समस्या
In C#, an IComparer<T> is an interface: any class that implements its Compare method can tell a sort which item comes first. Choosing a different comparer changes the order without changing the sort.
Sort the people by key:
name: by name (ordinal character order, so uppercase letters come before lowercase), and by age when names are equal.age: by age, youngest first, and by name when ages are equal.
Implement sort_by_key(names, ages, key).
Input
names, ages: The people (same length).
key: "name" or "age".
Output
Return the names in sorted order.
सीमाएँ
0 ≤ number of people ≤ 100,000
0 ≤ age ≤ 150
उदाहरण
इनपुट:
key = "age"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
अपेक्षित आउटपुट:
expected = ["Bo", "Lin", "Ada"]इनपुट:
std::string key = "age";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
अपेक्षित आउटपुट:
std::vector<std::string> expected = {"Bo", "Lin", "Ada"};इनपुट:
string key = "age";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
अपेक्षित आउटपुट:
string[] expected = new string[] {"Bo", "Lin", "Ada"};व्याख्या
Bo and Lin are both 30, so they are ordered by name before Ada (36): ["Bo", "Lin", "Ada"].
इनपुट:
key = "name"
ages = [30, 36, 30]
names = ["Lin", "Ada", "Bo"]
अपेक्षित आउटपुट:
expected = ["Ada", "Bo", "Lin"]इनपुट:
std::string key = "name";
std::vector<long long> ages = {30, 36, 30};
std::vector<std::string> names = {"Lin", "Ada", "Bo"};
अपेक्षित आउटपुट:
std::vector<std::string> expected = {"Ada", "Bo", "Lin"};इनपुट:
string key = "name";
long[] ages = new long[] {30, 36, 30};
string[] names = new string[] {"Lin", "Ada", "Bo"};
अपेक्षित आउटपुट:
string[] expected = new string[] {"Ada", "Bo", "Lin"};व्याख्या
Alphabetical order: ["Ada", "Bo", "Lin"].
कैसे हल करें
पहले खुद कोशिश करें, फिर एक-एक करके संकेत खोलें।
संकेत 1
Sort pairs of (name, age) instead of two separate lists.
संकेत 2
The chosen key gives the main order; the other field breaks ties.
संकेत 3
Use ordinal string comparison so uppercase sorts before lowercase.
अपेक्षित जटिलता
O(n log n) time, O(n) extra space
संदर्भ
यह चुनौती CodeUtility के लिए अनुकूलित की गई है। ऊपर दिए गए स्टेटमेंट और सीमाओं का पालन करें।