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5,485 c software interview questions shared by candidates

First round: Hackerrank Test. Question: Given a string which contains only A and B, remove all the occurrences of AB and BB until the string doesn't contain any of these 2 substrings. Return the length of the original string after removing all the occurrences of the above substring. eg: "ABAABBA" answer: 1 2nd Round: One-on-One Interview(Online) Questions: What is a dangling pointer? In what scenario does a pointer become dangling? What was your role in your previous company? What features did you work on for the project in previous company? What were some mistakes you made and how did you correct them? 2 predict the output questions: int a=1; print("%d\n",(a++)*(++a)); char *p = 'a'; delete a; 2 coding questions: Print a singly linked list in reverse order(A:Recursion) How to detect a loop in a linked list(A: Hare and Tortoise Algo) (Avoid using extra space) What are some projects you did and explain the one you liked. After receiving the offer: HR Assesment: Cognitive Analysis and Psychological analysis test(30 mins non-proctored simple test)
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C++ Software Developer

Interviewed at Accelya Group

3.3★
Sep 28, 2022

First round: Hackerrank Test. Question: Given a string which contains only A and B, remove all the occurrences of AB and BB until the string doesn't contain any of these 2 substrings. Return the length of the original string after removing all the occurrences of the above substring. eg: "ABAABBA" answer: 1 2nd Round: One-on-One Interview(Online) Questions: What is a dangling pointer? In what scenario does a pointer become dangling? What was your role in your previous company? What features did you work on for the project in previous company? What were some mistakes you made and how did you correct them? 2 predict the output questions: int a=1; print("%d\n",(a++)*(++a)); char *p = 'a'; delete a; 2 coding questions: Print a singly linked list in reverse order(A:Recursion) How to detect a loop in a linked list(A: Hare and Tortoise Algo) (Avoid using extra space) What are some projects you did and explain the one you liked. After receiving the offer: HR Assesment: Cognitive Analysis and Psychological analysis test(30 mins non-proctored simple test)

Shortest path in graph: /// /// graph is a function that accepts the node name, and int*. It returns the /// neighbors of the node and the length of the neighbors in the int* /// /// Name of start node /// Name of goal node /// /// The shortest path with format start->...->goal char *shortestPath(Inner *(*graph)(const char *, int *), const char *start, const char *goal) { // int neighborsLength; // Inner *neighbors = graph(start, &neighborsLength); // std::cout << "name: " << neighbors[0].name << neighbors[0].cost << " // length: " << neighborsLength << std::endl; std::cout << "name: " << // neighbors[1].name << neighbors[1].cost << " length: " << neighborsLength // << std::endl; // neighbors = graph(neighbors[0].name.c_str(), &neighborsLength); // std::cout << "name: " << neighbors[0].name << neighbors[0].cost << " // length: " << neighborsLength << std::endl; std::cout << "name: " << // neighbors[1].name << neighbors[1].cost << " length: " << neighborsLength // << std::endl;
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Junior Software Developer (C++)

Interviewed at Accelya Group

3.3★
Aug 15, 2024

Shortest path in graph: /// /// graph is a function that accepts the node name, and int*. It returns the /// neighbors of the node and the length of the neighbors in the int* /// /// Name of start node /// Name of goal node /// /// The shortest path with format start->...->goal char *shortestPath(Inner *(*graph)(const char *, int *), const char *start, const char *goal) { // int neighborsLength; // Inner *neighbors = graph(start, &neighborsLength); // std::cout << "name: " << neighbors[0].name << neighbors[0].cost << " // length: " << neighborsLength << std::endl; std::cout << "name: " << // neighbors[1].name << neighbors[1].cost << " length: " << neighborsLength // << std::endl; // neighbors = graph(neighbors[0].name.c_str(), &neighborsLength); // std::cout << "name: " << neighbors[0].name << neighbors[0].cost << " // length: " << neighborsLength << std::endl; std::cout << "name: " << // neighbors[1].name << neighbors[1].cost << " length: " << neighborsLength // << std::endl;

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