java
Let's walk through each iteration of the given code with input s = "100011001" and k = 3, step by step, showing how substrings are formed, evaluated, and update
Let's walk through **each iteration** of the given code with input `s = "100011001"` and `k = 3`, step by step, showing how substrings are formed, evaluated, and updated.

---

### **Outer Loop Iterations**
The outer loop runs from `i = 0` to `s.length() - k = 6`. Let's break down each iteration.

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#### **Iteration 1: `i = 0`, `s[0] = '1'` (valid starting point)**
- Initialize `str = ""`, `count = 0`
  
##### **Inner Loop:**
1. `j = 0`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "1"`, `count = 1`
2. `j = 1`: `s[j] = '0'`
   - Append `'0'` → `str = "10"`
3. `j = 2`: `s[j] = '0'`
   - Append `'0'` → `str = "100"`
4. `j = 3`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "1001"`, `count = 2`
5. `j = 4`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "10001"`, `count = 3`
   - **Condition met:** `count == k`.  
   - **First valid substring found:** Update `res = "10001"`.

---

#### **Iteration 2: `i = 1`, `s[1] = '0'` (invalid starting point)**
- Skip this iteration.

---

#### **Iteration 3: `i = 2`, `s[2] = '0'` (invalid starting point)**
- Skip this iteration.

---

#### **Iteration 4: `i = 3`, `s[3] = '1'` (valid starting point)**
- Initialize `str = ""`, `count = 0`

##### **Inner Loop:**
1. `j = 3`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "1"`, `count = 1`
2. `j = 4`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "11"`, `count = 2`
3. `j = 5`: `s[j] = '0'`
   - Append `'0'` → `str = "110"`
4. `j = 6`: `s[j] = '0'`
   - Append `'0'` → `str = "1100"`
5. `j = 7`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "11001"`, `count = 3`
   - **Condition met:** `count == k`
   - Compare `"11001"` with current `res ("10001")`.
   - Call `isSmaller("11001", "10001")`, which returns `true`.  
   - **Update `res = "11001"`**.

---

#### **Iteration 5: `i = 4`, `s[4] = '1'` (valid starting point)**
- Initialize `str = ""`, `count = 0`

##### **Inner Loop:**
1. `j = 4`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "1"`, `count = 1`
2. `j = 5`: `s[j] = '0'`
   - Append `'0'` → `str = "10"`
3. `j = 6`: `s[j] = '0'`
   - Append `'0'` → `str = "100"`
4. `j = 7`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "1001"`, `count = 2`
5. `j = 8`: `s[j] = '1'`
   - Append `'1'`, increment `count` → `str = "10011"`, `count = 3`
   - **Condition met:** `count == k`
   - Compare `"10011"` with current `res ("11001")`
   - Call `isSmaller("10011", "11001")`, which returns `false`
   - **No update to `res`.**

---

#### **Iteration 6: `i = 5`, `s[5] = '0'` (invalid starting point)**
- Skip this iteration.

---

#### **Iteration 7: `i = 6`, `s[6] = '0'` (invalid starting point)**
- Skip this iteration.

---

### **Final Result**
After all iterations:
- **Shortest Length:** 5  
- **Lexicographically Smallest Beautiful Substring:** `"11001"`

---

### **Key Insights**
1. The previous error was due to incorrectly keeping `"10001"` instead of `"11001"`.
2. The `isSmaller()` function ensures that the lexicographically smallest substring is correctly identified.


# Code
```java []
class Solution {
    public String shortestBeautifulSubstring(String s, int k) {
        StringBuilder res = new StringBuilder();

        for (int i = 0; i <= s.length() - k; i++) {
            if (s.charAt(i) == '0') continue;
            StringBuilder str = new StringBuilder();
            int count = 0;
            for (int j = i; j < s.length(); j++) {
                char c = s.charAt(j);
                if (c == '1') count++;
                str.append(c);
              if (count == k && (res.isEmpty() || isSmaller(str, res)) ) {
                res = new StringBuilder(str);
                break;
              }
            }
        }
        return res.toString();
    }

    public static boolean isSmaller(StringBuilder s1, StringBuilder s2) {
        int length1 = s1.length();
        int length2 = s2.length();
        
        if (length1 == length2) {
            int i = 0;
            while (i < length1-1 && s1.charAt(i) == s2.charAt(i)) {
               i++;
            }
            return s1.charAt(i) < s2.charAt(i);
        }
        
        return length1 < length2;
    }
}
```
['Java']