[100%] Nesting (codility lessons)

Lesson7

Stacks and Queues

Open reading material (PDF)

Easy

Nesting

Determine whether a given string of parentheses (single type) is properly nested.

Task description

---

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.

For example, given S = "(()(())())", the function should return 1 and given S = "())", the function should return 0, as explained above.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [0..1,000,000];
• string S consists only of the characters "(" and/or ")".

---

Source code

class Solution {
    public int solution(String S) {
        java.util.Stack<String> stack = new java.util.Stack();
        for (char c : S.toCharArray()) {
            String currentFlag = String.valueOf(c);
            if ("(".equals(currentFlag)) {
                stack.push(currentFlag);
            } else {
                if (stack.isEmpty() || !"(".equals(stack.pop())) return 0;
            }
        }
        return stack.isEmpty() ? 1 : 0;
    }
}

Detected time complexity:

O(N)

Report

training5KXWDU-R3T

---

See also: CodilityのLessonsをすべて解く（更新中）