32. Longest Valid Parentheses

LeetCode
, ,

Two-pass Scan (T2R & R2T)

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class Solution {
    public int longestValidParentheses(String s) {
        if (s.length() < 2) return 0;

        int maxLen = 0;
        int leftCount = 0, rightCount = 0;

        // left to right
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                leftCount++;
            } else {
                rightCount++;
            }

            if (leftCount == rightCount) {
                maxLen = Math.max(maxLen, 2 * rightCount);
            } else if (rightCount > leftCount) {
                // reset
                leftCount = 0;
                rightCount = 0;
            }
        }

        // reset
        leftCount = 0;
        rightCount = 0;

        // right to left
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            if (c == '(') {
                leftCount++;
            } else {
                rightCount++;
            }

            if (leftCount == rightCount) {
                maxLen = Math.max(maxLen, 2 * leftCount);
            } else if (leftCount > rightCount) {
                // reset
                leftCount = 0;
                rightCount = 0;
            }
        }

        return maxLen;
    }
}

Remarks:

  1. TC: $O(2n)$ or $O(n)$, SC: $O(1)$

  2. Why Two pass?

    Example: ((), left to right: 0, right to left: 1*2

  3. L2R: left should always be equal or more than right parenthesis; R2L: right should always be equal or more than left parenthesis.

Stack

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class Solution {
    public int longestValidParentheses(String s) {
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);  // base
        int maxLen = 0;

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                stack.push(i);
            } else {
                stack.pop();  // try to match the parenthesis

                if (stack.isEmpty()) {
                    stack.push(i);  // update the base
                } else {
                    maxLen = Math.max(maxLen, i - stack.peek()); // check the index of the previous `)`
                }
            }
        }

        return maxLen;
    }
}

Remarks:

  1. TC: $O(n)$, SC: $O(n)$

Dynamic Programming

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class Solution {
    public int longestValidParentheses(String s) {
        int n = s.length();
        if (n < 2) return 0;

        int[] dp = new int[n];
        int maxLen = 0;

        for (int i = 1; i < n; i++) {
            if (s.charAt(i) == ')') {
                if (s.charAt(i - 1) == '(') {
                    dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
                } else if (i - dp[i - 1] - 1 >= 0 &&
                           s.charAt(i - dp[i - 1] - 1) == '(') {
                    dp[i] = dp[i - 1] + 2 +
                            (i - dp[i - 1] - 2 >= 0 ? dp[i - dp[i - 1] - 2] : 0);
                }

                maxLen = Math.max(maxLen, dp[i]);
            }
        }

        return maxLen;
    }
}

Remarks:

  1. Base: dp[0]=0

  2. if s[i] == ')' and last one s[i-1]=='(' then dp[i] = dp[i-2] + 2

  3. if s[i] == ')' and last one s[i-1]==')' then check s[i - dp[i-1] - 1],

    dp[i] = dp[i-1] + 2 + dp[i- dp[i-1] -2]