# Data Structure Questions and Answers-Wagner-Fischer Algorithm

## Click on any option to know the CORRECT ANSWERS

 Question 1
Wagner-Fischer is a ..... algorithm.
 A Brute force B Greedy C Dynamic programming D Recursive
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Question 1 Explanation:
Wagner-Fischer belongs to the dynamic programming type of algorithms.

 Question 2
Wagner-Fischer algorithm is used to find .....
 A Longest common subsequence B Longest increasing subsequence C Edit distance between two strings D All of the mentioned
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Question 2 Explanation:
Wagner-Fischer algorithm is used to find the edit distance between two strings.

 Question 3
What is the edit distance between the strings "abcd" and "acbd" when the allowed operations are insertion, deletion and substitution?
 A 1 B 2 C 3 D 4
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Question 3 Explanation:
The string "abcd" can be changed to "acbd" by substituting "b" with "c" and "c" with "b". Thus, the edit distance is 2.

 Question 4
For which of the following pairs of strings is the edit distance maximum?
 A sunday & monday B monday & tuesday C tuesday & wednesday D wednesday & thursday
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Question 4 Explanation:
The edit distances are 2, 4, 4 and 5 respectively. Hence, the maximum edit distance is between the strings wednesday and thursday.

 Question 5
Consider the following implementation of the Wagner-Fischer algorithm:

int get....min(int a,  int b) { if(a < b) return a; return b; } int edit....distance(char *s1,  char *s2) { int len1, len2, i, j, min; len1 = strlen(s1); len2 = strlen(s2); int arr[len1 + 1][len2 + 1]; for(i = 0;i <= len1; i++) arr[i][0] = i; for(i = 0; i <= len2; i++) arr[0][i] = i; for(i = 1; i <= len1; i++) { for(j = 1; j <= len2; j++) { min = get....min(arr[i-1][j], arr[i][j-1]) + 1; if(s1[i - 1] == s2[j - 1]) { if(arr[i-1][j-1] < min) .....; } else { if(arr[i-1][j-1] + 1 < min) min = arr[i-1][j-1] + 1; } arr[i][j] = min; } } return arr[len1][len2]; }

Which of the following lines should be inserted to complete the above code?

 A arr[i][j] = min B min = arr[i-1][j-1] - 1; C min = arr[i-1][j-1]. D min = arr[i-1][j-1] + 1;
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Question 5 Explanation:
The line min = arr[i-1][j-1] completes the above code.

There are 5 questions to complete.