1. Data Structure and Algorithms - S.S. Deshmukh

2. Linear Search algorithm 1.[Intialize] Set K:=1, LOC:= 0 2.Repeat Step 3 & 4 while LOC:=0 & K<=N 3.If ITEM = DATA[K], then: Set LOC:=K 4.Set K:=K+1 [Increment counter] [End of Step 2 loop] 5. [LOC found?] If LOC=0, then: Write: ITEM not found Else Write: LOC is location of ITEM [End of If structure] 6. Exit

3. Largest Element in a array 1.[Initialize] Set K:= 1, LOC:=1, MAX:= DATA[1] 2. Repeat Step 3 and 4 while K<=N 3. If MAX< DATA[K], then: Set LOC:=K, MAX:= DATA[K] [End of if structure] 4. Set K:=K+1 [End of Step 2 loop] 5. Write: LOC, MAX 6. Exit

4. Types of string storage Fixed length storage: SGBAU,AM AMRAVATI, MAHARASHTR T, A. 200 280 360 Suppose input contains sting as, ‘SGBAU, AMT, ’ ‘AMRAVATI,’ ‘MAHARASHTRA.’ And first character address is 200 and record length is 80 Record are stored in memory sequentially as above

5. Types of string storage Fixed length storage using the pointers 200 280 360 SGBAU,AM MAHARASHTR AMRAVATI, T, A. POINTER 1 2 3 4 5 200 280 360 Suppose input contains sting as, ‘SGBAU, AMT, ’ ‘AMRAVATI,’ ‘MAHARASHTRA.’

6. Types of string storage Variable length storage: 1) Use of marker to signal end of string 200 280 360 SGBAU,AM MAHARASHTR AMRAVATI,$$ T, A. POINTER 1 2 3 4 5 200 280 360 Suppose input contains sting as, ‘SGBAU, AMT, ’ ‘AMRAVATI,’ ‘MAHARASHTRA.’

7. Types of string storage Variable length storage: 2) Record length is listed 200 280 360 SGBAU,AM MAHARASHTR AMRAVATI, T, A. 200 280 360 Suppose input contains sting as, ‘SGBAU, AMT, ’ ‘AMRAVATI,’ ‘MAHARASHTRA.’ 12 9 LGH PTR

8. Types of string storage Linked list storage Suppose the string is ‘TO BE OR NOT’ Using 1 character per node T OBEOR NOT0

9. Types of string storage Linked list storage Suppose the string is ‘TO BE OR NOT’ Using 4 character per node TOB EORNOT0

10. String operation Substring Finding part of the string is done in substring SUBSTRING(String, Initial, Length) String- Name of string Initial- Position of the first character Length- Length of the substring For ex: In string ‘PRMCEAM Badnera’ we find the substring ‘Badnera’ by using SUBSTRING(‘PRMCEAM Badnera’, 9,7)  Badnera 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7

11. String operation Indexing Finding position where the pattern P first appears in the string T INDEX(String,Pattern) String- Name of string Pattern – Pattern which we want to find If the pattern does not found then INDEX is assigned the value 0 For ex: In string ‘PRMCEAM Badnera’ we find the INDEX of ‘Bad’ by using INDEX(‘PRMCEAM Badnera’, ‘Bad’)  9 1 2 3 4 5 6 7 8 9 10 11

12. String operation Concatenation For string S1 & S2, finding new string consisting of char of S1 followed by char of S2. Denoted by S1//S2 For ex: S1= ‘PRMCEAM’, S2= ‘Badnera’ we find the S1//S2  ‘PRMCEAM Badnera’

13. String operation Length The number of character in string is called its length Denoted by LENGTH(String) For ex: LENGTH(‘PRMCEAM Badnera’)  15

14. Word Processing Types of word processing: 1)Insertion: Inserting string in the middle of the given string is called the insertion operation. Given text T we want to insert string S so that S begins from postion K. Denoted by: INSERT(TEXT,POSITION,STRING) Ex: INSERT(‘ABCDEFG’, 3, ’XYZ’)  ‘ABXYZCDEFG’ INSERT(T,K,S) = SUBSTRING(T,1,K-1) // S // SUBSTRING(T,K,(LENGTH(T)-K+1))

15. Word Processing Types of word processing: 2) Deletion: Deleting substring from the given string which begins in position K & has length L. Denoted by: DELETE(TEXT,POSITION,LENGTH) Ex: DELETE(‘ABCDEFG’,4,2)  ‘ABCFG’ DELETE (T,K,S) = SUBSTRING(T,1,K-1) // SUBSTRING(T, K+L, LENGTH(T)-K-L+1)

16. Word Processing Types of word processing: 3) Replacement: In given text T we want to replace the first occurrence of pattern P1 by pattern P2 Denoted by: REPLACE(TEXT, PATTERN1, PATTERN2) Ex: REPLACE(‘XABYABZ’, ’AB’, ’C’)  ‘XCYABZ’ No replacement if P1 not found in the given string K := INDEX(T, P1) T :=DELETE(T,K,LENGTH(P1)) INSERT (T,K,P2)

17. Algorithm: A text T & pattern P & Q are in memory. This algorithm replaces every occurrence of P in T by Q Step 1: [Find index of P] Set K:= INDEX(T,P) Step 2: Repeat while K = 0 [Replace P by Q] Set T:= REPLACE(T,P,Q) [Update Index] Set K:= INDEX(T,P) [End of loop] Step 3: Write: T Step 4: Exit

18. Pattern Matching Algorithm Pattern matching algorithm is the problem of deciding whether or not the given string pattern P appears in the string text T. Characters are denoted by lowercase, can also use exponents to show repetition. ex:a 2 b 3 == aabbb Empty string is denoted by ^, lambda

19. Pattern Matching Algorithm First pattern matching We compare given pattern P with each of substring of T, moving from left to right until we get a match or dont reach end of the T. W k = substring of T having same length as P MAX = LENGTH(T) – LENGTH(P) + 1

20. Pattern Matching Algorithm First pattern matching Ex: If P is 4 char string, T is 20 char string P = P[1]P[2]P[3]P[4] T = T[1]T[2]T[3]T[4]T[5]T[6]….T[18]T[19]T[20] MAX= 20- 4 + 1 = 17 ie total 17 substring of T W1=T[1]T[2]T[3]T[4] W2=T[2]T[3]T[4]T[5]. W17=T[17]T[18]T[19]T[20] Example

21. First pattern matching algorithm Step 1. [Initialize] Set K:=1, MAX := S – R + 1 Step 2. Repeat Step 3 to 5 while K<= MAX Step 3. Repeat for L = 1 to R [Test each character of P] If P[L] = T [K+L-1], then Goto Step 5 [End of inner loop] Step 4. [Success] Set INDEX:= K and Exit Step 5. Set K:=K+1 [End of Step 2 outer loop] Step 6. [Failure] Set INDEX :=0 Step 7. Exit

22. First pattern matching algorithm Ex 3.9, Pg no 3.17 Use slow pattern matching algorithm to find number of comparisons C and also find INDEX of P in text T. a) P = aaba and T = cdcd..cd = (cd) 10 b) P = aaba and T = abababab…. c) P = aaab and T = aa….aaa= a 20

23. Second pattern matching algorithm