CS1010E Programming Methodology Tutorial 6 Arrays and Search

1. CS1010E Programming MethodologyTutorial 6Arrays and Search C14,A15,D11,C08,C11,A02

2. Arrays • Homogenous Collection of Data • Declaration: • TYPE name [NumberofElement > 0] • int scores[10]; char name[5]; double prices[7]; etc • Initialization: • Initialize before use !!!! • Initializing each element separately • for(i=0;i<10;i++) arr[i] = something; • Initializing array at the time of declaration • int a[] = {1, 2, 3,4}; • What if int a[10] = {1,2,3} ?;

3. Arrays • Access by index: • a[10], a[0] etc. • Access by address: • *(a+ 10) Why? • Array Specials : • Array Name: • stores the address of the first element • Array Length: • Always keep the length of array!! This is vital! • Array is pass-by-reference • Why & how ? • No variable size: • inti = 10; inta[i]; IS WRONG!! What is the output? What is the output?

4. Question 1 • printArray(list1, 5); • 11 22 33 44 55 list1 list2

5. Question 1 • passElement(list1[0]); • printArray(list1, 5); • 11 22 33 44 55 list1 list2

6. Question 1 • changeElements(list2); • printArray(list2, 5); • ?? list1 list2 list[2] list[4]

7. Question 1 • changeElements(list2); • printArray(list2, 5); • 99 99 77 99 88 list1 list2 list[2] list[4]

8. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); • ??? list1 list2

9. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); list1 list2 List1[0] List2[0]

10. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); list1 list2 List1[1] List2[1]

11. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); list1 list2 List1[2] List2[2]

12. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); list1 list2 List1[3] List2[3]

13. Question 1 • copyArray(list2, list1, 5) ); • printArray(list2, 5); • 11 22 33 44 55 list1 list2 List1[4] List2[4]

14. Question 1 • printPattern(list1,5); • ??? list1 list2

15. Question 1 • printPattern(list1,5); • printArray(list+4, 1) • 55 list1 list2 list+ 5 – i = list + 4 i = 1

16. Question 1 • printPattern(list1,5); • printArray(list+3, 1) • 44 55 list1 list2 list+ 5 – i = list + 3 i = 2

17. Question 1 • printPattern(list1,5); • printArray(list+2, 1) • 33 44 55 list1 list2 list+ 5 – i = list + 2 i = 3

18. Question 1 • printPattern(list1,5); • printArray(list+1, 1) • 22 33 44 55 list1 list2 list+ 5 – i = list + 1 i = 4

19. Question 1 • printPattern(list1,5); • printArray(list+0, 1) • 11 22 33 44 55 list1 list2 list+ 5 – i = list + 0 i = 5

27. Question 3 (a) • FindIndex(int list[], intnumElem, int key) • Return the position where the key should be inserted • Algorithm I : • Scan the list, [0..numElem -1] • Find the first element whose value is greater than key • if(list[k] key) continue • if(list[k] > key) break; • Return the position of that element key Stops, return 4

28. Question 3 (a) • Translate Algorithm I into Code: Direct Translate Cleaner version

29. Question 3 (b) • shiftRight(int list[], int start, int n) • Right shift 1 position in the range of [ start, start + n -1] • Right most element will become the first one • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp;

30. Question 3(b) • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp; Start = 1, n = 4 Start + n -1 = 4 [1, 4] tmp = 44

31. Question 3(b) • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp; Start = 1, n = 4 [1, 4] tmp = 44

32. Question 3(b) • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp; Start = 1, n = 4 [1, 4] tmp = 44

33. Question 3(b) • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp; Start = 1, n = 4 [1, 4] tmp = 44 Stops scan

34. Question 3(b) • Algorithm II: • Store the last element • tmp = list[start+n-1] • Reverse Scan list start+n-1  start +1 • list[k] = list[k-1] • Set start element to • list[start] = temp; Start = 1, n = 4 [1, 4] tmp = 44

35. Question 3 (b) • Translate Algorithm II into Code Code for Algorithm II

36. Question 3 (c) • InsertionSort(int list[], int n) • Sort elements in list in increasing order • Algorithm III: • Scan list from [1, n-1] • Find insertion position • Index = findIndex(k + 1, list[k]); • Insert list[k] to that position • rightshift(index, k-index+1) When scanning list[k],list[0…k-1] is sorted

37. Question 3 (c) • Algorithm III: • For element list[k] • Find insertion position • Index = findIndex(k + 1, list[k]); • Insert list[k] to that position • rightshift(index, k-index+1) When scanning list[k],list[0…k-1] is sorted list list[0…3] is sorted list[4]

38. Question 3 (c) • Algorithm III: • For element list[k] • Find insertion position • Index = findIndex(k + 1, list[k]); • Insert list[k] to that position • rightshift(index, k-index+1) list Insertion position list[0…3] is sorted list[4] Index = 2

39. Question 3 (c) • Algorithm III: • For element list[k] • Find insertion position • Index = findIndex(k + 1, list[k]); • Insert list[k] to that position • rightshift(index, k-index+1) list Insertion position Rightshift[2, 3] Index = 2 list[4]

40. Question 3 (c) • Algorithm III: • For element list[k] • Find insertion position • Index = findIndex(k + 1, list[k]); • Insert list[k] to that position • rightshift(index, k-index+1) list list[0, 4] is sorted move to list[5]

41. Question 3 (c) • Translate Algorithm III into code: Code for Algorithm III

42. Thank you See you next week!!