19. Case-specific Sorting of Strings

✅ GFG solution to the Case-specific Sorting of Strings problem: sort uppercase and lowercase characters separately while maintaining their original positions. 🚀

The problem can be found at the following link: 🔗 Question Link

🧩 Problem Description

Given a string s consisting of only uppercase and lowercase characters. The task is to sort uppercase and lowercase letters separately such that if the ith place in the original string had an Uppercase character then it should not have a lowercase character after being sorted and vice versa.

An element maintains its case but gets sorted within its case group while preserving the original case pattern of the string.

📘 Examples

Example 1

Input: s = "GEekS"
Output: EGekS
Explanation:
- Uppercase positions: 0(G), 1(E), 4(S) → sorted: E, G, S
- Lowercase positions: 2(e), 3(k) → sorted: e, k
- Result: EGekS (maintaining original case positions)

Example 2

Input: s = "XWMSPQ"
Output: MPQSWX
Explanation: Since all characters are of the same case, we can simply perform a sorting operation on the entire string.

🔒 Constraints

• $1 \le s.length() \le 10^5$

✅ My Approach

The optimal approach uses Frequency Counting with Bitwise Operations for ultra-fast case detection:

Bitwise Optimized Frequency Counting

1. Case Detection:

   • Use bitwise operation c & 32 to detect case instantly.

   • If c & 32 is true, it's lowercase; otherwise uppercase.

2. Frequency Counting:

   • Count frequency of each character using separate arrays for lowercase and uppercase.

   • Use direct ASCII arithmetic: c - 97 for lowercase, c - 65 for uppercase.

3. In-place Reconstruction:

   • Iterate through original string maintaining case positions.

   • Replace each character with the next available sorted character of the same case.

   • Use pointer tracking to avoid redundant searches.

4. Pointer Optimization:

   • Maintain separate pointers for lowercase and uppercase arrays.

   • Skip zero-frequency positions efficiently.

📝 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), where n is the length of the string. We make two passes through the string - one for counting frequencies and one for reconstruction, both taking O(n) time.

• Expected Auxiliary Space Complexity: O(1), as we only use fixed-size arrays of 26 elements each for lowercase and uppercase character frequencies, which is constant space regardless of input size.

🧑‍💻 Code (C++)

class Solution {
public:
    string caseSort(string &s) {
        int l[26] = {0}, u[26] = {0};
        for (char c : s) (c & 32) ? l[c - 97]++ : u[c - 65]++;
        int i = 0, j = 0;
        for (char &c : s) {
            if (c & 32) {
                while (!l[i]) i++;
                c = i + 97;
                l[i]--;
            } else {
                while (!u[j]) j++;
                c = j + 65;
                u[j]--;
            }
        }
        return s;
    }
};

⚡ View Alternative Approaches with Code and Analysis

📊 2️⃣ Frequency Array with Position Tracking

💡 Algorithm Steps:

1. Count frequency of each character using arrays.

2. Maintain separate pointers for lowercase and uppercase.

3. Replace characters in-place with sorted equivalents.

class Solution {
public:
    string caseSort(string &s) {
        int lower[26] = {0}, upper[26] = {0};
        for (char ch : s) {
            if (islower(ch)) lower[ch - 'a']++;
            else upper[ch - 'A']++;
        }
        int l = 0, u = 0;
        for (int i = 0; i < s.length(); i++) {
            if (islower(s[i])) {
                while (lower[l] == 0) l++;
                s[i] = 'a' + l;
                lower[l]--;
            } else {
                while (upper[u] == 0) u++;
                s[i] = 'A' + u;
                upper[u]--;
            }
        }
        return s;
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(n + 26) = O(n)

• Auxiliary Space: 💾 O(1) - Fixed size arrays

✅ Why This Approach?

• Standard readable implementation.

• Clear separation of concerns.

