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September 1, 2024Switching
September 1, 2024GATE 2014 [Set-3]
| Question 49 |
Suppose we have a balanced binary search tree T holding n numbers. We are given two numbers L and H and wish to sum up all the numbers in T that lie between L and H. Suppose there are m such numbers in T. If the tightest upper bound on the time to compute the sum is O(nalogbn + mclogdn), the value of a + 10b + 100c + 1000d is ____.
| 110 | |
| 111 | |
| 112 | |
| 113 |
Question 49 Explanation:
It takes (log n) time to determine numbers n1 and n2 in balanced binary search tree T such that
1. n1 is the smallest number greater than or equal to L and there is no predecessor n1‘ of >n1 such that n1‘ is equal to n1.
2. n22′ of n2 such that is equal to n2.
Since there are m elements between n1 and n2, it takes ‘m’ time to add all elements between n1 and n2.
So time complexity is O(log n+m)
So the given expression becomes O(n0log’n+m’log0n)
And a+10b+100c+1000d = 0+10*1+100*1+1000*1 = 10+100 = 110
Because a=0, b=1, c=1 and d=0
1. n1 is the smallest number greater than or equal to L and there is no predecessor n1‘ of >n1 such that n1‘ is equal to n1.
2. n22′ of n2 such that is equal to n2.
Since there are m elements between n1 and n2, it takes ‘m’ time to add all elements between n1 and n2.
So time complexity is O(log n+m)
So the given expression becomes O(n0log’n+m’log0n)
And a+10b+100c+1000d = 0+10*1+100*1+1000*1 = 10+100 = 110
Because a=0, b=1, c=1 and d=0
Correct Answer: A
Question 49 Explanation:
It takes (log n) time to determine numbers n1 and n2 in balanced binary search tree T such that
1. n1 is the smallest number greater than or equal to L and there is no predecessor n1‘ of >n1 such that n1‘ is equal to n1.
2. n22′ of n2 such that is equal to n2.
Since there are m elements between n1 and n2, it takes ‘m’ time to add all elements between n1 and n2.
So time complexity is O(log n+m)
So the given expression becomes O(n0log’n+m’log0n)
And a+10b+100c+1000d = 0+10*1+100*1+1000*1 = 10+100 = 110
Because a=0, b=1, c=1 and d=0
1. n1 is the smallest number greater than or equal to L and there is no predecessor n1‘ of >n1 such that n1‘ is equal to n1.
2. n22′ of n2 such that is equal to n2.
Since there are m elements between n1 and n2, it takes ‘m’ time to add all elements between n1 and n2.
So time complexity is O(log n+m)
So the given expression becomes O(n0log’n+m’log0n)
And a+10b+100c+1000d = 0+10*1+100*1+1000*1 = 10+100 = 110
Because a=0, b=1, c=1 and d=0