> For the complete documentation index, see [llms.txt](https://syjohnson11.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://syjohnson11.gitbook.io/leetcode/7_dp/yi-wei-dp.md).

# 一维DP

## 划分型：f\[] means XXXX ending with i

word break

longest valid parentheses

decode ways

## Longest Valid Parentheses

* 不能算Hard, 但是转移方程需要考虑坐标edge case, bug free不容易
* 划分型类似Longest Palindromic Substring，word break思路

```
public int longestValidParentheses(String s) {
        if (s == null || s.length() == 0) return 0;
        int n = s.length();
        int[] f = new int[n];
        int max = 0;
        for (int i = 1; i < n; i++) {
            if (s.charAt(i) == ')') {
                if (s.charAt(i - 1) == '(') { //“（）”
                    f[i] = 2 + (i >= 2 ? f[i - 2] : 0);
                } else if (i - f[i-1] - 1 >= 0 && s.charAt(i - f[i-1] - 1) == '(') {
                    f[i] = 2 + f[i-1] + (i - f[i-1] - 2 >= 0 ? f[i - f[i-1] - 2] : 0);
                }
            }
            max = Math.max(max, f[i]);
        }
        return max;
    }
```

## Decode Ways

* 注意corner case, any valid substring must starts with non-zero
* how to handle "0000" ?
* state:f\[i] means num of decode ways for s.substring(i)&#x20;
* function:&#x20;
  * f\[i] = isValid(s.substring(i, i + 1)) && f\[i + 1] + isValid(s.substring(i, i + 2)) && f\[i + 2]&#x20;
* oneDigit != 0&#x20;
* 10 <= twoDigits <= 26&#x20;
* res: f\[0]&#x20;
* initialize:&#x20;
  * f\[n] = 0&#x20;
  * f\[n - 1] = isValid(s.substring(n - 1)) ? 1 : 0&#x20;
  * f\[n - 2] = isValid(s.substring(n - 2)) ? 1 + f\[n - 1] : 0
* **面试还是写从dfs -> memo的思路更make sense一点**
* **除非面试官要求optimal space complexity-> DP with rolling array**

```
class Solution {
    /*    
    brute force: DFS
    decode("226") = decode("26") + isValid("22") && decode("6")
    
    state:f[i] means num of decode ways for s.substring(i)
    function: 
    f[i] = isValid(s.substring(i, i + 1)) && f[i + 1] + isValid(s.substring(i, i + 2)) && f[i + 2]
    oneDigit != 0
    10 <= twoDigits <= 26
    res: f[0]
    initialize: 
    f[n] = 0
    f[n - 1] = isValid(s.substring(n - 1)) ? 1 : 0
    f[n - 2] = isValid(s.substring(n - 2)) ? 1 + f[n - 1] : 0
    
    O(N) Time and O(N) Space
    can use rolling array to optimize space complexity to O(1)
    
    */
    public int numDecodings(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int n = s.length();
        int[] f = new int[n + 1];
        //initialize
        f[n] = 1;
        f[n - 1] =  s.charAt(n - 1) != '0' ? 1 : 0;
        for (int i = n - 2; i >= 0; i--) {
            char c = s.charAt(i);
            //one digit, need to make sure it's not '0'
            if (c != '0') {
                f[i] += f[i + 1];
            }
            int twoDigits = (s.charAt(i) - '0') * 10 + s.charAt(i + 1) - '0';
            if (twoDigits >= 10 && twoDigits <= 26) {
                f[i] += f[i + 2];
            }
        }
        return f[0];
    }
}
```


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