Given an array arr[] of strings of equal lengths. The task is to calculate the total number of pairs of strings which differ in exactly one position.
Examples:
Input: arr[] = {“abc”, “abd”, “bbd”}
Output: 2
(abc, abd) and (abd, bbd) are the only valid pairs.Input: arr[] = {“def”, “deg”, “dmf”, “xef”, “dxg”}
Output: 4
Method 1: For every possible pair, check if both the strings differ in exactly a single index position with a single traversal of the strings.
Method 2: Two string can be compared in the following way in order to check whether they differ in a single index position:
Let str1 = “abc” and str2 = “adc”
For str1, add “#bc”, “a#c” and “ab#” to a set.
Now for str2, generate the string in the similar manner and if any of the generated string
is already present in the set then both the strings differ in exactly 1 index position.
For example, “a#c” is one of the generated strings.
Note that “#” is used because it will not be a part of any of the original strings.
Below is the implementation of the above approach:
C++
// CPP implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count of same pairs int pairCount(map<string, int > &d) { int sum = 0; for ( auto i : d) sum += (i.second * (i.second - 1)) / 2; return sum; } // Function to return total number of strings // which satisfy required condition int difference(vector<string> &array, int m) { // Dictionary changed will store strings // with wild cards // Dictionary same will store strings // that are equal map<string, int > changed, same; // Iterating for all strings in the given array for ( auto s : array) { // If we found the string then increment by 1 // Else it will get default value 0 same[s]++; // Iterating on a single string for ( int i = 0; i < m; i++) { // Adding special symbol to the string string t = s.substr(0, i) + "//" + s.substr(i + 1); // Incrementing the string if found // Else it will get default value 0 changed[t]++; } } // Return counted pairs - equal pairs return pairCount(changed) - pairCount(same) * m; } // Driver Code int main() { int n = 3, m = 3; vector<string> array = { "abc" , "abd" , "bbd" }; cout << difference(array, m) << endl; return 0; } // This code is contributed by // sanjeev2552 |
Python3
# Python3 implementation of the approach # Function to return the count of same pairs def pair_count(d): return sum ((i * (i - 1 )) / / 2 for i in d.values()) # Function to return total number of strings # which satisfy required condition def Difference(array, m): # Dictionary changed will store strings # with wild cards # Dictionary same will store strings # that are equal changed, same = {}, {} # Iterating for all strings in the given array for s in array: # If we found the string then increment by 1 # Else it will get default value 0 same[s] = same.get(s, 0 ) + 1 # Iterating on a single string for i in range (m): # Adding special symbol to the string t = s[:i] + '#' + s[i + 1 :] # Incrementing the string if found # Else it will get default value 0 changed[t] = changed.get(t, 0 ) + 1 # Return counted pairs - equal pairs return pair_count(changed) - pair_count(same) * m # Driver code if __name__ = = "__main__" : n, m = 3 , 3 array = [ "abc" , "abd" , "bbd" ] print (Difference(array, m)) |
2
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