Check for balanced parentheses in an expression

Given an expression string exp, write a program to examine whether the pairs and the orders of “{“,”}”,”(“,”)”,”[","]” are correct in exp. For example, the program should print true for exp = “[()]{}{[()()]()}” and false for exp = “[(])”

Algorithm:
1) Declare a character stack S.
2) Now traverse the expression string exp.
a) If the current character is a starting bracket (‘(‘ or ‘{‘ or ‘[') then push it to stack.
b) If the current character is a closing bracket (')' or '}' or ']‘) then pop from stack and if the popped character is the matching starting bracket then fine else parenthesis are not balanced.
3) After complete traversal, if there is some starting bracket left in stack then “not balanced”


#include
#include
#define bool int

/* structure of a stack node */
struct sNode
{
char data;
struct sNode *next;
};

/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);

/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);

/* Returns 1 if character1 and character2 are matching left
and right Parenthesis */
bool isMatchingPair(char character1, char character2)
{
if(character1 == '(' && character2 == ')')
return 1;
else if(character1 == '{' && character2 == '}')
return 1;
else if(character1 == '[' && character2 == ']')
return 1;
else
return 0;
}

/*Return 1 if expression has balanced Parenthesis */
bool areParenthesisBalanced(char exp[])
{
int i = 0;

/* Declare an empty character stack */
struct sNode *stack = NULL;

/* Traverse the given expression to check matching parenthesis */
while(exp[i])
{
/*If the exp[i] is a starting parenthesis then push it*/
if(exp[i] == '{' || exp[i] == '(' || exp[i] == '[')
push(&stack, exp[i]);

/* If exp[i] is a ending parenthesis then pop from stack and
check if the popped parenthesis is a matching pair*/
if(exp[i] == '}' || exp[i] == ')' || exp[i] == ']')
{

/*If we see an ending parenthesis without a pair then return false*/
if(stack == NULL)
return 0;

/* Pop the top element from stack, if it is not a pair
parenthesis of character then there is a mismatch.
This happens for expressions like {(}) */
else if ( !isMatchingPair(pop(&stack), exp[i]) )
return 0;
}
i++;
}

/* If there is something left in expression then there is a starting
parenthesis without a closing parenthesis */
if(stack == NULL)
return 1; /*balanced*/
else
return 0; /*not balanced*/
}

/* UTILITY FUNCTIONS */
/*driver program to test above functions*/
int main()
{
char exp[100] = "{()}[]";
if(areParenthesisBalanced(exp))
printf("\n Balanced ");
else
printf("\n Not Balanced "); \
getchar();
}

/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));

if(new_node == NULL)
{
printf("Stack overflow \n");
getchar();
exit(0);
}

/* put in the data */
new_node->data = new_data;

/* link the old list off the new node */
new_node->next = (*top_ref);

/* move the head to point to the new node */
(*top_ref) = new_node;
}

/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
char res;
struct sNode *top;

/*If stack is empty then error */
if(*top_ref == NULL)
{
printf("Stack overflow \n");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}

Source Geeks4Geeks

WAP to Find Longest Comman Subsequence

The longest common subsequence (or LCS) of groups A and B is the longest group of elements from A and B that are common between the two groups and in the same order in each group. For example, the sequences "1234" and "1224533324" have an LCS of "1234":

1234
1224533324

For a string example, consider the sequences "thisisatest" and "testing123testing". An LCS would be "tsitest":

thisisatest
testing123testing

In this puzzle, your code only needs to deal with strings. Write a function which returns an LCS of two strings (case-sensitive). You don't need to show multiple LCS's.

More Info Wiki

C Code of LCS

#include
#include
#include

#define MAX(A,B) (((A)>(B))? (A) : (B))

char * lcs(const char *a,const char * b) {
int lena = strlen(a)+1;
int lenb = strlen(b)+1;

int bufrlen = 40;
char bufr[40], *result;

int i,j;
const char *x, *y;
int *la = calloc(lena*lenb, sizeof( int));
int **lengths = malloc( lena*sizeof( int*));
for (i=0; i
for (i=0,x=a; *x; i++, x++) {
for (j=0,y=b; *y; j++,y++ ) {
if (*x == *y) {
lengths[i+1][j+1] = lengths[i][j] +1;
}
else {
int ml = MAX(lengths[i+1][j], lengths[i][j+1]);
lengths[i+1][j+1] = ml;
}
}
}

result = bufr+bufrlen;
*--result = '\0';
i = lena-1; j = lenb-1;
while ( (i>0) && (j>0) ) {
if (lengths[i][j] == lengths[i-1][j]) i -= 1;
else if (lengths[i][j] == lengths[i][j-1]) j-= 1;
else {
// assert( a[i-1] == b[j-1]);
*--result = a[i-1];
i-=1; j-=1;
}
}
free(la); free(lengths);
return strdup(result);
}

int main()
{
printf("%s\n", lcs("thisisatest", "testing123testing")); // tsitest
return 0;
}

