给出的字符串STR []的长度为N的阵列，由相同长度的字符串，任务是找到字符串仅由一个字符从所有给定的字符串不同。如果无法生成这样的字符串，请打印-1 。如果有多个可能的答案，请打印其中的任何一个。
例子：
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
方法：请按照以下步骤解决问题：
- 将第一个字符串设置为答案。
- 现在，将字符串的第一个字符替换为所有可能的字符，并检查它是否与其他字符串相差一个字符。
- 对第一个字符串中的所有字符重复此过程。
- 如果在上述步骤中找到了所需类型的任何此类字符串，请打印字符串。
- 如果没有出现这样的情况，即替换第一个字符串的单个字符生成所需类型的字符串，请打印-1。
下面是上述方法的实现。
C++
// C++ Program to implement
// the above approach
#include
#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& 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& 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 s = { "geeks", "teeds" };
// Function call
cout << findString(s) << endl;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check if a given string
// differs by a single character from
// all the strings in an array
static boolean check(String ans, 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.charAt(j) != s[i].charAt(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
static String findString(String[] s)
{
// Size of the array
int n = s.length;
String ans = s[0];
// Length of a string
int m = ans.length();
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 = x.replace(x.charAt(i), (char)(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
public static void main(String []args)
{
String s[] = { "geeks", "teeds" };
// Function call
System.out.println(findString(s));
}
}
// This code is contributed by chitranayalPython3
# Python3 program to implement
# the above approach
# Function to check if a given string
# differs by a single character from
# all the strings in an array
def check(ans, s, n, m):
# Traverse over the srings
for i in range(1, n):
# Stores the count of characters
# differing from the strings
count = 0
for j in range(m):
if(ans[j] != s[i][j]):
count += 1
# 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
def findString(s):
# Size of the array
n = len(s)
# Length of a string
m = len(s[0])
ans = s[0]
flag = 0
for i in range(m):
for j in range(26):
x = list(ans)
# Replace i-th character by all
# possible characters
x[i] = chr(j + ord('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 ''.join(ans)
# Driver Code
# Given array of strings
s = [ "geeks", "teeds" ]
# Function call
print(findString(s))
# This code is contributed by Shivam SinghC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check if a given string
// differs by a single character from
// all the strings in an array
static bool check(String ans, 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
static String findString(String[] s)
{
// Size of the array
int n = s.Length;
String ans = s[0];
// Length of a string
int m = ans.Length;
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 = x.Replace(x[i], (char)(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
public static void Main(String []args)
{
String []s = { "geeks", "teeds" };
// Function call
Console.WriteLine(findString(s));
}
}
// This code is contributed by Rajput-Ji输出：
teeks
时间复杂度： O(N * M 2 * 26)
辅助空间： O(M)