← Back to all recaps
VO Linkedin

LinkedIn High-Freq Questions: String Replace (No Built-in) + Connection Distance

1 min read

Organized by oavoservice: Two common questions covering "Linear Scan String Processing" and "Social Graph Shortest Path".

Question 1: String Replace without Built-in Function

Problem

Given string s, pattern pat, replacement rep, return a new string where all occurrences of pat in s are replaced by rep, without using built-in replace.

Idea (Two Pointers / Sliding Window)

Manually implement "left-to-right matching":

  • Scan s with index i.
  • If s[i:i+len(pat)] == pat, append rep and i += len(pat).
  • Otherwise append s[i] and i += 1.

Python Implementation

def replace_without_builtin(s: str, pat: str, rep: str) -> str:
    if pat == "":
        return s

    out = []
    i = 0
    m = len(pat)
    while i < len(s):
        if i + m <= len(s) and s[i:i+m] == pat:
            out.append(rep)
            i += m
        else:
            out.append(s[i])
            i += 1
    return "".join(out)

Interview Questions

  • Overlapping matches? (Standard replace semantics are non-overlapping, greedy left-to-right).
  • Complexity? (Worst case O(n*m), KMP can improve to linear).

Question 2: LinkedIn Connection Distance

Problem

Given an undirected graph of user connections and two users src and dst, find the shortest distance (min edges) between them.

  • Return -1 if unreachable.

Idea (BFS Shortest Path)

Standard BFS for unweighted graph shortest path:

  • Start BFS from src, distance 0.
  • Expand neighbors, first time hitting dst gives shortest distance.

Python Template

from collections import deque
from typing import Dict, List


def connection_distance(graph: Dict[str, List[str]], src: str, dst: str) -> int:
    if src == dst:
        return 0
    if src not in graph or dst not in graph:
        return -1

    q = deque([src])
    dist = {src: 0}

    while q:
        u = q.popleft()
        for v in graph.get(u, []):
            if v in dist:
                continue
            dist[v] = dist[u] + 1
            if v == dst:
                return dist[v]
            q.append(v)

    return -1

Interview Questions

  • Large graph? (Bi-directional BFS / Limit depth).
  • Return path? (Store parent pointers).

If preparing for LinkedIn VO/OA, oavoservice trains you to quickly implement and explain these fundamental questions and extensions.

Need real interview questions? Contact WeChat Coding0201 immediately to get real questions.

联系方式

Email: [email protected] Telegram: @OAVOProxy