Run Here https://ideone.com/tCMa0

Java Code of LCS

class LCS
{

public static String lcs_dp(String a, String b) {
int[][] lengths = new int[a.length()+1][b.length()+1];

// row 0 and column 0 are initialized to 0 already

for (int i = 0; i < a.length(); i++)
for (int j = 0; j < b.length(); j++)
if (a.charAt(i) == b.charAt(j))
lengths[i+1][j+1] = lengths[i][j] + 1;
else
lengths[i+1][j+1] =
Math.max(lengths[i+1][j], lengths[i][j+1]);

// read the substring out from the matrix
StringBuffer sb = new StringBuffer();
for (int x = a.length(), y = b.length();
x != 0 && y != 0; ) {
if (lengths[x][y] == lengths[x-1][y])
x--;
else if (lengths[x][y] == lengths[x][y-1])
y--;
else {
assert a.charAt(x-1) == b.charAt(y-1);
sb.append(a.charAt(x-1));
x--;
y--;
}
}

return sb.reverse().toString();
}


public static String lcs_rec(String a, String b){
int aLen = a.length();
int bLen = b.length();
if(aLen == 0 || bLen == 0){
return "";
}else if(a.charAt(aLen-1) == b.charAt(bLen-1)){
return lcs_rec(a.substring(0,aLen-1),b.substring(0,bLen-1))
+ a.charAt(aLen-1);
}else{
String x = lcs_rec(a, b.substring(0,bLen-1));
String y = lcs_rec(a.substring(0,aLen-1), b);
return (x.length() > y.length()) ? x : y;
}
}

public static void main(String a[])
{
System.out.println(lcs_rec("ABCDEFG", "FACHDGB"));
System.out.println(lcs_dp("ABCDEFG", "FACHDGB"));

}

}

Source http://www.ics.uci.edu/~eppstein/161/960229.html

Run Here https://ideone.com/lsc8b

WAP to Print All Possible Combination of Balanced Parenthesis

# include
# define MAX_SIZE 100

void _printParenthesis(int pos, int n, int open, int close);

/* Wrapper over _printParenthesis()*/
void printParenthesis(int n)
{
if(n > 0)
_printParenthesis(0, n, 0, 0);
return;
}

void _printParenthesis(int pos, int n, int open, int close)
{
static char str[MAX_SIZE];

if(close == n)
{
printf("%s \n", str);
return;
}
else
{
if(open > close) {
str[pos] = '}';
_printParenthesis(pos+1, n, open, close+1);
}
if(open < n) {
str[pos] = '{';
_printParenthesis(pos+1, n, open+1, close);
}
}
}

/* driver program to test above functions */
int main()
{
int n = 3;
printParenthesis(n);
getchar();
return 0;
}

WAP to Find Next Greater Palindrom Then the Given Number

#include

int ten(int s)
{
int i=1,product=1;
for(i=1;i<=s;i++)
product=product*10;
return product;
}
int main()
{
int n=0,num=0,b=0,c=0,d=0,i=1,input=99999,upper=0,lower=0,output=0;

num=input;
while(num!=0)
{
n++;
num/=10;
}
num=input;
printf("\n n=%d",n);
lower=num%ten(n/2);
printf("\nlower=%d",lower); //34 45
c=num/ten(n/2); //12 123
d=c;
if(n%2!=0)//if not even digits
d=c/10; // 12 12
printf("\n%d%d",c,d);
while(d!=0)
{
upper=upper*10+(d%10);
d=d/10;
}
printf("\nupper=%d",upper);//21 21
if(upper>lower)
{
output=c*ten(n/2)+upper;
}
else
{
c=c+1; //124
d=c;
upper=0;
if(n%2!=0)
d=d/10; //12
while(d!=0)
{
upper=upper*10+(d%10);
printf("\nd=%d",d);
d=d/10;
}
output=c*ten(n/2)+upper;
}
printf("\noutput=%d",output); //1331 12421

}

WAP to Removing characters from an input string in place.

#include
#include

WAP To Implement The Bitonic Sorting Can Be extended to K-tonic Sorting

public class BitonicSorterForArbitraryN
{
private static int[] a;
private final static boolean ASCENDING=true; // sorting direction

public static void sort(int[] ar)
{
a=ar;
bitonicSort(0, a.length, ASCENDING);
}

private static void bitonicSort(int lo, int n, boolean dir)
{
if (n>1)
{
int m=n/2;
bitonicSort(lo, m, !dir);
bitonicSort(lo+m, n-m, dir);
bitonicMerge(lo, n, dir);
}
}

private static void bitonicMerge(int lo, int n, boolean dir)
{
if (n>1)
{
int m=greatestPowerOfTwoLessThan(n);
for (int i=lo; i compare(i, i+m, dir);
bitonicMerge(lo, m, dir);
bitonicMerge(lo+m, n-m, dir);
}
}

private static void compare(int i, int j, boolean dir)
{
if (dir==(a[i]>a[j]))
exchange(i, j);
}

private static void exchange(int i, int j)
{
int t=a[i];
a[i]=a[j];
a[j]=t;
}

private static int greatestPowerOfTwoLessThan(int n)
{
int k=1;
while (k k=k<<1;
return k>>1;
}

public static void main(String ar[])
{
int a[]={1,2,3,4,5,9,8,7,6};
sort(a);
for(int i=0;i<9;i++)
System.out.println(a[i]);
}

