248. Strobogrammatic Number III

 (3)找出1-650中所有的ambiguous number,定义：如果把一个数upside down还是一个有效数的话，这个数就是ambiguous，比如19 -> 61, follow up找到1-n中的这种数



A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).
Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.
Example:
Input: low = "50", high = "100"
Output: 3 
Explanation: 69, 88, and 96 are three strobogrammatic numbers.
Note:
Because the range might be a large number, the low and high numbers are represented as string.

// solution 1 , use dfs . 
// 明天自己写
https://leetcode.com/problems/strobogrammatic-number-iii/discuss/67378/Concise-Java-Solution





Construct char arrays from low.length() to high.length()

Add stro pairs from outside

When left > right, add eligible count

private static final char[][] pairs = {{'0', '0'}, {'1', '1'}, {'6', '9'}, {'8', '8'}, {'9', '6'}};

public int strobogrammaticInRange(String low, String high) {
    int[] count = {0};
    for (int len = low.length(); len <= high.length(); len++) {
        char[] c = new char[len];
        dfs(low, high, c, 0, len - 1, count);
    }
    return count[0];
}

public void dfs(String low, String high , char[] c, int left, int right, int[] count) {
    if (left > right) { // meet the middle 
        String s = new String(c);
        // check if out of the the low and high 
        if ((s.length() == low.length() && s.compareTo(low) < 0) ||
            (s.length() == high.length() && s.compareTo(high) > 0)) {
            return;
        }
        count[0]++;
        return;
    }
    for (char[] p : pairs) {
        c[left] = p[0];
        c[right] = p[1];
        if (c.length != 1 && c[0] == '0') { // 0 can not be on the outside 
            continue;
        }
        if (left == right && p[0] != p[1]) { // the middle element has to be the same as itself, like 0, 8, 1
            continue;
        }
        dfs(low, high, c, left + 1, right - 1, count);
    }
}






// use strobe 2 
// Memory Limit Exceeded

https://leetcode.com/problems/strobogrammatic-number-iii/discuss/67406/Clear-Java-AC-solution-using-Strobogrammatic-Number-II-method


class Solution{
        
        
        
    public int strobogrammaticInRange(String low, String high) {
        // can use stro 2 to get all the stro numbers of certain length and check each number , if exists in the range 
        int res = 0;
        int start = low.length();
        int end = high.length();
        List<String> list = new ArrayList<>();
        for(int i = start; i <= end; start++){
            list.addAll(findStrobogrammatic(i));
        }
        for(String num : list){
            if((num.length() == low.length()&&num.compareTo(low) < 0 ) ||(num.length() == high.length() && num.compareTo(high) > 0)) continue;
            res++;
            
        }
        return res;
    }

    private List<String> findStrobogrammatic(int n) {
      return helper(n, n);
        
    }
  
    private List<String> helper(int n, int m){
      // base case
      // if(n == 0){
      //   return new ArrayList<>("");
      // }
      // if(n == 1){
      //   return new ArrayList<>("1","8", "0");
      // }
      if (n == 0){
        return new ArrayList<String>(Arrays.asList(""));
        // Arrays.asList returns a fixed sized list
      }
      if (n == 1){
        return new ArrayList<String>(Arrays.asList("1","0","8"));
      }
      
      List<String> prevResult = helper(n - 2, m);
      
      List<String> result = new ArrayList<>();
      
      // two cases: if we are filling numbers on the very outside, we can add 1 6 8 9 
      // another case is if we are not on the very outside, we can add 1 6 8 9 0 
      
      for(String s : prevResult){
        if(n != m){
          // we are not on the very outside, which means we can fill up with 0 
          result.add("0" + s + "0");
        }
        
        result.add("1" + s + "1");
        result.add("6" + s + "9");
        result.add("9" + s + "6");
        result.add("8" + s + "8");
      }
      return result;
    }
}