24. Merge k Sorted Lists
hardAsked at ByteDanceMerge k sorted linked lists into one sorted list — ByteDance uses it as a direct stand-in for merging k partitioned ranking streams into a single feed.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an array of k sorted linked lists, merge them into one sorted linked list and return its head. The total number of nodes across all lists may be large.
Constraints
0 <= k <= 10^40 <= total nodes <= 10^4-10^4 <= Node.val <= 10^4
Examples
Example 1
lists = [[1,4,5],[1,3,4],[2,6]][1,1,2,3,4,4,5,6]Example 2
lists = [][]Approaches
1. Flatten and sort
Push every value into an array, sort it, then rebuild a linked list.
- Time
- O(N log N)
- Space
- O(N)
// collect all values into an array, sort, then convert back into nodesTradeoff:
2. Min-heap of list heads
Push each list's head into a min-heap. Pop the smallest, attach it to the output, then push its next pointer.
- Time
- O(N log k)
- Space
- O(k)
function mergeKLists(lists) {
const heap = [];
const up = (i) => { while (i > 0) { const p = (i - 1) >> 1; if (heap[p].val > heap[i].val) { [heap[p], heap[i]] = [heap[i], heap[p]]; i = p; } else break; } };
const down = (i) => { const n = heap.length; for (;;) { const l = i * 2 + 1, r = l + 1; let s = i; if (l < n && heap[l].val < heap[s].val) s = l; if (r < n && heap[r].val < heap[s].val) s = r; if (s === i) break; [heap[s], heap[i]] = [heap[i], heap[s]]; i = s; } };
for (const head of lists) if (head) { heap.push(head); up(heap.length - 1); }
const dummy = { val: 0, next: null };
let cur = dummy;
while (heap.length) {
const node = heap[0];
cur.next = node;
cur = cur.next;
if (node.next) { heap[0] = node.next; down(0); }
else { const last = heap.pop(); if (heap.length) { heap[0] = last; down(0); } }
}
cur.next = null;
return dummy.next;
}Tradeoff:
ByteDance-specific tips
ByteDance interviewers grade explicit reasoning about why log-k merge wins at TikTok scale — exactly the multi-stream merge their ranking layer runs to produce a single ordered feed.
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