codeparrot/xlcost-text-to-code huggingface dataset:
Check if two expressions with brackets are same | C # program to check if two expressions evaluate to same . ; Return local sign of the operand . For example , in the expr a - b - ( c ) , local signs of the operands are + a , - b , + c ; Evaluate expressions into the count vector of the 26 alphabets . If add is true , then add count to the count vector of the alphabets , else remove count from the count vector . ; stack stores the global sign for operands . ; + means true global sign is positive initially ; global sign for the bracket is pushed to the stack ; global sign is popped out which was pushed in for the last bracket ; global sign is positive ( we use different values in two calls of functions so that we finally check if all vector elements are 0. ; global sign is negative here ; Returns true if expr1 and expr2 represent same expressions ; Create a vector for all operands and initialize the vector as 0. ; Put signs of all operands in expr1 ; Subtract signs of operands in expr2 ; If expressions are same , vector must be 0. ; Driver Code
using System ; using System . Collections . Generic ; public class GFG { static readonly int MAX_CHAR = 26 ; static bool adjSign ( String s , int i ) { if ( i == 0 ) return true ; if ( s [ i - 1 ] == ' - ' ) return false ; return true ; } static void eval ( String s , int [ ] v , bool add ) { Stack < Boolean > stk = new Stack < Boolean > ( ) ; stk . Push ( true ) ; int i = 0 ; while ( i < s . Length ) { if ( s [ i ] == ' + ' s [ i ] == ' - ' ) { i ++ ; continue ; } if ( s [ i ] == ' ( ' ) { if ( adjSign ( s , i ) ) stk . Push ( stk . Peek ( ) ) ; else stk . Push ( ! stk . Peek ( ) ) ; } else if ( s [ i ] == ' ) ' ) stk . Pop ( ) ; else { if ( stk . Peek ( ) ) v [ s [ i ] - ' a ' ] += ( adjSign ( s , i ) ? add ? 1 : - 1 : add ? - 1 : 1 ) ; else v [ s [ i ] - ' a ' ] += ( adjSign ( s , i ) ? add ? - 1 : 1 : add ? 1 : - 1 ) ; } i ++ ; } } static bool areSame ( String expr1 , String expr2 ) { int [ ] v = new int [ MAX_CHAR ] ; eval ( expr1 , v , true ) ; eval ( expr2 , v , false ) ; for ( int i = 0 ; i < MAX_CHAR ; i ++ ) if ( v [ i ] != 0 ) return false ; return true ; } public static void Main ( String [ ] args ) { String expr1 = " - ( a + b + c ) " , expr2 = " - a - b - c " ; if ( areSame ( expr1 , expr2 ) ) Console . WriteLine ( " Yes " ) ; else Console . WriteLine ( " No " ) ; } }
Find index of closing bracket for a given opening bracket in an expression | C # program to find index of closing bracket for given opening bracket . ; Function to find index of closing bracket for given opening bracket . ; If index given is invalid and is not an opening bracket . ; Stack to store opening brackets . ; Traverse through string starting from given index . ; If current character is an opening bracket push it in stack . ; If current character is a closing bracket , pop from stack . If stack is empty , then this closing bracket is required bracket . ; If no matching closing bracket is found . ; Driver Code ; should be 8 ; should be 7 ; should be 12 ; No matching bracket
using System ; using System . Collections ; public class GFG { static void test ( String expression , int index ) { int i ; if ( expression [ index ] != ' [ ' ) { Console . Write ( expression + " , ▁ " + index + " : ▁ - 1 STRNEWLINE " ) ; return ; } Stack st = new Stack ( ) ; for ( i = index ; i < expression . Length ; i ++ ) { if ( expression [ i ] == ' [ ' ) { st . Push ( ( int ) expression [ i ] ) ; } else if ( expression [ i ] == ' ] ' ) { st . Pop ( ) ; if ( st . Count == 0 ) { Console . Write ( expression + " , ▁ " + index + " : ▁ " + i + " STRNEWLINE " ) ; return ; } } } Console . Write ( expression + " , ▁ " + index + " : ▁ - 1 STRNEWLINE " ) ; } public static void Main ( ) { test ( " [ ABC [ 23 ] ] [89 ] " , 0 ) ; test ( " [ ABC [ 23 ] ] [89 ] " , 4 ) ; test ( " [ ABC [ 23 ] ] [89 ] " , 9 ) ; test ( " [ ABC [ 23 ] ] [89 ] " , 1 ) ; } }
