C. Marina and Vasya
Marina loves strings of the same length and Vasya loves when there is a third string, different from them in exactly t characters. Help Vasya find at least one such string.
More formally, you are given two strings s1, s2 of length n and number t. Let‘s denote as f(a, b) the number of characters in which strings a and b are different. Then your task will be to find any string s3 of length n, such that f(s1, s3) = f(s2, s3) = t. If there is no such string, print - 1.
Input
The first line contains two integers n and t (1 ≤ n ≤ 105, 0 ≤ t ≤ n).
The second line contains string s1 of length n, consisting of lowercase English letters.
The third line contain string s2 of length n, consisting of lowercase English letters.
Output
Print a string of length n, differing from string s1 and from s2 in exactly t characters. Your string should consist only from lowercase English letters. If such string doesn‘t exist, print -1.
Sample test(s)
Input
3 2abcxyc
Output
ayd
Input
1 0cb
1 #include<cstdio>
2 #include<vector>
3 #include<cmath>
4 #include<queue>
5 #include<map>
6 #include<cstring>
7 #include<algorithm>
8 using namespace std;
9 typedef long long ll;
10 typedef unsigned long long ull;
11 const int maxn=100005;
12 int main()
13 {
14 int t,n;
15 char s1[maxn],s2[maxn],s3[maxn];
16 scanf("%d%d",&n,&t);
17 scanf("%s%s",s1,s2);
18 t=n-t;
19 int k1=0,k2=0;
20 for(int i=0;i<n;i++)
21 {
22 if(s1[i]==s2[i]&&(k1<t&&k2<t))s3[i]=s1[i],k1++,k2++;
23 else if(s1[i]!=s2[i]||k1==t)
24 for(int j=0;j<26;j++)
25 if(s1[i]!=‘a‘+j&&s2[i]!=‘a‘+j)
26 {
27 s3[i]=(char)(‘a‘+j);
28 break;
29 }
30 }
31 if(k1==t)
32 {
33 puts(s3);
34 return 0;
35 }
36 for(int i=0;i<n;i++)
37 {
38 if(s1[i]==s2[i])continue;
39 if(k1<t)
40 s3[i]=s1[i],k1++;
41 else if(k2<t)
42 s3[i]=s2[i],k2++;
43 else
44 for(int j=0;j<26;j++)
45 if(s1[i]!=‘a‘+j&&s2[i]!=‘a‘+j)
46 {
47 s3[i]=(char)(‘a‘+j);
48 break;
49 }
50 }
51 if(k1<t||k2<t)puts("-1");
52 else puts(s3);
53 return 0;
54 }
Output
-1