C++ DP Solution

  • 0

    O(key * ring * ring) time, O(ring) memory , 10ms

    dp[curr char] = min (dp[previous char] + distance between curr and prev char in ring)

    For example, ring = "abcab", key = "ba"
    For second char 'a' in key, 'a' indices in ring is [0,3];
    previous char of 'a' is 'b' in key, 'b' indices in ring is [1,4].
    dp[0] = min( dp[1] + dist(0,1) , dp[4] + dist(0,4) );
    dp[3] = min( dp[1] + dist(3,1) , dp[4] + dist(3,4) );

    class Solution {
        int findRotateSteps(string ring, string key) {
            vector<int> ring_idx[26]; //for each character in a-z, record the indices in ring
            for(int i=0; i<ring.size(); i++)
            int rsize = ring.size();
            vector<int> dp(rsize,0); //record the minimum step for each character on ring
            for(int i=0; i<key.size(); i++)
                vector<int> prev_idcs = (i>0) ? ring_idx[key[i-1]-'a'] : vector<int>(1,0); //previous char indices on the ring
                for(int curr_idx : ring_idx[key[i]-'a']) //current char indices on the ring
                    int min_step = INT_MAX;
                    // for each current char on the ring, find the minimum step rotated from all the previous char indices
                    for(int prev_idx : prev_idcs)
                        min_step = min(min_step,dp[prev_idx]+dist(curr_idx,prev_idx,rsize));
                    //update the dp table
                    dp[curr_idx] = min_step;
            int ans = INT_MAX;
            for(int last_char_idx : ring_idx[key.back()-'a'])
                ans = min(ans,dp[last_char_idx]);
            return ans+key.size();
        //return the minimum distance between two char indices
        int dist(int i, int j, const int size){
            int d = abs(i-j);
            return min(d,size-d);

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