{"id":6691,"date":"2020-05-03T20:23:29","date_gmt":"2020-05-04T03:23:29","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=6691"},"modified":"2020-05-08T08:57:26","modified_gmt":"2020-05-08T15:57:26","slug":"leetcode-1434-number-of-ways-to-wear-different-hats-to-each-other","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-1434-number-of-ways-to-wear-different-hats-to-each-other\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1434. Number of Ways to Wear Different Hats to Each Other"},"content":{"rendered":"\n

There are n<\/code> people and 40 types of hats labeled from 1 to 40.<\/p>\n\n\n\n

Given a list of list of integers hats<\/code>, where hats[i]<\/code> is a list of all hats preferred by the i-th<\/code> person.<\/p>\n\n\n\n

Return the number of ways that the n people wear different hats to each other.<\/p>\n\n\n\n

Since the answer may be too large, return it modulo 10^9 + 7<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> hats = [[3,4],[4,5],[5]]\nOutput:<\/strong> 1\nExplanation: <\/strong>There is only one way to choose hats given the conditions. \nFirst person choose hat 3, Second person choose hat 4 and last one hat 5.<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> hats = [[3,5,1],[3,5]]\nOutput:<\/strong> 4\nExplanation: <\/strong>There are 4 ways to choose hats\n(3,5), (5,3), (1,3) and (1,5)\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]\nOutput:<\/strong> 24\nExplanation: <\/strong>Each person can choose hats labeled from 1 to 4.\nNumber of Permutations of (1,2,3,4) = 24.\n<\/pre>\n\n\n\n
Example 4:<\/strong><\/p>\n\n\n\n
Input:<\/strong> hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]\nOutput:<\/strong> 111\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • n == hats.length<\/code><\/li>
  • 1 <= n <= 10<\/code><\/li>
  • 1 <= hats[i].length <= 40<\/code><\/li>
  • 1 <= hats[i][j] <= 40<\/code><\/li>
  • hats[i]<\/code> contains a list of unique<\/strong> integers.<\/li><\/ul>\n\n\n\n
    Solution: DP<\/strong><\/h2>\n\n\n\n
    dp[i][j] := # of ways using first i hats, j is the bit mask of people wearing hats.<\/p>\n\n\n\n
    e.g. dp[3][101] == # of ways using first 3 hats that people 1 and 3 are wearing hats.<\/p>\n\n\n\n
    init dp[0][0] = 1<\/p>\n\n\n\n
    dp[i][mask |  (1 << p)] = dp[i-1][mask | (1 << p)] + dp[i-1][mask], where people p prefers hats i.<\/p>\n\n\n\n
    ans: dp[nHat][1…1] <\/p>\n\n\n\n
    Time complexity: O(2^n * h * n)
    Space complexity: O(2^n * h) -> O(2^n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int numberWays(vector>& hats) {\n    constexpr int kHat = 40;\n    constexpr int kMod = 1e9 + 7;\n    const int n = hats.size();    \n    vector> people(kHat + 1); \/\/ hat -> {people}\n    for (int i = 0; i < n; ++i)\n      for (int hat : hats[i])\n        people[hat].push_back(i);\n    \/\/ dp[i][j] := # of ways using first i hats where j is bit mask of people wearing hats.\n    vector> dp(kHat + 1, vector(1 << n));\n    dp[0][0] = 1;\n    \n    for (int i = 1; i <= kHat; ++i) {\n      dp[i] = dp[i - 1];\n      for (int mask = (1 << n) - 1; mask >= 0; --mask)\n        for (int p : people[i]) {\n          if (mask & (1 << p)) continue;\n          dp[i][mask | (1 << p)] += dp[i - 1][mask];\n          dp[i][mask | (1 << p)] %= kMod;          \n        }\n    }\n    return dp.back().back();\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    O(2^n) memory<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int numberWays(vector>& hats) {\n    constexpr int kHat = 40;\n    constexpr int kMod = 1e9 + 7;\n    const int n = hats.size();    \n    vector> people(kHat + 1); \/\/ hat -> {people}\n    for (int i = 0; i < n; ++i)\n      for (int hat : hats[i])\n        people[hat].push_back(i);\n    \/\/ dp([i])[j] := # of ways using first i hats where j is bit mask of people wearing hat.\n    vector dp(1 << n);\n    dp[0] = 1;\n    \n    for (int i = 1; i <= kHat; ++i)      \n      for (int mask = (1 << n) - 1; mask >= 0; --mask)\n        for (int p : people[i]) {\n          if (mask & (1 << p)) continue;\n          dp[mask | (1 << p)] += dp[mask];\n          dp[mask | (1 << p)] %= kMod;          \n      }\n    return dp.back();\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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