1. Trace the array for Bubble sort 34, 8, 64, 51, 32, 21. And fill in the following table

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1 1. Trac th array for Bubbl sort 34, 8, 64, 51, 3, 1. And fill in th following tabl bubbl(intgr Array x, Intgr n) Stp 1: Intgr hold, j, pass; Stp : Boolan switchd = TRUE; Stp 3: for pass = 0 to (n - 1 && switchd = TRUE) incrmnting pass by 1 Stp 3.1: switchd = FALSE; Stp 3.: for j = 0 to n-pass-1 incrmnting j by 1 Stp 3..1: if (x[j] > x[j+1]) Stp : switchd = TRUE; Stp 3..1.: hold = x[j]; Stp : x[j] = x[j+1]; Stp : x[j+1] = hold; Aftr Pass # Altrd squnc aftr an itration # of swaps Answr

2 . Suppos you hav th following sortd list [3, 5, 6, 8, 11, 1, 14, 15, 17, 18] and ar using th binary sarch algorithm. Show th stps for sarching th ky 8 and 16. Also idntify which group of numbrs form th corrct squnc of comparisons usd to find th kys. Answr (A) 1, 6, 11, 8 (B) 1, 17, 15 3.Givn an array of n intgrs, w nd to find th no. of ways of choosing pairs with maximum diffrnc. Exampls: Input : a[] = {3,, 1, 1, 3 Output : 4 Explanation:- Hr, th maximum diffrnc you can find is which is from (1, 3). No. of ways of choosing it: 1) Choosing th first and third lmnts, ) Choosing th first and fourth lmnts, 3) Choosing th third and fifth lmnts, 4) Choosing th fourth and fifth lmnts. Hnc ans is 4. Input : a[] = {, 4, 1, 1 Output : Explanation:- Hr, th maximum diffrnc is 3 from (1, 4). No. of ways choosing it: 1) Choosing th ond and third lmnts, ) Choosing th ond and fourth lmnts. Hnc ans is. Answr. A Simpl solution is to find th minimum lmnt and maximum lmnt to find th maximum diffrnc. Thn w can find th no. of ways of choosing a pair by running two loops. In th innr loop, chck if th two lmnts(on in outr loop and othr in innr loop) ar making maximum diffrnc, if ys incras th count.at last output th count.

3 SET 1. Trac th givn cod String itms = hgadwnoqlbc String sarchitm Output("What ar you sarching for: ") Input sarchitm For x = 0 to 10 If itms[x] = sarchitm Thn Output("Found itm " & sarchitm & " at position " & x) Exit For End If Nxt Output(-1) a. Mak a trac tabl for th cod abov, whr sarchitm = d. b. Mak a trac tabl for th cod abov, whr sarchitm = z". Also writ numbr of comparisons rquird for sarching th itm.. Sort an array of popl's hights. Th following data st is bing passd: hight Sub bubblsort(intgr array hight) Boolan swappd intgr tmp Do swappd = Fals For Count = 1 To MaxSiz - 1 If hight[count + 1] < hight[count] Thn tmp = hight[count] hight[count] = hight[count + 1] hight[count + 1] = tmp swappd = Tru End If

4 Nxt Loop Until swappd = Fals 'Print out th lmnts For Count = 1 To MaxSiz output(count & ": " & hight[count]) Nxt End Sub Construct a trac tabl for th abov cod: Swappd Count MaxSiz Tmp hight null Answr: Swappd Count MaxSiz Tmp 0 4 null hight Rarrang array such that vn indx lmnts ar smallr and odd indx lmnts ar gratr

5 Givn an array, rarrang th array such that : 1. If indx i is vn, arr[i] <= arr[i+1]. If indx i is odd, arr[i] >= arr[i+1] Answr A simpl solution is to sort th array in dcrasing ordr, thn starting from ond lmnt, swap th adjacnt lmnts. SET 3 1. Ali, Brni, Clair, Mohammd, Ptr, Simon, Yvonn a. For th list of 7 s shown abov, can you think of a cas whr linar sarch is fastr than binary sarch? b. For linar sarch of a larg list, th bst cas is if th sought itm is in th first position. What is th bst cas for binary sarch of a larg list? c. Show th stps to sarch for th Mils using Binary Sarch Answr: a. If w sarch for th first itm in th list, Ali, binary sarch still taks 3 comparisons (against Mohammd, Brni and Ali) but linar sarch only nds 1 comparison. b. Binary sarch only rquirs 1 comparison If th sought itm is in th middl of th list.. Nam and Scor ar two diffrnt arrays of typ String and Intgr rspctivly. Trac th following cod for this input arrays Nam Scor 1 Michal 45 Dav 78 3 Grald 3 4 Colin 75 Sub bubblsort(nam[], Scor[]) Boolan swappd String tmpnam Intgr tmpscor Do swappd = Fals For Count = 1 To MaxSiz - 1

6 If Scor[Count + 1] > Scor[Count] Thn tmpnam = Nam[Count] tmpscor = Scor[Count] Scor[Count] = Scor[Count + 1] Nam[Count] = Nam[Count + 1] Nam[Count + 1] = tmpnam Scor[Count + 1] = tmpscor swappd = Tru End If Nxt Loop Until swappd = Fals 'Print out th lmnts For Count = 1 To MaxSiz Output(Count & ": " & Nam[Count] & " " & Scor[Count]) Nxt End Sub Tmp TopScors Swapp d Cou nt MaxS iz or or na m o r o r na m o r null null Micha l 45 Dav 78 Gra ld 3 Coli n 75 Answr: Tmp TopScors Swap pd Co unt Max Siz or or or or na m or

