If ask for spefific plan combinations 求具体方案 O(n^3)

Input: nums = [2,2,3,4]
Explanation: Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3

Have to count one by one. C^3_n.
E.g. ask for all valid triangles and specific combination, it's O(n^3)

If ask for plans count 求方案总数可批量处理降维到 O(n^2)

Input: nums = [2,2,3,4]
Output: 3

Can batch process similar situations.
If array[A] + array[B] > array[C] then array[any elem after A] + array[B] > array[C]

Summary 求具体方案(DFS) vs 求方案数(DP)

If ask for specific plan combinations O(n^3). Think of DFS immediately.

求具体方案只能一个个数出来，时间复杂度 O(n^3). 后期要求直接反应到DFS.

If ask for plans count. Thsnk of DP immediately. 求方案总数，立刻想到动态规划 dynamic programming.
- If ask for plans count and cannot repeat O(n^3)
  
  求不可以重复的方案数也只能一个个数
- If ask for plans count but can repeat O(n^2)
  
  求可以重复的方案数可以批量累加，时间复杂度 O(n^2)