Skip to main content

Stack

22. No of permutations with n numbers. (1/(n+1)2ncn).
23. A string with a’s and b’s and with one special character ‘x’. Find out whether its palindrome or not.
24. Implement queue by stack with enque and deque of order O(1);
25. Convert infix expression to postfix.
Labels:

Comments

  1. AshuMay 24, 2009 at 3:58 PM

    22. No of permutations with n numbers. (1/(n+1)2ncn).
    a) No of permutations are 1/(n+1) (2n)cn.
    b) No of stack permutation possible.
    c) No of binary trees possible with n nodes.
    d) No of ways to parenthesise with n data element.

    23. A string with a’s and b’s and with one special character ‘x’. Find out whether its palindrome or not.
    a) Take two pointer one from start and another from end. Start comparing till pointers don’t cross each other.

    24. Implement queue by stack with enque and deque of order O(1);
    Pop all the element and push in second stack2. The top2 will point to the beginning of
    Deque and top1 will be at end. Maintain size1 and size2 for the bottom removing with both the cases.
    Add at beg: push in stack2, size1 and size2 incremented.
    Add at end: push in stack1, size1 and size2 incremented.
    Delete at beg: pop in stack2 and decrement size1 and size2.
    Delete at end: pop in stack2 and decrement size1 and size2.

    25. Convert infix expression to postfix.
    While ( ch )
    {
    If ( ch is alphanumeric ) push (ch);
    If ( ch is ‘)’ ) pop from operatorstack and push(ch);
    If ( ch is operator) pushoperator(ch);
    }

    ReplyDelete

Post a Comment

Popular posts from this blog

[PUZZLE] ELEPHANT AND BANANA PUZZLE

A merchant has bough some 3000 banana from market and want to bring them to his home which is 1000 km far from Market. He have a elephant which can carry maximum of 1000/- banana at time. That could have been easy but this elephant is very shrewd and he charges 1 banana for every one kilometer he travel. Now we have to maximise number of banana which should reach to home. Solution: At present we are at 0th km with 3000 banana and elephant. Lets see if elephant have to shift all the 3000 banana and elephant by 1 km. For 1 km shift: Take first 1000 banana and drop at 1st km. 2 banana consumed. One banana each for to and fro. Second  1000 banana and drop at 1st km. 2 banana consumed. Third 1000 banana and reach at 1st km. 1 banana consumed. So all in all total 5 banana will be consumed if we shift a kilometer. But that's not all, our total banana number are also reducing so we may have to reduce the number of turns too. And this will happen once we have reduced the b
Post a Comment
Read more

[JAVA] Evil Null

Elvis operator could have been a good use here. But unfortunately have been decline till now. Its used to say "?." Operate only if not null. Class A {  public B getB() {    return new B(); } } public void test( A a) {     if ( a != null  )         return a.getB();  } Above method would be replaced with public void test ( A a) {      return a?.getB();  } Unfortunately its still some time we can see some think like that. So till that we have to live with two choices: 1. Check for null. 2. Use the Null Object Pattern. So we all know that how to check for null objects and believe me i had real long chain of checking null. Second way can be useful but if this chain is pretty long its can be tiresome to have null object for all the hierarchy Null Object means there will be definition of class A (mostly derived from same interface.) and will have dummy methods which will nullify any call to these methods.
Post a Comment
Read more

Two numbers with minimum difference

Find the two numbers whose difference is minimum among the set of numbers. For example the sequence is 5, 13, 7, 0, 10, 20, 1, 15, 4, 19 The algorithm should return min diff = 20-19 = 1. Make an efficient algorithm, yeah best could be O(n). Not sure if i could use DP here, will post a solution tommorow. Let me know if you find a answer. Algorithm:MiniMumDiffN Get the size of array , say n. Get the range of array, range. Bucket Sort the above array. Search the above sorted array for minimum difference. This will be the minimum difference numbers. Now question comes where the range of above given array is very big. Algorithm:DivideArray Take an random element. Search for its position in array. Return the index of element. Algorithm:MiniMumDiff if array_range > MAXRANGE int mid = divideArray (arr,start,end); d1,max1 = MiniMumDiff( arr,start,mid-1); d2,max2 = MiniMumDIff(arr,mid+1,end); d3 = difference max1 and arr[mid] d4 = difference max2 and arr[mid] return minimum( d1,d2
1 comment
Read more