Convert a given Binary tree to a tree that holds Logical AND property | C # code to convert a given binary tree to a tree that holds logical AND property . ; Structure of binary tree ; function to create a new node ; Convert the given tree to a tree where each node is logical AND of its children The main idea is to do Postorder traversal ; first recur on left child ; then recur on right child ; first recur on left child ; then print the data of node ; now recur on right child ; Driver code ; Create following Binary Tree 1 / \ 1 0 / \ / \ 0 1 1 1
using System ; class GfG { class Node { public int data ; public Node left ; public Node right ; } static Node newNode ( int key ) { Node node = new Node ( ) ; node . data = key ; node . left = null ; node . right = null ; return node ; } static void convertTree ( Node root ) { if ( root == null ) return ; convertTree ( root . left ) ; convertTree ( root . right ) ; if ( root . left != null && root . right != null ) root . data = ( root . left . data ) & ( root . right . data ) ; } static void printInorder ( Node root ) { if ( root == null ) return ; printInorder ( root . left ) ; Console . Write ( root . data + " ▁ " ) ; printInorder ( root . right ) ; } public static void Main ( ) { Node root = newNode ( 0 ) ; root . left = newNode ( 1 ) ; root . right = newNode ( 0 ) ; root . left . left = newNode ( 0 ) ; root . left . right = newNode ( 1 ) ; root . right . left = newNode ( 1 ) ; root . right . right = newNode ( 1 ) ; Console . Write ( " Inorder ▁ traversal ▁ before ▁ conversion ▁ " ) ; printInorder ( root ) ; convertTree ( root ) ; Console . WriteLine ( ) ; Console . Write ( " Inorder ▁ traversal ▁ after ▁ conversion ▁ " ) ; printInorder ( root ) ; } }
Find if an expression has duplicate parenthesis or not | C # program to find duplicate parenthesis in a balanced expression ; Function to find duplicate parenthesis in a balanced expression ; create a stack of characters ; Iterate through the given expression ; if current character is close parenthesis ' ) ' ; pop character from the stack ; stores the number of characters between a closing and opening parenthesis if this count is less than or equal to 1 then the brackets are redundant else not ; push open parenthesis ' ( ' , operators and operands to stack ; No duplicates found ; Driver code ; input balanced expression
using System ; using System . Collections . Generic ; class GFG { static Boolean findDuplicateparenthesis ( String s ) { Stack < char > Stack = new Stack < char > ( ) ; char [ ] str = s . ToCharArray ( ) ; foreach ( char ch in str ) { if ( ch == ' ) ' ) { char top = Stack . Peek ( ) ; Stack . Pop ( ) ; int elementsInside = 0 ; while ( top != ' ( ' ) { elementsInside ++ ; top = Stack . Peek ( ) ; Stack . Pop ( ) ; } if ( elementsInside < 1 ) { return true ; } } else { Stack . Push ( ch ) ; } } return false ; } public static void Main ( String [ ] args ) { String str = " ( ( ( a + ( b ) ) + ( c + d ) ) ) " ; if ( findDuplicateparenthesis ( str ) ) { Console . WriteLine ( " Duplicate ▁ Found ▁ " ) ; } else { Console . WriteLine ( " No ▁ Duplicates ▁ Found ▁ " ) ; } } }
Find next Smaller of next Greater in an array | C # Program to find Right smaller element of next greater element ; function find Next greater element ; create empty stack ; Traverse all array elements in reverse order order == ' G ' we compute next greater elements of every element order == ' S ' we compute right smaller element of every element ; Keep removing top element from S while the top element is smaller then or equal to arr [ i ] ( if Key is G ) element is greater then or equal to arr [ i ] ( if order is S ) ; store the next greater element of current element ; If all elements in S were smaller than arr [ i ] ; Push this element ; Function to find Right smaller element of next greater element ; stores indexes of next greater elements ; stores indexes of right smaller elements ; Find next greater element Here G indicate next greater element ; Find right smaller element using same function nextGreater ( ) Here S indicate right smaller elements ; If NG [ i ] = = - 1 then there is no smaller element on right side . We can find Right smaller of next greater by arr [ RS [ NG [ i ] ] ] ; Driver code
