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Diffstat (limited to 'freedom_trail.cpp')
-rw-r--r-- | freedom_trail.cpp | 55 |
1 files changed, 0 insertions, 55 deletions
diff --git a/freedom_trail.cpp b/freedom_trail.cpp deleted file mode 100644 index 2763f8c..0000000 --- a/freedom_trail.cpp +++ /dev/null @@ -1,55 +0,0 @@ -// 514. Freedom Trail -#include "leetcode.h" - -/* - * in the video game fallout 4, the quest "road to freeom" requires players to - * reach a metal dial call the "freedom trail ring" and use the dial to spell a - * specific keyword to open the door. given a string 'rint' that represents the - * code engraved on the outer ring and another string 'key' that represents the - * keyword that needs to be spelled, return the minimum number of steps to spell - * all of the characters in the keyword. initially, the first character of the - * ring is aligned at "12:00" diretion. you should spell all the characters in - * 'key' one by rotating 'ring' clockwise or anticlockwise to make each - * character of the string aligned at the "12:00" direction and then by pressing - * the center button. at the stage of rotating the ring to spell the key - * character 'key[i]' you can rotate the ring clockwise or anticlockwise by one - * place, which counts as one step. the final purpose of the rotation is to - * align one of 'ring's characters at "12:00" direction, where this character - * must equal 'key[i]' - */ - -class Solution { -public: - int findRotateSteps(string ring, string key) { - int n = ring.size(); - unordered_map<char, vector<int>> idx_map; - for (int i = 0; i < n; ++i) - idx_map[ring[i]].push_back(i); - vector<vector<pair<int, int>>> dp(key.size()); - for (int i : idx_map[key[0]]) { - int dist = min(i, n - i); - dp[0].push_back({i, dist}); - } - int min_sum = INT_MAX; - for (int i = 1; i < key.size(); ++i) { - for (int idx : idx_map[key[i]]) { - int min_dist = INT_MAX; - for (auto &pair : dp[i - 1]) { - int abs_dist = abs(pair.first - idx); - int dist = min(abs_dist, n - abs_dist); - min_dist = min(min_dist, dist + pair.second); - } - dp[i].push_back({idx, min_dist}); - if (i == key.size() - 1) - min_sum = min(min_dist, min_sum); - } - } - return min_sum + key.size(); - } -}; - -int main() { - Solution obj; - printf("%d\n", obj.findRotateSteps("godding", "gd")); // expect: 4 - printf("%d\n", obj.findRotateSteps("godding", "godding")); // expect: 13 -} |