Minimum Difficulty of a Job Schedule LeetCode solution in C++

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class Solution {
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  vector<int> v;
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  int len;
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  int dp[301][11];
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  int dfs(int idx, int d) { // n*d
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    if (idx >= len) return 0;
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    if (d == 1) return *max_element(v.begin() + idx, v.end());
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    if (dp[idx][d] != -1) return dp[idx][d];
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    int maxx = INT_MIN;
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    int res = INT_MAX;
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    for (int i = idx; i <= len - d; i++) { // n
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      maxx = max(v[i], maxx);
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      res = min(maxx + dfs(i + 1, d - 1), res);
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    }
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    return dp[idx][d] = res;
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  }
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 public:
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  int minDifficulty(vector<int> &jobDifficulty, int d) {
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    v = jobDifficulty;
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    len = v.size();
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    if (d > len) return -1;
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    memset(dp, -1, sizeof(dp));
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    return dfs(0, d);
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  }
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};