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Copy pathLesson07(StacksAndQueues)-Nesting.cpp
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43 lines (38 loc) · 1.4 KB
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// 3. Nesting.
/**
* A string S consisting of N characters is called properly nested if:
* • S is empty;
* • S has the form "(U)" where U is a properly nested string;
* • S has the form "VW" where V and W are properly nested strings.
*
* For example, string "(()(())())" is properly nested but string "())" isn't.
*
* Write a function:
* class Solution { public int solution(String S); }
* that, given a string S consisting of N characters,
* returns 1 if string S is properly nested and 0 otherwise.
*
* Write an efficient algorithm for the following assumptions:
* • N is an integer within the range [0..1,000,000];
* • string S is made only of the characters '(' and/or ')'.
*/
#include <string>
#include <stack>
int nesting(std::string& S)
{
std::stack<char> brackets; // A stack to store opening brackets.
for (char c : S) {
// If the character is an opening bracket, push it onto the stack.
if (c == '(') {
brackets.push(c);
} else {
// If the character is a closing bracket, check if it matches the top of the stack.
if (brackets.empty() || (brackets.top() == '(' && c != ')')) {
return 0;
}
// Pop the top from the stack when a closing bracket is encountered.
brackets.pop();
}
}
return brackets.empty();
}