Given an array of strings str[] of length N, consisting of strings of the same length, the task is to find the string which only differs by a single character from all the given strings. If no such string can be generated, print -1. In case of multiple possible answers, print any of them.
Example:
Input:_ str[] = { “abac”, “zdac”, “bdac”} _
Output:_ adac _
Explanation:
The string “adac” differs from all the given strings by a single character.
Input:_ str[] = { “geeks”, “teeds”} _
Output:_ teeks_
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach: Follow the steps below to solve the problem:
- Set the first string as the answer.
- Now, replace the first character of the string to all possible characters and check if it differs by a single character from the other strings or not.
- Repeat this process for all the characters in the first string.
- If any such string of the required type is found from the above step, print the string.
- If no such situation arises where replacing a single character of the first string generates a string of the required type, print -1.
Below is the implementation of the above approach.
- C++
- Python3
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
#define ll long long
**using** **namespace** std;
// Function to check if a given string
// differs by a single character from
// all the strings in an array
**bool** check(string ans, vector<string>& s,
**int** n, **int** m)
{
// Traverse over the srings
**for** (``**int** i = 1; i < n; ++i) {
// Stores the count of characters
// differing from the strings
**int** count = 0;
**for** (``**int** j = 0; j < m; ++j) {
**if** (ans[j] != s[i][j])
count++;
}
// If differs by more than one
// character
**if** (count > 1)
**return** **false**``;
}
**return** **true**``;
}
// Function to find the string which only
// differ at one position from the all
// given strings of the array
string findString(vector<string>& s)
{
// Size of the array
**int** n = s.size();
// Length of a string
**int** m = s[0].size();
string ans = s[0];
**int** flag = 0;
**for** (``**int** i = 0; i < m; ++i) {
**for** (``**int** j = 0; j < 26; ++j) {
string x = ans;
// Replace i-th character by all
// possible characters
x[i] = (j + 'a'``);
// Check if it differs by a
// single character from all
// other strings
**if** (check(x, s, n, m)) {
ans = x;
flag = 1;
**break**``;
}
}
// If desired string is obtained
**if** (flag == 1)
**break**``;
}
// Print the answer
**if** (flag == 0)
**return** "-1"``;
**else**
**return** ans;
}
// Driver code
**int** main()
{
vector<string> s = { "geeks"``, "teeds" };
// Function call
cout << findString(s) << endl;
}
Output:
teeks
Time Complexity:_ O(N * M2 * 26)_
Auxiliary Space:_ O(M)_
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#arrays #greedy #searching #strings #frequency-counting #permutation and combination
