2013
11-12

# AGTC

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• Deletion: a letter in x is missing in y at a corresponding position.
• Insertion: a letter in y is missing in x at a corresponding position.
• Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C
| | |       |   |   | |
A G T * C * T G A C G C

Deletion: * in the bottom line
Insertion: * in the top line
Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C
|  |  |        |     |     |  |
A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where nm.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

An integer representing the minimum number of possible operations to transform any string x into a string y.

10 AGTCTGACGC
11 AGTAAGTAGGC

4

import java.io.BufferedReader;
import java.util.StringTokenizer;

public class Main{

public static void main(String[] args) throws Exception{
String str ;
StringTokenizer toke1 = new StringTokenizer(str);
int m = Integer.parseInt(toke1.nextToken());
String str1 = toke1.nextToken();
int n = Integer.parseInt(toke2.nextToken());
String str2 = toke2.nextToken();

int[][]array = new int[m+1][n+1];

for(int i=0;i<=m;i++){
array[i][0] = i;
}
for(int i=0;i<=n;i++){
array[0][i] = i;
}

for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(str1.charAt(i-1)==str2.charAt(j-1))
array[i][j] = Math.min(Math.min(array[i-1][j-1], array[i-1][j]+1), array[i][j-1]+1);
else
array[i][j] = Math.min(Math.min(array[i-1][j-1]+1, array[i-1][j]+1), array[i][j-1]+1);
}
}

System.out.println(array[m][n]);
}
}

}

1. #include <cstdio>
#include <cstring>

const int MAXSIZE=256;
//char store[MAXSIZE];
char str1[MAXSIZE];
/*
void init(char *store) {
int i;
store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
for(i=’F';i<=’Z';++i) store =i-5;
}
*/
int main() {
//freopen("input.txt","r",stdin);
//init(store);
char *p;
while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
if(p=fgets(str1,MAXSIZE,stdin)) {
for(;*p;++p) {
//*p=store[*p]
if(*p<’A’ || *p>’Z') continue;
if(*p>’E') *p=*p-5;
else *p=*p+21;
}
printf("%s",str1);
}
fgets(str1,MAXSIZE,stdin);
}
return 0;
}