Java solution to Codility Nesting problem (Lesson 7 – Stacks and Queues) which scored 100%. The problem is to determine whether a given string of single type of parentheses is properly nested.
The strategy (very similar to the Brackets problem) is to store opening brackets on a stack and pop elements off stack when encounter closing bracket and check that it’s a proper open bracket-close bracket match.
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import java.util.Stack;
public class Nesting {
public int solution(String S) {
if(S.length() % 2 != 0) return 0; //all odd numbered strings can't be properly nested
Stack charStack = new Stack();
for(int i=0; i<S.length(); i++) {
char currentChar = S.charAt(i);
if(isOpeningBracket(currentChar)) {
charStack.push(currentChar);
}
else {
//adding closing bracket to empty stack will never become properly nested
if(charStack.size() == 0) {
return 0;
}
char poppedChar = charStack.pop();
if(isBracketMatch(poppedChar, currentChar)) {
continue;
}
else return 0;
}
}
if(charStack.isEmpty()) return 1;
return 0;
}
boolean isOpeningBracket(char pC) {
if(pC == '(') {
return true;
}
return false;
}
boolean isBracketMatch(char pC1, char pC2) {
if(pC1 == '(' && pC2 == ')') return true;
return false;
}
}
TestNG test cases for this problem which all passed:
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import org.testng.Assert;
import org.testng.annotations.BeforeTest;
import org.testng.annotations.DataProvider;
import org.testng.annotations.Test;
import com.codility.lesson07.stacksqueues.Nesting;
public class NestingTests {
private Nesting solution;
@BeforeTest
public void setUp() {
solution = new Nesting();
}
@DataProvider(name = "test1")
public Object [][] createData1() {
return new Object [][] {
new Object [] { "()()", 1 },
new Object [] { "((()(()))())", 1 },
new Object [] { "()", 1 },
new Object [] { "((()))", 1 },
new Object [] { "((()())())", 1 },
new Object [] { ")(", 0 },
new Object [] { "", 1 },
new Object [] { "((((", 0 },
new Object [] { "))", 0 },
};
}
@Test(dataProvider = "test1")
public void verifySolution(String pS, int pExpected) {
Assert.assertEquals(solution.solution(pS), pExpected);
}
}