using System ; using System . Collections . Generic ; public class GFG { public static void nextGreater ( int [ ] arr , int [ ] next , char order ) { Stack < int > stack = new Stack < int > ( ) ; for ( int i = arr . Length - 1 ; i >= 0 ; i -- ) { while ( stack . Count != 0 && ( ( order == ' G ' ) ? arr [ stack . Peek ( ) ] <= arr [ i ] : arr [ stack . Peek ( ) ] >= arr [ i ] ) ) stack . Pop ( ) ; if ( stack . Count != 0 ) next [ i ] = stack . Peek ( ) ; else next [ i ] = - 1 ; stack . Push ( i ) ; } } public static void nextSmallerOfNextGreater ( int [ ] arr ) { int [ ] NG = new int [ arr . Length ] ; int [ ] RS = new int [ arr . Length ] ; nextGreater ( arr , NG , ' G ' ) ; nextGreater ( arr , RS , ' S ' ) ; for ( int i = 0 ; i < arr . Length ; i ++ ) { if ( NG [ i ] != - 1 && RS [ NG [ i ] ] != - 1 ) Console . Write ( arr [ RS [ NG [ i ] ] ] + " ▁ " ) ; else Console . Write ( " - 1 ▁ " ) ; } } public static void Main ( ) { int [ ] arr = { 5 , 1 , 9 , 2 , 5 , 1 , 7 } ; nextSmallerOfNextGreater ( arr ) ; } }
Convert Ternary Expression to a Binary Tree | C # program to convert a ternary expreesion to a tree . ; Class to represent Tree node ; Class to convert a ternary expression to a Tree ; Function to convert Ternary Expression to a Binary Tree . It return the root of tree ; Base case ; store current character of expression_string [ ' a ' to ' z ' ] ; Move ahead in str ; if current character of ternary expression is ' ? ' then we add next character as a left child of current node ; else we have to add it as a right child of current node expression . at ( 0 ) == ' : ' ; function print tree ; Driver Code
using System ; public class Node { public char data ; public Node left , right ; public Node ( char item ) { data = item ; left = null ; right = null ; } } public class BinaryTree { public virtual Node convertExpression ( char [ ] expression , int i ) { if ( i >= expression . Length ) { return null ; } Node root = new Node ( expression [ i ] ) ; ++ i ; if ( i < expression . Length && expression [ i ] == ' ? ' ) { root . left = convertExpression ( expression , i + 1 ) ; } else if ( i < expression . Length ) { root . right = convertExpression ( expression , i + 1 ) ; } return root ; } public virtual void printTree ( Node root ) { if ( root == null ) { return ; } Console . Write ( root . data + " ▁ " ) ; printTree ( root . left ) ; printTree ( root . right ) ; } public static void Main ( string [ ] args ) { string exp = " a ? b ? c : d : e " ; BinaryTree tree = new BinaryTree ( ) ; char [ ] expression = exp . ToCharArray ( ) ; Node root = tree . convertExpression ( expression , 0 ) ; tree . printTree ( root ) ; } }
Count natural numbers whose all permutation are greater than that number | C # program to count the number less than N , whose all permutation is greater than or equal to the number . ; Return the count of the number having all permutation greater than or equal to the number . ; Pushing 1 to 9 because all number from 1 to 9 have this property . ; take a number from stack and add a digit smaller than last digit of it . ; Driver Code
using System ; using System . Collections . Generic ; class GFG { static int countNumber ( int n ) { int result = 0 ; Stack < int > s = new Stack < int > ( ) ; for ( int i = 1 ; i <= 9 ; i ++ ) { if ( i <= n ) { s . Push ( i ) ; result ++ ; } while ( s . Count != 0 ) { int tp = s . Peek ( ) ; s . Pop ( ) ; for ( int j = tp % 10 ; j <= 9 ; j ++ ) { int x = tp * 10 + j ; if ( x <= n ) { s . Push ( x ) ; result ++ ; } } } } return result ; } public static void Main ( String [ ] args ) { int n = 15 ; Console . WriteLine ( countNumber ( n ) ) ; } }
Delete consecutive same words in a sequence | C # program to remove consecutive same words ; Method to find the size of manipulated sequence ; Start traversing the sequence ; Compare the current string with next one Erase both if equal ; Erase function delete the element and also shifts other element that 's why   i is not updated  ; Update i , as to check from previous element again ; Reduce sequence size ; Increment i , if not equal ; Return modified size ; Driver Code
