For the following four program fragments, give an analysis of the running time. (Use Big-O) Sum = 0; for i := 1 to N do for j := i to N do for k := j to N do Sum := Sum + 1; Sum = 0; for i := 1 to N * N do for j := 1 to 10000 do Sum = Sun + 1; Sum = 0; for i := 1 to N do if i > 10 for j := 1 to i do Sum := Sum + 1; int product(int n) {if n = 0 then return 0; else return product(n / 2);} Solution a)The Time complexity for this is O(n^3) because it is having 3 for loops last statement execute for the n*n*n times b)The Time complexity for this is O(n^3) because first loop runs for N^2 times and second loop runs for N so total O(n^3) c)The complexity for that is O*(n(n-10)) so the complexity is O(N^2) .

1. For the following four program fragments, give an analysis of the running time. (Use Big-O)
Sum = 0; for i := 1 to N do for j := i to N do for k := j to N do Sum := Sum + 1; Sum = 0; for i :=
1 to N * N do for j := 1 to 10000 do Sum = Sun + 1; Sum = 0; for i := 1 to N do if i > 10 for j :=
1 to i do Sum := Sum + 1; int product(int n) {if n = 0 then return 0; else return product(n / 2);}
Solution
a)The Time complexity for this is O(n^3) because it is having 3 for loops last statement execute
for the n*n*n times
b)The Time complexity for this is O(n^3) because first loop runs for N^2 times and second loop
runs for N so total O(n^3)
c)The complexity for that is O*(n(n-10)) so the complexity is O(N^2)