}

The new sorting network for arbitrary n can be embedded into an original bitonic sorting network for 2k. Therefore, it also consists of ceiling 1log(n)ceiling 2 · (ceiling 1log(n)ceiling 2 + 1) / 2 stages. Each stage consists of at most n/2 comparators. This results in a complexity of O(n log(n)2) comparators, just as the original bitonic sorting network.


Time Complexity O(n(logn)^2))
Space O(1)

WAP remove(Delete Node From Binary Tree

WAP to SumTwo Number Which are represented by node in linked list

#include<stdio.h>
#include<malloc.h>
 
/* Link list node */
struct node
{
int data;
struct node* next;
};
 
/* Function to reverse the linked list */
static void reverse(struct node** head_ref)
{
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*head_ref = prev;
}
 
/* Function to push a node */
void push(struct node** head_ref, int new_data)
{
/* allocate node */
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
 
/* put in the data */
new_node->data = new_data;
 
/* link the old list off the new node */
new_node->next = (*head_ref); 
 
/* move the head to point to the new node */
(*head_ref) = new_node;
}
 
 
 
struct node* addition (struct node* temp1, struct node* temp2)
{
 
struct node* prev = NULL;
int carry = 0,a,b,result;
 
while (temp1 || temp2) //while both lists exist
{
 
if(temp1) a=temp1->data;
else a=0;
 
if(temp2) b=temp2->data;
else b=0;
 
 
 
 
result = carry;
if(temp1)
result+=temp1->data;
if(temp2)
result+=temp2->data;
 
 
carry=result/10;
 
struct node * newnode = (struct node*) malloc (sizeof(struct node));
newnode->data=(result)%10;
newnode->next = prev;
 
prev=newnode;
if(temp1)
temp1=temp1->next;
if(temp2)
temp2=temp2->next;
}
return prev;
 
}
void printList(struct node *node)
{
while(node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
} 
 
/* Drier program to test above function*/
int main(void)
{
/* Start with the empty list */
struct node* head = NULL;
struct node* head1 = NULL;
struct node* head2 = NULL;
 
 
/* Created Linked list is 1->2->3->4->5->6->7->8 */
 
push(&head1, 4);
push(&head1, 3);
push(&head1, 2);
push(&head1, 1);
 
reverse(&head1); 
 
push(&head2, 6);
push(&head2, 5);
push(&head2, 4);
 
reverse(&head2); 
 
head=addition(head1,head2); 
 
 
 
printList(head);
 
return(0);
}

TC O(n+m) where n,m are the number of digits in given numbers.
SC O(1)
Run Here https://ideone.com/j4Z8Z

WAP Find Minimu Number of Jumps to Reach In The End , In An UnSorted Array

#include
#include

int main()
{

int arr[] = {1 ,3, 5, 8 ,9 ,2, 6,7, 6, 8, 9};//{1, 3, 6, 0, 0, 3, 2, 3, 6, 8, 9};
int size=11;

int step=0,jump=0;
int i=0,j=0,max=0;

while( i< size)
{
jump++;
max=0;
step=0;

/*computes max distance it can cover in this jump*/
for(j=1;j<=arr[i];j++)
{
if(arr[i+j]+j>max)
{
max=arr[i+j]+j;
step=j;
}
}
i=i+step;
}

printf ( " %d ",jump);



return 0;
}

Run Here https://ideone.com/bvQph
Still DP is Best Will Post Soln Soon Using DP

WAP Find Minimu Number of Jumps to Reach In The End , In An UnSorted Array

#include
#include

int main()
{

int arr[] = {1 ,3, 5, 8 ,9 ,2, 6,7, 6, 8, 9};//{1, 3, 6, 0, 0, 3, 2, 3, 6, 8, 9};
int size=11;

int step=0,jump=0;
int i=0,j=0,max=0;

while( i< size)
{
jump++;
max=0;
step=0;

/*computes max distance it can cover in this jump*/
for(j=1;j<=arr[i];j++)
{
if(arr[i+j]+j>max)
{
max=arr[i+j]+j;
step=j;
}
}
i=i+step;
}

printf ( " %d ",jump);



return 0;
}