# Count of pairs of strings which differ in exactly one position

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Function to return the count of same pairs ` `int` `pairCount(map &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 &array, ``int` `m) ` `{ ` `    ``// Dictionary changed will store strings ` `    ``// with wild cards ` `    ``// Dictionary same will store strings ` `    ``// that are equal ` `    ``map 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 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)) `

Output:

```2
```

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Improved By : sanjeev2552

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