Longest Valid Parentheses
Created: November 30, 2019 by [lek-tin]
Last updated: November 30, 2019
Given a string containing just the characters '(' and ')', find the length of the longest valid (well-formed) parentheses substring.
Example 1
Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"
Example 2
Input: ")()())"
Output: 4
Explanation: The longest valid parentheses substring is "()()"
Solution
Dynamic programming
// Time: `O(n)`
// Space: `O(n)`
public class Solution {
public int longestValidParentheses(String s) {
int maxans = 0;
// dp array: ith element of dp represents the length of the longest valid substring ending at ith index.
int dp[] = new int[s.length()];
for (int i = 1; i < s.length(); i++) {
// substrings ending with '(' are not valid.
// So we only care about substrings ending with ')'
if (s.charAt(i) == ')') {
// string looks like ".......()"
if (s.charAt(i - 1) == '(') {
dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
// string looks like ".......))"
} else if (i - dp[i - 1] > 0 && s.charAt(i - dp[i - 1] - 1) == '(') {
dp[i] = dp[i - 1] + ((i - dp[i - 1]) >= 2 ? dp[i - dp[i - 1] - 2] : 0) + 2;
}
maxans = Math.max(maxans, dp[i]);
}
}
return maxans;
}
}
Stack
// Time: `O(n)`
// Space: `O(n)`
public class Solution {
public int longestValidParentheses(String s) {
int maxans = 0;
Stack<Integer> stack = new Stack<>();
stack.push(-1);
// ")()())"
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') {
stack.push(i);
} else {
stack.pop();
if (stack.empty()) {
stack.push(i);
} else {
// stack.peek(): last invalid position.
maxans = Math.max(maxans, i - stack.peek());
}
}
}
return maxans;
}
}
without extra space
// Time: `O(n)`
// Space: O(1)
public class Solution {
public int longestValidParentheses(String s) {
int left = 0, right = 0, maxlength = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') {
left++;
} else {
right++;
}
if (left == right) {
maxlength = Math.max(maxlength, 2 * right);
} else if (right > left) {
left = right = 0;
}
}
left = right = 0;
for (int i = s.length() - 1; i >= 0; i--) {
if (s.charAt(i) == '(') {
left++;
} else {
right++;
}
if (left == right) {
maxlength = Math.max(maxlength, 2 * left);
} else if (left > right) {
left = right = 0;
}
}
return maxlength;
}
}