EDIT DISTANCE (DYNAMIC PROGRAMMING)

SPOJ Problem Set (classical)

6219. Edit distance

Problem code: EDIST



You are given two strings, A and B. Answer, what is the smallest number of operations you need to transform A to B.Operations are: 1) Delete one letter from one of strings 2) Insert one letter into one of strings 3) Replace one of letters from one of strings with another letter InputT - number of test casesFor each test case: String A String BBoth strings will contain only lowercase characters and they won'tbe longer than 2000 characters. OutputFor each test case, one line, minimum number of operations.ExampleInput:1FOODMONEY Output:4
You are given two strings, A and B. Answer, what is the smallest number of operations you need to
transform A to B?

Operations are:

 1) Delete one letter from one of strings
 2) Insert one letter into one of strings
 3) Replace one of letters from one of strings with another letter 

Input

T - number of test cases

For each test case:

 String A
 String B

Both strings will contain only uppercase characters and they won't
be longer than 2000 characters. 

There will be 10 test cases in data set.

Output

For each test case, one line, minimum number of operations.

Example

Input:

1
FOOD
MONEY 

Output:

4


SOURCE CODE--


#include
#include
int min(int a,int b)
{
    if(a>b)
        return b;
    else
        return a;
}
int main()
{
    int t,len1,len2,i,j,m,temp1,temp2;
    char str1[2100];
    char str2[2100];
    static int a[20001][2001];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",str1);
        scanf("%s",str2);
        len1=strlen(str1);
        len2=strlen(str2);
        for(i=0;i<=len1;i++)
            a[i][0]=i;
        for(i=1;i<=len2;i++)
            a[0][i]=i;
        for(i=1;i<=len1;i++)
        {
            for(j=1;j<=len2;j++)
            {
                m=(str1[i-1]==str2[j-1]) ? 0:1;
                temp1=min(a[i-1][j]+1,a[i][j-1]+1);
              temp2=a[i-1][j-1]+m;
              a[i][j]=min(temp1,temp2);
            }
        }
        printf("%d\n",a[len1][len2]);
    }
    return 0;
}

time complexity-O(|str1|*|str2|)