7 0 1 4 null null Mich al 45 Dav 78 Gr ald 3 Coli n Mich al 45 Dav 78 Mich al Gr ald 3 Coli n 75 Gr ald Mich al 45 Coli n 75 Mich al Th output should b: 1: Dav 78 : Colin 75 3: Michal 45 4: Grald 3 3. Givn an array of positiv intgrs and many quris for divisibility. In vry qury, w ar givn an intgr k ( > 0), w nd to count all lmnts in th array which ar prfctly divisibl by k. Input: k = k = 3 k = 5 Output: 3 3 Explanation: Multipls of '' in array ar:- {, 4, 0 Multipls of '3' in array ar:- {9, 15, 1 Multipls of '5' in array ar:- {15, 0

8 SET 4 1. Givn th following algorithm of slction sort and th input list of numbrs, writ / trac th outcom of slction sort aftr vry pass. Assum th maximum numbr is chosn at vry pass. Stp 1: St n to lngth of th list A Stp : Dclar Intgr smallst Stp 3: Rpat from i = 1 to n Stp 3.1: st smallst to i Stp 3.: Rpat from j = i + 1 to n Stp 3..1: if list A[j] < list A[smallst] thn Stp : st smallst to j Stp 3.3: xchang list A[i] and list A[smallst] Input List: [ 14, 33, 7, 10, 35, 19, 4, 44 ]. Givn an array of intgrs, print th array in such a way that th first lmnt is first maximum and ond lmnt is first minimum and so on. Considr th algorithm givn blow and writ th dfinition for slctionsort() and altrnatsort(). Algorithm { Stp 1: Dclar intgr array arr Stp : Assign arr[]={1,1,4,6,7,10 Stp 3: Slctionsort (arr[], intgr n) Stp 4: Altrnatsort (arr[], intgr n) Exampls: Input : arr[] = {7, 1,, 3, 4, 5, 6 Output : Input : arr[] = {1, 6, 9, 4, 3, 7, 8, Output : Answr: void altrnatsort(int arr[], int n) { // Sorting th array sort(arr, arr+n); // Printing th last lmnt of array // first and thn first lmnt and thn // ond last lmnt and thn ond // lmnt and so on. int i = 0, j = n-1; whil (i < j) { cout << arr[j--] << " "; cout << arr[i++] << " "; // If th total lmnt in array is odd // thn print th last middl lmnt. if (n %!= 0) cout << arr[i];

9 3. Givn th following partial binary sarch function which works wll for th numbrs arrangd in anding ordr, mak changs in th algorithm suitabl for th numbrs givn in dnding ordr and fill th mpty placs with suitabl xprssions. Algorithm BinarySarch Intgr BinarySarch(Intgr low,intgr high,intgr ky) Stp 1: whil(low<=high) Stp 1.1: Intgr mid=( ) Stp 1.: if(a[mid]<ky) Stp 1..1: low= Stp 1.3: ls if(a[mid]>ky) Stp 1.3.1: high= Stp 1.4: ls Stp 1.4.1: rturn mid; Stp : rturn -1; //ky not found SET 5 1. Givn th algorithm of slction sort, what changs ar rquird in th algorithm to sort th lmnts by slcting th maximum numbr. Stp 1: St n to lngth of th list A Stp : Dclar Intgr smallst Stp 3: Rpat from i = 1 to n Stp 3.1: st smallst to i Stp 3.: Rpat from j = i + 1 to n Stp 3..1: if list A[j] < list A[smallst] thn Stp : st smallst to j Stp 3.3: xchang list A[i] and list A[smallst]. Writ a psudo cod / algorithm to chck whthr th givn two strings ar sam. If not, display th indx at which thy diffr. Mak us of th functions of flowgorithm. 3. Givn a binary string s and two intgrs x and y ar givn. Task is to arrang th givn string in such a way so that 0 coms X-tim thn 1 coms Y-tim and so on until on of th 0 or 1 is finishd. Thn concatnat rst of th string and print th final string. Givn : x or y cannot b 0. Considr th algorithm givn blow and writ th dfinition for arrangstring (str, x, y) Main Algorithm Stp 1: Dclar string str Stp : Assign str to " " Stp 3: Dclar Intgr x Stp 4: Assign x to 1 Stp 5: Dclar Intgr y Stp 6: Assign y to Stp 7: arrangstring(str, x, y) Exampls:

10 Input : s = "0011" x = 1 y = 1 Output : 0101 x is 1 and y is 1. So first w print '0' on tim th '1' on tim and thn w print '0', aftr printing '0', all 0's ar vanishd from th givn string so w concatnat rst of th string which is '1'. Input : s = '101101' x = 1 y = Output : Answr: void arrangstring(string str, int x, int y) { int count_0 = 0; int count_1 = 0; int ln = str.lngth(); // Counting numbr of 0's and 1's in th // givn string. for (int i = 0; i < ln; i++) { if (str[i] == '0') count_0++; ls count_1++; // Printing first all 0's x-tims // and dcrmnt count of 0's x-tims // and do th similar task with '1' whil (count_0 > 0 count_1 > 0) { for (int j = 0; j < x && count_0 > 0; j++) { if (count_0 > 0) { cout << "0"; count_0--; for (int j = 0; j < y && count_1 > 0; j++) { if (count_1 > 0) { cout << "1"; count_1--;