Linked List Partitioner - Problem

Given a linked list and a value x, partition it such that all nodes with values less than x come before all nodes with values greater than or equal to x.

Important: You should preserve the original relative order of the nodes in each of the two partitions.

Return the head of the modified linked list.

Input & Output

Example 1 — Basic Partition

$ Input: head = [1,4,3,2,5,2], x = 3

› Output: [1,2,2,4,3,5]

💡 Note: Values less than 3 (1,2,2) come first, followed by values >= 3 (4,3,5). Original relative order is preserved within each group.

Example 2 — All Values Greater

$ Input: head = [2,1], x = 2

› Output: [1,2]

💡 Note: Only value 1 is less than 2, so it comes first, followed by value 2.

Example 3 — Single Node

$ Input: head = [1], x = 2

› Output: [1]

💡 Note: Single node case: 1 < 2, so the list remains unchanged.

Constraints

The number of nodes in the list is in the range [0, 200]
-100 ≤ Node.val ≤ 100
-200 ≤ x ≤ 200

Visualization

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Asked in

a Amazon 45 M Microsoft 32 🍎 Apple 28 G Google 25

The key insight is to use two dummy head nodes to build separate chains for values less than x and values greater than or equal to x. The optimal approach uses two dummy heads, traverses the list once, and connects the chains at the end. Time: O(n), Space: O(1).

Common Approaches

✓ Two Dummy Heads

⏱️ Time: O(n) Space: O(1)

Create two dummy head nodes - one for values less than x, another for values >= x. Traverse original list once, adding each node to appropriate chain. Finally connect the two chains.

Two-Pass Collection

⏱️ Time: O(n) Space: O(n)

First pass collects all node values into an array. Then separate values into 'less than x' and 'greater/equal to x' arrays. Finally, create a new linked list from the combined arrays.

Two Dummy Heads — Algorithm Steps

Create two dummy heads: beforeHead and afterHead
Traverse original list, directing each node to appropriate chain
Connect tail of 'before' chain to head of 'after' chain
Return the head of combined result

Visualization

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Step-by-Step Walkthrough

Setup Dummy Heads

Create before and after dummy nodes

Route Each Node

Direct nodes to appropriate chain based on value vs x

Connect Chains

Link tail of before chain to head of after chain

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct ListNode {
    int val;
    struct ListNode* next;
};

struct ListNode* solution(struct ListNode* head, int x) {
    // Create two dummy heads
    struct ListNode beforeHead = {0, NULL};
    struct ListNode afterHead = {0, NULL};
    
    // Pointers to build chains
    struct ListNode* before = &beforeHead;
    struct ListNode* after = &afterHead;
    
    // Traverse original list
    struct ListNode* current = head;
    while (current) {
        if (current->val < x) {
            before->next = current;
            before = before->next;
        } else {
            after->next = current;
            after = after->next;
        }
        current = current->next;
    }
    
    // Connect the two chains
    after->next = NULL;  // End the after chain
    before->next = afterHead.next;  // Connect before chain to after chain
    
    return beforeHead.next;
}

int listToArray(struct ListNode* head, int* arr) {
    int count = 0;
    while (head) {
        arr[count++] = head->val;
        head = head->next;
    }
    return count;
}

struct ListNode* arrayToList(int* arr, int size) {
    if (size == 0) return NULL;
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->val = arr[0];
    head->next = NULL;
    struct ListNode* current = head;
    for (int i = 1; i < size; i++) {
        current->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        current->next->val = arr[i];
        current->next->next = NULL;
        current = current->next;
    }
    return head;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int x;
    scanf("%d", &x);
    
    int headArr[1000];
    int headSize = 0;
    
    if (strcmp(line, "[]") != 0) {
        char* token = strtok(line + 1, ",]");
        while (token) {
            headArr[headSize++] = atoi(token);
            token = strtok(NULL, ",]");
        }
    }
    
    struct ListNode* head = arrayToList(headArr, headSize);
    struct ListNode* result = solution(head, x);
    
    int resultArr[1000];
    int resultSize = listToArray(result, resultArr);
    
    printf("[");
    for (int i = 0; i < resultSize; i++) {
        if (i > 0) printf(",");
        printf("%d", resultArr[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single traversal through all n nodes in the linked list

✓ Linear Growth

Space Complexity

O(1)

Only uses a constant number of extra pointers (dummy heads and current pointers)

✓ Linear Space

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Linked List Partitioner - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two Dummy Heads — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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