LeetCode "473. Matchsticks to Square"

A trickier DFS, with a little bit complex recursion param tweak, and what's more important is pruning strategies.. or your code will TLE.

class Solution {
    bool go(vector<bool> &m, int edge, int eleft, int etgt, vector<int> &in, int taken)
    {
        if(edge > ) return false;
        if(edge ==  && eleft == etgt && in.size() == taken)
        {
            return true;
        }
        int last = -;
        for(int i = in.size()-; i >=; i --)
        {
            if(m[i]) continue;
            if(in[i] > eleft) continue; // because it is sorted
            if(in[i] == last) continue; // important: critical pruning.
            if(in[i] <= eleft)
            {
                m[i] = true;
                bool fit = in[i] == eleft;
                if(go(m, edge + (fit?:), fit?etgt:(eleft-in[i]), etgt, in, taken + ))
                {
                    return true;
                }
                m[i] = false;
                last = in[i];
            }
        }

        return false;
    }
public:
    bool makesquare(vector<int>& nums) {
        if(nums.size() < ) return false;

        int sum = accumulate(nums.begin(), nums.end(), );
        if(!sum || (sum % )) return false;

        sort(nums.begin(), nums.end());
        if (nums.back() > (sum/)) return false;

        vector<bool> m(nums.size(), false);

        int tgt = sum / ;
        return go(m, , tgt, tgt, nums, );
    }
};