using System ; using System . Collections . Generic ; class GFG { public static int removeConsecutiveSame ( List < string > v ) { int n = v . Count ; for ( int i = 0 ; i < n - 1 ; ) { if ( v [ i ] . Equals ( v [ i + 1 ] ) ) { v . RemoveAt ( i ) ; v . RemoveAt ( i ) ; if ( i > 0 ) { i -- ; } n = n - 2 ; } else { i ++ ; } } return v . Count ; } public static void Main ( string [ ] args ) { List < string > v = new List < string > ( ) ; v . Add ( " tom " ) ; v . Add ( " jerry " ) ; v . Add ( " jerry " ) ; v . Add ( " tom " ) ; Console . WriteLine ( removeConsecutiveSame ( v ) ) ; } }
Delete consecutive same words in a sequence | C # implementation of above method ; Method to find the size of manipulated sequence ; Start traversing the sequence ; Push the current string if the stack is empty ; compare the current string with stack top if equal , pop the top ; Otherwise push the current string ; Return stack size ; Driver Code
using System ; using System . Collections . Generic ; class GFG { public static int removeConsecutiveSame ( List < string > v ) { Stack < string > st = new Stack < string > ( ) ; for ( int i = 0 ; i < v . Count ; i ++ ) { if ( st . Count == 0 ) { st . Push ( v [ i ] ) ; } else { string str = st . Peek ( ) ; if ( str . Equals ( v [ i ] ) ) { st . Pop ( ) ; } else { st . Push ( v [ i ] ) ; } } } return st . Count ; } public static void Main ( string [ ] args ) { List < string > v = new List < string > ( ) ; v . Add ( " ab " ) ; v . Add ( " aa " ) ; v . Add ( " aa " ) ; v . Add ( " bcd " ) ; v . Add ( " ab " ) ; Console . WriteLine ( removeConsecutiveSame ( v ) ) ; } }
Decode a string recursively encoded as count followed by substring | C # program to decode a string recursively encoded as count followed substring ; Returns decoded string for ' str ' ; Traversing the string ; If number , convert it into number and push it into integerstack . ; If closing bracket ' ] ' , pop elemment until ' [ ' opening bracket is not found in the character stack . ; Repeating the popped string ' temo ' count number of times . ; Push it in the character stack . ; If ' [ ' opening bracket , push it into character stack . ; Pop all the elmenet , make a string and return . ; Driver Code
using System ; using System . Collections . Generic ; class GFG { public static string decode ( string str ) { Stack < int > integerstack = new Stack < int > ( ) ; Stack < char > stringstack = new Stack < char > ( ) ; string temp = " " , result = " " ; for ( int i = 0 ; i < str . Length ; i ++ ) { int count = 0 ; if ( char . IsDigit ( str [ i ] ) ) { while ( char . IsDigit ( str [ i ] ) ) { count = count * 10 + str [ i ] - '0' ; i ++ ; } i -- ; integerstack . Push ( count ) ; } else if ( str [ i ] == ' ] ' ) { temp = " " ; count = 0 ; if ( integerstack . Count > 0 ) { count = integerstack . Peek ( ) ; integerstack . Pop ( ) ; } while ( stringstack . Count > 0 && stringstack . Peek ( ) != ' [ ' ) { temp = stringstack . Peek ( ) + temp ; stringstack . Pop ( ) ; } if ( stringstack . Count > 0 && stringstack . Peek ( ) == ' [ ' ) { stringstack . Pop ( ) ; } for ( int j = 0 ; j < count ; j ++ ) { result = result + temp ; } for ( int j = 0 ; j < result . Length ; j ++ ) { stringstack . Push ( result [ j ] ) ; } result = " " ; } else if ( str [ i ] == ' [ ' ) { if ( char . IsDigit ( str [ i - 1 ] ) ) { stringstack . Push ( str [ i ] ) ; } else { stringstack . Push ( str [ i ] ) ; integerstack . Push ( 1 ) ; } } else { stringstack . Push ( str [ i ] ) ; } } while ( stringstack . Count > 0 ) { result = stringstack . Peek ( ) + result ; stringstack . Pop ( ) ; } return result ; } public static void Main ( string [ ] args ) { string str = "3 [ b2 [ ca ] ] " ; Console . WriteLine ( decode ( str ) ) ; } }
