Minimum Unique Word Abbreviation in C++

Suppose we have a string such as "word" and that contains the following abbreviations: ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]. We also have a target string and a set of strings in a dictionary, we have to find an abbreviation of this target string with the smallest possible length such that it does not conflict with abbreviations of the strings in the dictionary. Here each number or letter in the abbreviation is considered as length = 1. So as an example, the abbreviation "a32bc" is of length = 4.

So, if the input is like "apple", and dictionary is ["blade"], then the output will be "a4"

To solve this, we will follow these steps −

Define an array dict
Define a function abbrLen(), this will take mask,
ret := n
for initialize b := 3, when b < bn, update b <<= 1, do −
- if (mask AND b) is same as 0, then −
  - (decrease ret by 1)
return ret
Define a function dfs(), this will take bit, mask,
len := abbrLen(mask)
if len >= minLen, then −
- return
match := true
for each d in dict, do −
- if (mask AND d) is same as 0, then −
  - match := false
  - Come out from the loop
if match is non-zero, then −
- minLen := len
- minab := mask
Otherwise −
- for initialize b := bit, when b < bn, update b := b*2 , do −
  - if (cand AND b) is not equal to 0, then −
    - dfs(b*2, mask OR b)
From the main method do the following −
ret := blank string
n := size of target
bn := 2^n
cand := 0
minLen := inf
for each s in dictionary −
- if size of s is not equal to n, then −
  - Ignore following part, skip to the next iteration
- word := 0
- for initialize i := 0, when i < size of s, update (increase i by 1), do −
  - if s[i] is not equal to target[i], then −
    - word := word OR (2^i)
- insert word at the end of dict
- cand := cand OR word
dfs(1, 0)
for initialize i := 0, when i < n, do −
- if (minab AND 2^i) is not equal to 0, then −
  - ret := ret + target[i]
  - (increase i by 1)
- Otherwise
  - j := i
  - while (i < n and (minab AND 2^i) is same as 0), do −
    - (increase i by 1)
  - ret := ret concatenate (i - j)
return ret

Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int n, cand, bn, minLen, minab;
   vector<int> dict;
   int abbrLen(int mask) {
      int ret = n;
      for (int b = 3; b < bn; b <<= 1) {
         if ((mask & b) == 0)
            ret--;
      }
      return ret;
   }
   void dfs(int bit, int mask) {
      int len = abbrLen(mask);
      if (len >= minLen)
         return;
      bool match = true;
      for (int d : dict) {
         if ((mask & d) == 0) {
            match = false;
         break;
      }
   }
   if (match) {
      minLen = len;
      minab = mask;
   }
   else {
      for (int b = bit; b < bn; b <<= 1) {
         if ((cand & b) != 0)
            dfs(b << 1, mask | b);
         }
      }
   }
   string minAbbreviation(string target, vector<string> &dictionary) {
      string ret = "";
      n = target.size();
      bn = 1 << n;
      cand = 0;
      minLen = INT_MAX;
      for (string &s : dictionary) {
         if (s.size() != n)
            continue;
         int word = 0;
         for (int i = 0; i < s.size(); i++) {
            if (s[i] != target[i])
               word |= (1 << i);
         }
         dict.push_back(word);
         cand |= word;
      }
      dfs(1, 0);
      for (int i = 0; i < n;) {
         if ((minab & (1 << i)) != 0) {
            ret += target[i];
            i++;
         }
         else {
            int j = i;
            while (i < n && (minab & (1 << i)) == 0)
               i++;
            ret += to_string(i - j);
         }
      }
      return ret;
   }
};
main() {
   Solution ob;
   vector<string> v = {"blade"};
   cout << (ob.minAbbreviation("apple",v));
}