Question

Find the worst case runtime f(n) for the following algorithms.

• Specify the number of operations executed for an input size n, for the worst case run time as a function of n.
• Circle statement(s) and draw a   line to the right side specifying the number of operations.
• If statement(s) are a part of an iteration of n, specify the total number of iterations as a function of n.
• 
  Algorithm-01 int sum = 0; int j = 1; while ( <= n) { sum++; j-j.2; } Algorithm-02 int sum - 0; for (int i = 1; i <= n * 2; i++) { for (int j = 1; j <= n; j++) { sum++; 3 7 for (int j = 1; j < 100; j++) { sum++; sum++; w Algorithm-03 int sum - ; for (int i - 1; i <n; i++) { for (int j - 1; j1008000; j++) { sum + 10 3 } sum +- 9999; Algorithm-04 int sum = 0; for (int i - 1; i<n; i++) { for (int j - 1; i <-i; j++) { sum - sum + i + j; اسی for (int i = 1; i <n; i++) { for (int j - 1; j - 20; j++) { sume+: }
  

Answer 1

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The solution to the above problem is given in the attached images.Please refer.

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Solution- 1.2 int sum = 0; int while lean) # No of operations log(n) sum tt; to tlagn Tatab no-ofoperations. It log.com) +log (m) LOC login) +2 In flegint # No. of operations 2 n 2) int sum = 0; for (int i=1; il i c= n2, i++) { for(int j = 1; jean; j++) sum tt Y { n/ *nty n/ * (n2+1) Y { 100 100 100 for (int j = 1; i < 100; j + +) sum tt; diom tt; 5 = Total no. of operations I + m2 + m *for+1)+ (n +4 + 100 +100 +100 2 n 2 2. * 2 +300 LEN 4 n 2 tn +301 + 2n

3 #No of operations int sum = 0 fort int i=1; icin; i++) fortint <= 1000000; j++) 106m n asf sum += 100 106m sun to 9999 Itnit 106m + 106n tl & Total no. of operations + 10 (10bn)

4.) #No. of operations 1 int sum=0; fortint i = 1; i =n; i++) fort int jal; jc = i; ji 4 m mm له n(n+1)/2 suma sum titji n(n+1) 12 } n for fint i=1; alin; i++) farlint j = 1; j<= 20; j++) € 2on فيه som tt; 20n y & Total no. of operations = Itn mn+1)/2 Ene +1)/2 tn 620n +20m

Х (= (429 | + - + + + " () t 41. . , . . . + 1 + + |