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Quantitative Aptitude

Fundamental Principle of Counting

Suppose you a...

Question

Suppose you are given an array s[1...n] and a procedure reverse (s, i, j) which reverses the order of elements in between positions i and j (both inclusive). What does the following sequence do:

$w h e r e 1 \leq k \leq n :$
reverse (s, 1, k);
reverse (s, k + 1, n);
reverse (s, 1, n);

Rotate s left by k position

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None of the above

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Leave s unchanged

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Reverse all elements of s

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Solution

The correct option is A Rotate s left by k position
Effect of the given 3 reversals for any k is equivalent to left rotation of the array of size n by k.

Let S[ 1.......7]

1 2 3 4 5 6 7

$∴$ n = 7, k = 2

reverse (S, 1, 2)
we get [ 2, 1, 3, 4, 5, 6, 7]

reverse (S, 3, 7)
we get [2, 1, 7, 6, 5, 4, 3 ]

reverse (S, 1, 7)
we get [3, 4, 5, 6, 7, 1, 2]

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Similar questions

Q. Consider the following implementations for rotating an array A[1, ....n] of size n by K positions , where 1 < = K < = n . Note that the procedure reverse (A, i, j) reverses the order of elements in A between positions i and j in A (both inclusive )
which of the above implementation(s) work correctly for all inputs ?
Implementation I:
reverser(A, 1, K);
reverse(A, 1, n) ;
reverse (A, k+1, n) ;
Implementation II:
reverser(a, K+1, n);
reverse(a, 1, K) ;
reverse (a, 1, n) ;
Implementation III:
reverser(a, 1, K);
reverse(a, K+1, n) ;
reverse (a, 1, n) ;

Q. Consider the following C-function in which a [n] and b [m] are two sorted integer arrays and c [n+m] be another integer array.

void xyz (int a[ ], int b[ ], int c[ ])
{
int i, j, k;
i = j = k = 0;
while ((i < n) && (j < m))
if (a[i] < b[j])
c[k++] = a[i ++];
else
c[k++] = b[j ++];
}

Which of the following condition(s) hold after the termination of the while loop?

(i) j < m, k = n + j - 1 & a[n- 1] < b[j] if i = n
(ii) i < n, k = m + i - 1 & b[m- 1]

\leq

a[i] if j = m

Q. Suppose

A_{1}, A_{2},, A_{30}

are thirty sets each having

5

elements and

B_{1}, B_{2}, . ., B_{n}

are

n

sets each with

3

elements, let

30 ⋃ i = 1 A_{i} = n ⋃ j = 1 B_{j} = S

and each element of

S

belongs to exactly

10

of the

A_{i} s

and exactly

9

of the

B_{j} s .

Then

n

is equal to

Suppose $A_{1}, A_{2}, . . ., A_{30}$ are thirty sets, each with five elements and $B_{1}, B_{2}, . . ., B_{30}$ are $n$ sets ecah with three elements. Let $30 ⋃ i = 1 A_{i} = n ⋃ j = 1 B_{j} = S$

If each element of $S$ belongs to exactly ten of the $A_{i}^{'} s$ and exactly none of the $B_{j}^{'} s$ then $n =$

Q. The entries of an

n \times n

array of numbers are denoted by

x_{i j},

with

1 \leq i, j \leq n .

For all i, j, k,

i, j, k, 1 \leq i, j, k \leq n

the following equality holds

x_{i j} + x_{j k} + x_{k i} = 0.

Prove that there exist numbers

t_{1}, t_{2}, . . ., t_{n}

such that

x_{i j} = t_{i} - t j

for all

i, j, k, 1 \leq i, j, k \leq n .

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Suppose you are given an array s[1...n] and a procedure reverse (s, i, j) which reverses the order of elements in between positions i and j (both inclusive). What does the following sequence do: where 1≤k≤n: reverse (s, 1, k); reverse (s, k + 1, n); reverse (s, 1, n);

Suppose you are given an array s[1...n] and a procedure reverse (s, i, j) which reverses the order of elements in between positions i and j (both inclusive). What does the following sequence do:

$w h e r e 1 \leq k \leq n :$
reverse (s, 1, k);
reverse (s, k + 1, n);
reverse (s, 1, n);