Consider the following algorithm:
j = 1 ;
while ( j <= n/2) {
i = 1 ;
while ( i <= j ) {
cout << j << i ;
i++;
}
j++;
}
(a) What is the output when n = 6, n = 8, and n = 10?
(b) What is the time complexity T(n)? You may assume that the input n is divisible by 2.
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Answer / gokul s
n=6:
(1, 1)
(2, 1)
(2, 2)
(3, 1)
(3, 2)
(3, 3)
n=8:
(1, 1)
(2, 1)
(2, 2)
(3, 1)
(3, 2)
(3, 3)
(4, 1)
(4, 2)
(4, 4)
(4, 4)
n=10:
(1, 1)
(2, 1)
(2, 2)
(3, 1)
(3, 2)
(3, 3)
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(5, 1)
(5, 2)
(5, 3)
(5, 4)
(5, 5)
b) Time complexity is ((n/2) * (n/2)+1)/2
| Is This Answer Correct ? | 2 Yes | 3 No |
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