has been used for subsquence.
} else if (nextMap.getOrDefault(num, 0) > 0) {
// add to an exist subsquence
nextMap.put(num, nextMap.getOrDefault(num, 0) - 1);
nextMap.put(num + 1, nextMap.getOrDefault(num + 1, 0) + 1);
} else if (freqMap.getOrDefault(num + 1, 0) > 0 && freqMap.getOrDefault(num + 2, 0) > 0) {
// create a new subsquence
freqMap.put(num + 1, freqMap.get(num + 1) - 1);
freqMap.put(num + 2, freqMap.get(num + 2) - 1);
nextMap.put(num + 3, nextMap.getOrDefault(num + 3, 0) + 1);
} else {
return false;
}
freqMap.put(num, freqMap.get(num) - 1);
}
return true;
}
}
```