📊 3️⃣ Two-Pass Sorting with Vectors

💡 Algorithm Steps:

1. Separate lowercase and uppercase characters.

2. Sort both groups independently.

3. Merge back maintaining original positions.

class Solution {
public:
    string caseSort(string &s) {
        vector<char> lower, upper;
        for (char c : s) {
            if (islower(c)) lower.push_back(c);
            else upper.push_back(c);
        }
        sort(lower.begin(), lower.end());
        sort(upper.begin(), upper.end());
        int l = 0, u = 0;
        for (char &c : s) {
            if (islower(c)) c = lower[l++];
            else c = upper[u++];
        }
        return s;
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(n log n) - Due to sorting

• Auxiliary Space: 💾 O(n) - For storing characters

✅ Why This Approach?

• Intuitive separation and sorting.

• Works with any character set.

📊 4️⃣ Counting Sort with String Builder

💡 Algorithm Steps:

1. Count frequency using counting sort technique.

2. Build result string character by character.

3. Use string builder for efficiency.

class Solution {
public:
    string caseSort(string &s) {
        int freq[128] = {0};
        string result = s;
        for (char c : s) freq[c]++;

        for (int i = 0; i < s.length(); i++) {
            if (islower(s[i])) {
                for (int j = 'a'; j <= 'z'; j++) {
                    if (freq[j] > 0) {
                        result[i] = j;
                        freq[j]--;
                        break;
                    }
                }
            } else {
                for (int j = 'A'; j <= 'Z'; j++) {
                    if (freq[j] > 0) {
                        result[i] = j;
                        freq[j]--;
                        break;
                    }
                }
            }
        }
        return result;
    }
};

📝 Complexity Analysis:

• Time: ⏱️ O(n * 26) = O(n)

• Auxiliary Space: 💾 O(128) = O(1)

✅ Why This Approach?

• Works with extended ASCII.

• Flexible for different character ranges.

🆚 🔍 Comparison of Approaches

🚀 Approach	⏱️ Time Complexity	💾 Space Complexity	✅ Pros	⚠️ Cons
🔍 Bitwise Optimized	🟢 O(n)	🟢 O(1)	⚡ Ultra-fast, minimal operations	🧮 Less readable
🔄 Frequency Array	🟢 O(n)	🟢 O(1)	🔧 Clear and readable	🐢 More function calls
🔺 Two-Pass Sorting	🟡 O(n log n)	🟡 O(n)	🚀 Intuitive logic	💾 Higher complexity
📊 Counting Sort	🟢 O(n)	🟢 O(1)	🏎️ Flexible character support	🔄 More iterations per char

🏆 Best Choice Recommendation

🎯 Scenario	🎖️ Recommended Approach	🔥 Performance Rating
⚡ Maximum performance, competitive programming	🥇 Bitwise Optimized	★★★★★
🔧 Production code, readability important	🥈 Frequency Array	★★★★☆
📊 Educational purposes, clear logic	🥉 Two-Pass Sorting	★★★☆☆
🏎️ Extended character set support	🏅 Counting Sort	★★★★☆

☕ Code (Java)

class Solution {
    public static String caseSort(String s) {
        int[] l = new int[26], u = new int[26];
        char[] c = s.toCharArray();
        for (char ch : c) {
            if (ch >= 'a') l[ch - 97]++;
            else u[ch - 65]++;
        }
        int i = 0, j = 0;
        for (int k = 0; k < c.length; k++) {
            if (c[k] >= 'a') {
                while (l[i] == 0) i++;
                c[k] = (char)(i + 97);
                l[i]--;
            } else {
                while (u[j] == 0) j++;
                c[k] = (char)(j + 65);
                u[j]--;
            }
        }
        return new String(c);
    }
}

🐍 Code (Python)

class Solution:
    def caseSort(self, s):
        l, u = [0] * 26, [0] * 26
        for c in s:
            if c.islower():
                l[ord(c) - 97] += 1
            else:
                u[ord(c) - 65] += 1
        r, i, j = list(s), 0, 0
        for k in range(len(s)):
            if s[k].islower():
                while l[i] == 0:
                    i += 1
                r[k] = chr(i + 97)
                l[i] -= 1
            else:
                while u[j] == 0:
                    j += 1
                r[k] = chr(j + 65)
                u[j] -= 1
        return ''.join(r)

🧠 Contribution and Support

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: 📬 Any Questions?. Let's make this learning journey more collaborative!

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