C Exercises: Check if a given string is valid or not
C Programming Practice: Exercise-10 with Solution
Write a C programming to check if a given string is valid or not, the string contains the characters '(', ')', '{', '}', '[' and ']'. The string is valid if the open brackets must be closed by the same type of brackets and in correct order.
C Code:
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
static bool is_valid_Parentheses(char *s)
{
int n = 0, cap = 100;
char *tstack = malloc(cap);
while (*s != '\0') {
switch(*s) {
case '(':
case '[':
case '{':
if (n + 1 >= cap) {
cap *= 2;
tstack = realloc(tstack, cap);
}
tstack[n++] = *s;
break;
case ')':
if (tstack[--n] != '(') return false;
break;
case ']':
if (tstack[--n] != '[') return false;
break;
case '}':
if (tstack[--n] != '{') return false;
break;
default:
return false;
}
s++;
}
return n == 0;
}
int main(void)
{
//char **vbracket = "()[]{}";
char **vbracket = "([)]";
printf("%s\n", is_valid_Parentheses(vbracket) ? "true" : "false");
return 0;
}
Sample Output:
false
Flowchart:
C Programming Code Editor:
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Previous: Write a C programming to find all unique quadruplets in a given array of integers whose sum equal to zero.
Next: Write a C program to generate all combinations of well-formed parentheses from n given pairs of parentheses.
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C Programming: Tips of the Day
Static variable inside of a function in C
The scope of variable is where the variable name can be seen. Here, x is visible only inside function foo().
The lifetime of a variable is the period over which it exists. If x were defined without the keyword static, the lifetime would be from the entry into foo() to the return from foo(); so it would be re-initialized to 5 on every call.
The keyword static acts to extend the lifetime of a variable to the lifetime of the programme; e.g. initialization occurs once and once only and then the variable retains its value - whatever it has come to be - over all future calls to foo().
Ref : https://bit.ly/3fOq7XP
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