Iterative method to find ancestors of a given binary tree | C # program to print all ancestors of a given key ; Structure for a tree node ; A utility function to create a new tree node ; Iterative Function to print all ancestors of a given key ; Create a stack to hold ancestors ; Traverse the complete tree in postorder way till we find the key ; Traverse the left side . While traversing , push the nodes into the stack so that their right subtrees can be traversed later ; push current node ; move to next node ; If the node whose ancestors are to be printed is found , then break the while loop . ; Check if right sub - tree exists for the node at top If not then pop that node because we don 't   need this node any more.  ; If the popped node is right child of top , then remove the top as well . Left child of the top must have processed before . ; if stack is not empty then simply set the root as right child of top and start traversing right sub - tree . ; If stack is not empty , print contents of stack Here assumption is that the key is there in tree ; Driver code ; Let us construct a binary tree
using System ; using System . Collections . Generic ; class GfG { public class Node { public int data ; public Node left , right ; } static Node newNode ( int data ) { Node node = new Node ( ) ; node . data = data ; node . left = null ; node . right = null ; return node ; } static void printAncestors ( Node root , int key ) { if ( root == null ) return ; Stack < Node > st = new Stack < Node > ( ) ; while ( 1 == 1 ) { while ( root != null && root . data != key ) { st . Push ( root ) ; root = root . left ; } if ( root != null && root . data == key ) break ; if ( st . Peek ( ) . right == null ) { root = st . Peek ( ) ; st . Pop ( ) ; while ( st . Count != 0 && st . Peek ( ) . right == root ) { root = st . Peek ( ) ; st . Pop ( ) ; } } root = st . Count == 0 ? null : st . Peek ( ) . right ; } while ( st . Count != 0 ) { Console . Write ( st . Peek ( ) . data + " ▁ " ) ; st . Pop ( ) ; } } public static void Main ( String [ ] args ) { Node root = newNode ( 1 ) ; root . left = newNode ( 2 ) ; root . right = newNode ( 7 ) ; root . left . left = newNode ( 3 ) ; root . left . right = newNode ( 5 ) ; root . right . left = newNode ( 8 ) ; root . right . right = newNode ( 9 ) ; root . left . left . left = newNode ( 4 ) ; root . left . right . right = newNode ( 6 ) ; root . right . right . left = newNode ( 10 ) ; int key = 6 ; printAncestors ( root , key ) ; } }
Tracking current Maximum Element in a Stack | C # program to keep track of maximum element in a stack ; main stack ; tack to keep track of max element ; If current element is greater than the top element of track stack , push the current element to track stack otherwise push the element at top of track stack again into it . ; Driver code
using System ; using System . Collections . Generic ; class GfG { public class StackWithMax { static Stack < int > mainStack = new Stack < int > ( ) ; static Stack < int > trackStack = new Stack < int > ( ) ; public void push ( int x ) { mainStack . Push ( x ) ; if ( mainStack . Count == 1 ) { trackStack . Push ( x ) ; return ; } if ( x > trackStack . Peek ( ) ) trackStack . Push ( x ) ; else trackStack . Push ( trackStack . Peek ( ) ) ; } public int getMax ( ) { return trackStack . Peek ( ) ; } public void pop ( ) { mainStack . Pop ( ) ; trackStack . Pop ( ) ; } } ; public static void Main ( ) { StackWithMax s = new StackWithMax ( ) ; s . push ( 20 ) ; Console . WriteLine ( s . getMax ( ) ) ; s . push ( 10 ) ; Console . WriteLine ( s . getMax ( ) ) ; s . push ( 50 ) ; Console . WriteLine ( s . getMax ( ) ) ; } }
Reverse a number using stack | C # program to reverse the number using a stack ; Stack to maintain order of digits ; Function to push digits into stack ; Function to reverse the number ; Function call to push number 's   digits to stack  ; Popping the digits and forming the reversed number ; Return the reversed number formed ; Driver Code ; Function call to reverse number
