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37 37 votes

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

           reverse (s, 1, k);
           reverse (s, k+1, n);
           reverse (s, 1, n);
  1. Rotates $s$ left by $k$ positions
  2. Leaves $s$ unchanged 
  3. Reverses all elements of $s$
  4. None of the above

5 Answers

Best answer
67 67 votes

Answer is A.

Effect of the above $3$ reversals for any $K$ is equivalent to left rotation of the array of size $n$ by $k$.

Let , $S[1\ldots7] = \begin{array}{|c|c|c|c|c|c|c|}\hline1&2&3&4&5&6&7\\ \hline\end{array}$
So, $n=7,k = 2$

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 $[3,4,5,6,7,1,2]$

Hence, option $(A)$ rotates $s$ left by $k$ positions and is correct.

• edited by
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k=4

string s:          3 5 1 9 2 13 7

reverse (s, 1, 4): 9 1 5 3 2 13 7

reverse (s, 5, 7): 9 1 5 3 7 13 2

reverse (s, 1, 7): 2 13 7 3 5 1 9

 

Roation to leftt by one position: 5 1 9 2 13 7 3

Roation to leftt by one position: 1 9 2 13 7 3 5

Roation to leftt by one position: 9 2 13 7 3 5 1

Roation to leftt by one position: 2 13 7 3 5 1 9

TRUE

 

Rotates s by 4 positions from towards L

OPTION C
0 0 votes
Let array S is [1...k, k+1...n]

reverse(S,1,k): [k...1, k+1...n]

reverse(S, k+1, n]:  [k...1, n...k+1]

reverse(S,1,n): [k+1...n, 1...k]

It's original array left rotate by k positions.

 
0 0 votes
let array : [ 1 .... k, k+1 .... n ]

1st rev : [ k .... 1, k+1 .... n ]

2nd rev : [ k .... 1, n .... k+1 ]

3rd rev : [ k+1 .... n, 1 .... k ] => clearly k elements shifted to left...
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