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🐛Spot the BugcppEasyArrays & Hashing

Two Sum - Spot the Bug (C++)

Find and fix the bug in this C++ solution

Problem Brief

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Puzzle Hints

Test the line against the actual promise of Two Sum, not just whether the syntax looks valid.
The repair changes only this statement shape: "for (int i = 1; i < n; i++) {". Keep the same role, but make it match the invariant.
Wrong initial value

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Two Sum — HashMap Lookup

hashmap

Input: [2,7,11,15], target=9. Use HashMap: complement → index

unordered_map & unordered_set in C++ref

C++ unordered_map and unordered_set use hash tables for O(1) average operations. Include <unordered_map> and <unordered_set>. For sorted order, use map/set (O(log n) red-black tree).

-unordered_map<K,V> — O(1) insert/find/erase

-unordered_set<T> — O(1) insert/count/erase

-operator[] auto-inserts default if missing

-count(key) returns 0 or 1 (use as boolean)

#include <unordered_map>
#include <unordered_set>

unordered_map<string, int> m;
m["a"] = 1;
m.count("a");  // 1 (exists)

unordered_set<int> s = {1, 2, 3};
s.count(2);    // 1

Official docs →

unordered_map & unordered_set in C++ref

C++ unordered_map and unordered_set use hash tables for O(1) average operations. Include <unordered_map> and <unordered_set>. For sorted order, use map/set (O(log n) red-black tree).

-unordered_map<K,V> — O(1) insert/find/erase

-unordered_set<T> — O(1) insert/count/erase

-operator[] auto-inserts default if missing

-count(key) returns 0 or 1 (use as boolean)

#include <unordered_map>
#include <unordered_set>

unordered_map<string, int> m;
m["a"] = 1;
m.count("a");  // 1 (exists)

unordered_set<int> s = {1, 2, 3};
s.count(2);    // 1

Official docs →

How to think: Hash Map / Setguide

You need O(1) lookups — checking if something exists, counting frequencies, or finding pairs.

1.Ask: "Am I searching for something repeatedly?" → Hash Map

2.Ask: "Do I need to check existence?" → Set

3.Ask: "Do I need to count occurrences?" → Map with value = count

4.Ask: "Do I need to find a pair that satisfies a condition?" → Store complement in Map

5.The tradeoff: O(n) extra space buys you O(1) per lookup

vs Nested loops (O(n²)): You're comparing every element against every other — a Map does it in one pass

vs Sorting (O(n log n)): You just need existence/frequency checks, not order

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

How to think: Hash Map / Setguide

You need O(1) lookups — checking if something exists, counting frequencies, or finding pairs.

1.Ask: "Am I searching for something repeatedly?" → Hash Map

2.Ask: "Do I need to check existence?" → Set

3.Ask: "Do I need to count occurrences?" → Map with value = count

4.Ask: "Do I need to find a pair that satisfies a condition?" → Store complement in Map

5.The tradeoff: O(n) extra space buys you O(1) per lookup

vs Nested loops (O(n²)): You're comparing every element against every other — a Map does it in one pass

vs Sorting (O(n log n)): You just need existence/frequency checks, not order

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

Inspect the code — click the line with the bug

1/*
2    Given int array & target, return indices of 2 nums that add to target
3    Ex. nums = [2,7,11,15] & target = 9 -> [0,1], 2 + 7 = 9
4    At each num, calculate complement, if exists in hash map then return
5    Time: O(n)
6    Space: O(n)
7*/
8class Solution {
9public:
10    vector<int> twoSum(vector<int>& nums, int target) {
11        int n = nums.size();
12        unordered_map<int, int> mp; // val -> index
13        for (int i = 1; i < n; i++) {
14            int complement = target - nums[i];
15            if (mp.find(complement) != mp.end()) {
16                return {mp[complement], i};

1/*

2 Given int array & target, return indices of 2 nums that add to target

3 Ex. nums = [2,7,11,15] & target = 9 -> [0,1], 2 + 7 = 9

4 At each num, calculate complement, if exists in hash map then return

5 Time: O(n)

6 Space: O(n)

7*/

8class Solution {

9public:

10 vector<int> twoSum(vector<int>& nums, int target) {

11 int n = nums.size();

12 unordered_map<int, int> mp; // val -> index

13 for (int i = 1; i < n; i++) {

14 int complement = target - nums[i];

15 if (mp.find(complement) != mp.end()) {

16 return {mp[complement], i};