Merge k Sorted Lists

Merge k Sorted Lists - Problem

You are given an array of k linked-lists lists, where each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Input & Output

Example 1 — Three Sorted Lists

$ Input: lists = [[1,4,5],[1,3,4],[2,6]]

› Output: [1,1,2,3,4,4,5,6]

💡 Note: The linked-lists are: [1→4→5], [1→3→4], [2→6]. Merging them results in [1→1→2→3→4→4→5→6]

Example 2 — Empty Lists Array

$ Input: lists = []

› Output: []

💡 Note: No lists to merge, return empty result

Example 3 — Single Empty List

$ Input: lists = [[]]

› Output: []

💡 Note: Only one empty list, result is empty

Constraints

k == lists.length
0 ≤ k ≤ 10⁴
0 ≤ lists[i].length ≤ 500
-10⁴ ≤ lists[i][j] ≤ 10⁴
lists[i] is sorted in ascending order
The sum of lists[i].length will not exceed 10⁴

Visualization

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Asked in

a Amazon 87 f Facebook 45 G Google 42 M Microsoft 38 🍎 Apple 25

The key insight is to avoid sorting all values from scratch. Instead, leverage the fact that individual lists are already sorted. The divide and conquer approach is optimal: merge lists pairwise until one remains. Time: O(n log k), Space: O(log k).

Common Approaches

✓ Brute Force - Collect All Values

⏱️ Time: O(n log n) Space: O(n)

Traverse all k linked lists to collect every node value into an array. Sort the array and construct a new linked list from the sorted values.

Divide and Conquer

⏱️ Time: O(n log k) Space: O(log k)

Recursively merge pairs of lists until only one remains. Use the efficient merge two sorted lists algorithm as the base operation.

Min Heap Approach

⏱️ Time: O(n log k) Space: O(k)

Maintain a min heap of the current heads of all non-empty lists. Extract minimum, add to result, and push the next node from that list back to heap.

Brute Force - Collect All Values — Algorithm Steps

Step 1: Traverse all k lists and collect values
Step 2: Sort the collected values
Step 3: Create new linked list from sorted values

Visualization

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Step-by-Step Walkthrough

Extract Values

Collect all node values from k lists

Sort Array

Sort the collected values

Build Result

Create new linked list from sorted array

Code -

solution.c — C

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

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

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

struct ListNode* solution(struct ListNode** lists, int listsSize) {
    if (lists == NULL || listsSize == 0) return NULL;
    
    // Collect all values
    int* values = malloc(10000 * sizeof(int));
    int valueCount = 0;
    
    for (int i = 0; i < listsSize; i++) {
        struct ListNode* curr = lists[i];
        while (curr) {
            values[valueCount++] = curr->val;
            curr = curr->next;
        }
    }
    
    if (valueCount == 0) {
        free(values);
        return NULL;
    }
    
    // Sort values
    qsort(values, valueCount, sizeof(int), compare);
    
    // Create new linked list
    struct ListNode* dummy = malloc(sizeof(struct ListNode));
    dummy->val = 0;
    dummy->next = NULL;
    struct ListNode* curr = dummy;
    
    for (int i = 0; i < valueCount; i++) {
        curr->next = malloc(sizeof(struct ListNode));
        curr->next->val = values[i];
        curr->next->next = NULL;
        curr = curr->next;
    }
    
    struct ListNode* result = dummy->next;
    free(dummy);
    free(values);
    return result;
}

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

void listToArray(struct ListNode* head, int* result, int* size) {
    *size = 0;
    while (head) {
        result[(*size)++] = head->val;
        head = head->next;
    }
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for the specific input format
    struct ListNode* lists[1000];
    int listsSize = 0;
    
    if (strstr(line, "[]") && !strstr(line, "[[]]")) {
        listsSize = 0;
    } else if (strstr(line, "[[]]")) {
        lists[0] = NULL;
        listsSize = 1;
    } else {
        // Parse nested arrays
        char* p = line;
        while (*p && *p != '[') p++;
        if (*p == '[') p++;
        
        while (*p && *p != ']') {
            while (*p == ' ' || *p == ',') p++;
            if (*p == ']') break;
            
            if (*p == '[') {
                char innerArray[1000] = {0};
                int depth = 0;
                char* start = p;
                while (*p) {
                    if (*p == '[') depth++;
                    else if (*p == ']') depth--;
                    p++;
                    if (depth == 0) break;
                }
                strncpy(innerArray, start, p - start);
                
                int arr[1000];
                int arrSize;
                parseArray(innerArray, arr, &arrSize);
                lists[listsSize++] = arrayToList(arr, arrSize);
            }
        }
    }
    
    struct ListNode* resultHead = solution(lists, listsSize);
    
    int result[10000];
    int resultSize;
    listToArray(resultHead, result, &resultSize);
    
    printf("[");
    for (int i = 0; i < resultSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Collecting n values takes O(n), sorting takes O(n log n) where n is total nodes

⚡ Linearithmic

Space Complexity

O(n)

Extra array stores all n node values

✓ Linear Space

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Very High Frequency

~25 min Avg. Time

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Related Problems

Common Approaches

Brute Force - Collect All Values — Algorithm Steps

Visualization

Code -

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