i am stuck with this dp problem , can any someone explain me the state of dp for this problem ? problem link is Question

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i am stuck with this dp problem , can any someone explain me the state of dp for this problem ? problem link is Question

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This is a difficult DP problem until you know the trick. When would a triangle be invalid? Only when one of the side lengths is >= 1/2 of the total sum. With this in mind we can count how many assignments are invalid triangles, rather than how many assignments are valid triangles. This problem is significantly easier — we have 2 parts to our state: current sum and current item. At each spot we have 2 choices — add to our sum or don't. Succeed when our value reaches 1/2 of the total array sum. Total time is O(2 choices * 300 indexes * 300^2 max sum). We can recover the answer by taking 3^N — 3 * the result of our DP.

What about the difference property. Difference of 2 sides must be smaller than third

That should be covered by the same logic.

What you said isn't much clear. But I understood is by subset sum problem we can find the number of ways to make every sum. Then we can add the counts for the numbers greater than n/2 Is that right

If you have |A-C|>B and A+B+C=X then A or C must be >= X/2

Thanks bro. I got it now. The main point is if we handle the sum property for a triangle then the difference property is handled automatically, right?