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https://github.com/TheAlgorithms/C
synced 2024-11-22 13:31:21 +03:00
Update balanced parenthesis using stack in C
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@ -8,15 +8,15 @@ struct node
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char data;
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struct node* link;
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};
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int c=0;
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struct node * head;
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int c=0; // c used as counter to check if stack is empty or not
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struct node * head; //declaring head pointer globally assigned to NULL
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void push(char x)
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void push(char x) //function for pushing
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{
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struct node*p,*temp;
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struct node *p,*temp;
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temp=(struct node*)malloc(sizeof(struct node));
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temp->data=x;
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if(head==NULL)
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if(head==NULL) //will be execute only one time i.e, 1st time push is called
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{
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head=temp;
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p=head;
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@ -32,7 +32,7 @@ struct node*p,*temp;
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}
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}
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char pop(void)
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char pop(void) //function for pop
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{
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char x;
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struct node*p=head;
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@ -44,15 +44,15 @@ char pop(void)
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}
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int isBalanced(char *s) { //{[()]}
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int isBalanced(char *s) {
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int i=0;char x;
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while(s[i]!='\0')
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while(s[i]!='\0') //loop for covering entire string of brackets
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{
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if(s[i]=='{'||s[i]=='('||s[i]=='[')
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if(s[i]=='{'||s[i]=='('||s[i]=='[') //if opening bracket then push
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push(s[i]);
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else
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{
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if(c<=0)
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if(c<=0) //i.e, stack is empty as only opening brackets are added to stack
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return 0;
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@ -65,7 +65,7 @@ int isBalanced(char *s) { //{[()]}
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return 0 ;
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}i++;
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}
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if(c==0)
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if(c==0) //at end if stack is empy which means whole process has been performed correctly so retuen 1
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return 1;
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else
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return 0;
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