using System ; using System . Collections . Generic ; class GFG { public static Stack < int > st = new Stack < int > ( ) ; public static void push_digits ( int number ) { while ( number != 0 ) { st . Push ( number % 10 ) ; number = number / 10 ; } } public static int reverse_number ( int number ) { push_digits ( number ) ; int reverse = 0 ; int i = 1 ; while ( st . Count > 0 ) { reverse = reverse + ( st . Peek ( ) * i ) ; st . Pop ( ) ; i = i * 10 ; } return reverse ; } public static void Main ( string [ ] args ) { int number = 39997 ; Console . WriteLine ( reverse_number ( number ) ) ; } }
Flip Binary Tree | C # program to flip a binary tree ; A binary tree node structure ; method to flip the binary tree ; recursively call the same method ; rearranging main root Node after returning from recursive call ; Iterative method to do level order traversal line by line ; Base Case ; Create an empty queue for level order traversal ; Enqueue Root and initialize height ; nodeCount ( queue size ) indicates number of nodes at current lelvel . ; Dequeue all nodes of current level and Enqueue all nodes of next level ; Driver code
using System ; using System . Collections . Generic ; public class FlipTree { public class Node { public int data ; public Node left , right ; public Node ( int data ) { this . data = data ; } } ; public static Node flipBinaryTree ( Node root ) { if ( root == null ) return root ; if ( root . left == null && root . right == null ) return root ; Node flippedRoot = flipBinaryTree ( root . left ) ; root . left . left = root . right ; root . left . right = root ; root . left = root . right = null ; return flippedRoot ; } public static void printLevelOrder ( Node root ) { if ( root == null ) return ; Queue < Node > q = new Queue < Node > ( ) ; q . Enqueue ( root ) ; while ( true ) { int nodeCount = q . Count ; if ( nodeCount == 0 ) break ; while ( nodeCount > 0 ) { Node node = q . Dequeue ( ) ; Console . Write ( node . data + " ▁ " ) ; if ( node . left != null ) q . Enqueue ( node . left ) ; if ( node . right != null ) q . Enqueue ( node . right ) ; nodeCount -- ; } Console . WriteLine ( ) ; } } public static void Main ( String [ ] args ) { Node root = new Node ( 1 ) ; root . left = new Node ( 2 ) ; root . right = new Node ( 1 ) ; root . right . left = new Node ( 4 ) ; root . right . right = new Node ( 5 ) ; Console . WriteLine ( " Level ▁ order ▁ traversal ▁ of ▁ given ▁ tree " ) ; printLevelOrder ( root ) ; root = flipBinaryTree ( root ) ; Console . WriteLine ( " Level ▁ order ▁ traversal ▁ of ▁ flipped ▁ tree " ) ; printLevelOrder ( root ) ; } }
Remove brackets from an algebraic string containing + and | C # program to simplify algebraic string ; Function to simplify the string ; resultant string of max length equal to length of input string ; create empty stack ; If top is 1 , flip the operator ; If top is 0 , append the same operator ; x is opposite to the top of stack ; push value equal to top of the stack ; If closing parentheses pop the stack once ; copy the character to the result ; Driver code
using System ; using System . Collections . Generic ; class GfG { static String simplify ( String str ) { int len = str . Length ; char [ ] res = new char [ len ] ; int index = 0 , i = 0 ; Stack < int > s = new Stack < int > ( ) ; s . Push ( 0 ) ; while ( i < len ) { if ( str [ i ] == ' + ' ) { if ( s . Peek ( ) == 1 ) res [ index ++ ] = ' - ' ; if ( s . Peek ( ) == 0 ) res [ index ++ ] = ' + ' ; } else if ( str [ i ] == ' - ' ) { if ( s . Peek ( ) == 1 ) res [ index ++ ] = ' + ' ; else if ( s . Peek ( ) == 0 ) res [ index ++ ] = ' - ' ; } else if ( str [ i ] == ' ( ' && i > 0 ) { if ( str [ i - 1 ] == ' - ' ) { int x = ( s . Peek ( ) == 1 ) ? 0 : 1 ; s . Push ( x ) ; } else if ( str [ i - 1 ] == ' + ' ) s . Push ( s . Peek ( ) ) ; } else if ( str [ i ] == ' ) ' ) s . Pop ( ) ; else res [ index ++ ] = str [ i ] ; i ++ ; } return new String ( res ) ; } public static void Main ( String [ ] args ) { String s1 = " a - ( b + c ) " ; String s2 = " a - ( b - c - ( d + e ) ) - f " ; Console . WriteLine ( simplify ( s1 ) ) ; Console . WriteLine ( simplify ( s2 ) ) ; } }