Problem Statement: Find the value of a given polynomial P(x) = a₀ + a₁x + a₂x² + a₃x³… a_n-1x^n-1 for a given value of x

Solution: Horner’s rule can be used to solve this problem in O(n) time where n is the maximum power of variable x in the polynomial.

this method works by evaluating the polynomial in a different form which requires reduced number of operations.

P(x) = a₀ + a₁x + a₂x² + a₃x³… a_n-1x^n-1

P(x) = a₀ + x(a₁ + x(a₂ + x(a₃+… +(a_n-2+ xa_n-1)  … )))

The input to this algorithm is an array of co-efficient  in the polynomial, i.e. coeff[i] = a_i, and a number c which is the given value of x at which we have to calculate P(x)

The algorithm maintains a loop invariant which we state below.

Loop Invariant: at the start of the loop the variable result contains a partially calculated polynomial Q_i(x)

Q_i(x) = a_i+1 + a_i+2x + a_i+3x² ….. a_n-1x^n-2-i

Intialization: i=n-1

Q_n-1(x) = 0

Maintenance:

Q_n-2(x) = a_n-1

Q_n-3(x) = a_n-2 + x*a_n-1

Q_i (x) = a_i+1 + a_i+2x + a_i+3x² + …. a_n-1x^n-2-i

Q_i-1(x) = a_i + x*Q_i(x)

= a_i + a_i+1x + a_i+2x² + a_i+3x³ + … a_n-1x^n-1-i

= Q_i-1(x)

Termination:

Q_-1(x) = a₀ + a₁x + a₂x² + …. a_n-1x^{n-2 +1}

Q_-1 (x) = a₀ + a₁x + a₂x² + …. a_n-1x^n-1

which is equivalent to P(c) if we substitute  c for x.

Thus this algorithm correctly calculates

Solution: We use a recursive variant of insertion sort to solve this problem. the main sort function recursively sorts first n-1 elements and inserts nth element in its proper location using insert routine. Running time of this algorithm can be expressed as T(n) = T(n-1) + O(n) OR, T(n) = O(n²)

Solution: Given n pairs of parentheses this solution uses a simple logic that  left parentheses can be place in the string as long number of left parentheses used so far is less than n, and a right parentheses can be place in the string as long as number of right parentheses used so far is less than the number of left parentheses used so far.

Solution: In order to under solution to this problem lets take an example. Assume that we are given three pairs of parentheses. A valid permutation of these parentheses will always have left number of left parentheses more than or equal to number of right p arentheses. Lets see an example

Let the pair [l,r] represent number of left parentheses and number of right parentheses encountered in the permutation so far.

a. Valid Parentheses ,

( [1,0]      ( [2,0]      ) [2,1]     ) [2,2]      ( [3,2]      ) [3,3]

b. Invalid Parentheses

([1,0]   )[1,1]   )[1,2  invalid    ]   (    (    )

Above examples demonstrate that for a permutation to be valid number of left parentheses in the permutation at any position should be more than or equal to number right parentheses

Now lets try to find out all possible permutation for 3 pairs of parentheses.  It can be noted that all permutations will have six characters , and any character can be either ‘(‘ left parentheses or ‘)’ right parentheses as long as it does not violate above property, let us represent left parentheses ‘(‘ by 0 and right parentheses ‘)’ by 1. So that a string of zeros and ones represent a permutation of parentheses

                                    

As you can see in the above tree that the tree is binary and that tree stops branching as long as number of left parentheses becomes 3 which is equal to the number of pairs of parenthesis. Thus number of valid parentheses is equal to number of node which have l ( number of left parentheses)  equal to n (number of pairs of parentheses). and for all other cases we have two choices we increment l and if r < l we can increment r also where r is number of right parentheses. The presented algorithm does exactly that

Solution:  Modified Preorder Traversal (MPTT) is an implementation of Nested Set Model. In this implementaion tree is traversed in preoder traveresal, i.e. root first then , its left child and finally its right child. In MPTT each node is visited twice and two numbers associated with each node. If first number is n then second number is [n+(2*no_children)+1].  Thus while numbering a node N, with set the first number to x, then we visit all its children from left to right assigning , first and second numbers to each and keep incrementing the counter in the process, then we visi the node N again and assign the current value of counter as second number. In the given algorithm. We first start with number1 with node root and assign root.start=1, then we recursively number its left children and the its right children , then we come back and set root.second to current value of the counter.

Solution: Solution is very simple. The only trick is to first test for 15 then for 5 and 3 because 15 is also a multiple of 3 and 5 and therefore all multiples of 15 are also multiple of 3 and/or 5.

This algorithm will generate all permutations of a string in lexicographic using an iterative algorithm. The input string to this algorithm must have all the characters in lexicographic order. For example, if we want to generate permutations of “bcad” string. Then input string should be “abcd” because a<b<c<d. This algorithm cannot handle duplicate characters so all characters must be unique.

The key routine of this algorithm is the routine “next”. Given a permutation “next” routine will find another permutation that comes next in lexicographic order.  This algorithm uses “next” to generate next permutation iteratively until all the characters in the string are in reverse lexicographic order. For example give “abcd” string , it will generate all permutaion until we reach “dcba” since this is the last permutation in lexicographic order. Now the working of “next” routine  is explained below with an example.

consider a permutaion of “abcdefghij”      p1  =  ”chdafijgeb”

now lets find out next permutation in lexicographic order.

Step1. Starting with last char i.e. ‘b’ , decrease  the counter as long as characters are in increasing order and find first character that breaks this trend. In this example we started ‘b’, thus ‘e’>’b',  ’g'>’e’  ,  ’j'>’e',  but ‘i’<’j’ , thus the character that break this trend is ‘i’.

Step2 . Find the largest element starting from last element i.e. ‘b’ such that it is smaller than the element found in last step i.e. ‘i’. ‘b’<’i', ‘e’<’i',’g'<’i',’j’ > ‘i’ , therefore the character is ‘g’;

Step 3  Now swap ‘g’ and ‘i’; after this step we get    p1′  =  ”chdafgjieb”;

Step 4 now reverse the string starting after ‘g’ to the end. which gives us p1”  = “chdafgbeij”, this gives us smallest string that is larger than the input string and thus next in the lexicographic order

2. Rotate part
as noted above in order to generate permutations we place all characters one by at position 0. The first case where “d” is moved to first place by rotating the string such that every character’s index is increased by one and last character i.e “d” is moved to first. At second step characters are moved by one place which bring “c” in first place. This is done four times since the length of string is four.

This solution does NOT generate permutations in lexicographic order and it does NOT take care of repeated characters.