# Recent posts tagged preparation-schedule

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Yes. You are correct. But not just the "max element" cause problem. All the elements in the last level can cause this problem. So, only binary search tree is the answer here.
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The total number of nodes accessed including root will be 5. The order is, (9)-->(5)-->(5,7)-->(9,11)-->(13,15).
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Probability that buckets other than $1$ are selected $= \dfrac{n-1}{n}$ This should happen $k$ times and each of the $k$ events are independent so $\dfrac{(n-1)^k}{n^k}$ When $k=1,$ probability of no collision $= \dfrac{n}{n}$ (for only one insert, there can ... $k>n+1$ probability of first collision $= 0$ (as it should have definitely happened in one of the previous $(n+1)$ insertions.
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if you solve this you will get XY' + Y'Z + XY (this can be simplified to X + Y'Z) with min terms as (1,4,5,6,7) and option A has the same min terms so option A is equivalent to XY' + Y'Z + XY Ans (A)
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option A Let P(x)=All are multiple of 4 Q(X)= All are even numbers if a &isin; P(X) then a must &isin; Q(X)
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No, heap should also do, just that we have to use min-heap. (C) should be the answer.
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The number of surjective functions defined from $A$ to $B$ where |A| = 5, |B| = 4, is _______
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$\text{Size of each segment} = \frac{2^{16}}{8}=2^{13}$ Let the size of page be $2^k$ bytes We need a page table entry for each page. For a segment of size $2^{13}$, number of pages required will be $2^{13-k}$ and so we need $2^{13-k}$ page table entries. Now ... $\text{Size of each page table entry} = 2$ bytes $=16$ bits $\text{Number of bits left for aging} = 16-12 = 4$ bits
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Ans: because we have given wavy edges form MST So, for verification of option A we have to check that with MST how many cost to reach at a->b then we will get a->e->d->b = -2+5+3 = 6 so in given option a with cost(a,b)>= 6 this is posible coz , cost ... making MST so cost must be >6 not equal to 6 so option A is Need Not HOLD. Like wise if you check for other option then enequality is holding...
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A language is a set of strings. Regularity property is for such sets. We cannot apply it to a single string. If you put that string in a set, it becomes a singleton set- which is finite and hence regular also.
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