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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // i: arr[i] we are currently looking at |
| 4 | + // mask: Bitmask that represents all the characters that have been added to the current string |
| 5 | + // If 0-th bit in mask is set, it means that we have added "a" in the current string |
| 6 | + int solve(vector<string> &arr, int i, int mask) { |
| 7 | + int n = arr.size(); |
| 8 | + |
| 9 | + if (i >= n) |
| 10 | + return 0; |
| 11 | + |
| 12 | + // Skip concatenating arr[i] |
| 13 | + int curRes = solve(arr, i+1, mask) |
| 14 | + |
| 15 | + // Mask to keep track of the characters that are present in arr[i] |
| 16 | + int curMask = 0; |
| 17 | + |
| 18 | + // Check whether any character in arr[i] is present in current string, i.e. |
| 19 | + // Check whether (arr[i]-'a')-th bit is set in mask |
| 20 | + // If any existing character's bit is set, it means that we cannot concatenate arr[i] |
| 21 | + // to the given string and so return curRes only which contains the result of skipping arr[i] |
| 22 | + // Also, use curMask to maintain the characters in arr[i] that have been seen. |
| 23 | + // It is possible that arr[i] itself has duplicate characters in which case, we will not be able to concatenate arr[i] |
| 24 | + // So check whether (c-'a')-th bit is set in curMask and after that set the (c-'a')-th bit in curMask |
| 25 | + for (char &c: arr[i]) { |
| 26 | + if (mask & (1 << (c - 'a'))) |
| 27 | + return curRes; |
| 28 | + |
| 29 | + if (curMask & (1 << (c - 'a'))) |
| 30 | + return curRes; |
| 31 | + |
| 32 | + curMask |= (1 << (c - 'a')); |
| 33 | + } |
| 34 | + |
| 35 | + // All the bits that were set in curMask will be now set in mask, |
| 36 | + // in order to add all characters of arr[i] to the current string |
| 37 | + mask |= curMask; |
| 38 | + |
| 39 | + // We make a call to i+1 with the updated mask and arr[i]'s length being added |
| 40 | + curRes = max(curRes, (int) arr[i].length() + solve(arr, i+1, mask)); |
| 41 | + return curRes; |
| 42 | + } |
| 43 | + |
| 44 | + int maxLength(vector<string>& arr) { |
| 45 | + return solve(arr, 0, 0); |
| 46 | + } |
| 47 | +}; |
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