Sort Linked List of 0, 1, 2
Problem Statement:
Given the head of a linked list where nodes can contain values 0s, 1s, and 2s only. Your task is to rearrange the list so that all 0s appear at the beginning, followed by all 1s, and all 2s are placed at the end.
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Example:
✅ Solution: Counting Sort (Two-pass Approach)
class Solution {
public:
Node* segregate(Node* head) {
int count[3] = {0, 0, 0};
// First pass: Count the number of 0s, 1s, and 2s
Node *temp = head;
while(temp != NULL){
count[temp->data]++;
temp = temp->next;
}
// Second pass: Overwrite values based on counts
int i = 0;
temp = head;
while(temp){
if(count[i] == 0){
i++;
} else {
temp->data = i;
count[i]--;
temp = temp->next;
}
}
return head;
}
};
📝 How It Works
- The list contains only 0s, 1s, and 2s.
- First, we count occurrences of each number using a frequency array.
- Then, we overwrite node values with 0s, 1s, and 2s based on the count — simulating sorting.
- No node creation or pointer manipulation needed — we reuse the existing nodes.
🧩 Key Logic
Step 1: Count frequency of 0, 1, 2
Step 2: Rewrite nodes sequentially with those values
⏱️ Time & Space Complexity
| Metric | Value |
|---|---|
| ⏱️ Time Complexity | O(N) — two traversals of the list |
| 🪄 Space Complexity | O(1) — uses only 3 counters |
⚠️ Edge Cases
- ✅ Empty list → return
NULL - ✅ All values are same (e.g., all 1s or all 2s) → remains unchanged
- ✅ Only one element → unchanged
- ✅ Random mix → will be sorted in-place as 0s → 1s → 2s
💡 Other Approaches
| Approach | Time | Space | Notes |
|---|---|---|---|
| Dummy Node List for 0s, 1s, 2s | O(N) | O(1) | Uses separate sub-lists and merges them |
| In-place data overwrite ✅ | O(N) | O(1) | Simple and efficient |
🔁 Related Problems
- LeetCode 75. Sort Colors – similar logic with arrays
- GFG: Given a linked list of 0s, 1s and 2s, sort it
- LC 86. Partition List – separate logic but related in idea
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