Given a list of names in an array arr[] of size N, display the longest name contained in it. If there are multiple longest names print all of that.
Examples:
Input: arr[] = {“neveropen”, “FreeCodeCamp”, “StackOverFlow”, “MyCodeSchool”}
Output: neveropen StackOverFlow
Explanation: size of arr[0] and arr[2] i.e., 13 > size of arr[1] and arr[3] i.e., 12Input: arr[] = {“Akash”, “Adr”}
Output: Akash
Approach: Follow the given idea to solve the problem:
Traverse the given array and store the names with the maximum length, if a name with greater length is found update max length and add that name to the final answer.
Follow the steps to solve this problem:
- If N = 0 then simply return.
- Create an array res to store the answer.
- Else, Initialize Max = size of arr[0] and insert arr[0] in the res.
- Now, Traverse the array and check
- If size of arr[i] = Max, then push back arr[i] in vector res.
- Else If size of arr[i] > Max, then
- Set, Max = size of arr[i]
- Empty the array res
- Insert arr[i] in res
- Return res as the final answer
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to display longest names// contained in the arrayvector<string> solve(string* arr, int N){ // Edge Case if (N == 0) return {}; // Initialize Max int Max = arr[0].size(); // Create an array res vector<string> res; // Insert first element in res res.push_back(arr[0]); // Traverse the array for (int i = 1; i < N; i++) { // If string with greater length // is found if (arr[i].size() > Max) { Max = arr[i].size(); res.clear(); res.push_back(arr[i]); } // If string with current max length else if (arr[i].size() == Max) { res.push_back(arr[i]); } } // Return the final answer return res;}// Driver Codeint main(){ string arr[] = { "neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool" }; int N = sizeof(arr) / sizeof(arr[0]); // Function call vector<string> v = solve(arr, N); // Printing the answer for (auto i : v) { cout << i << " "; } cout << endl; return 0;} |
Java
// Java code for the above approachimport java.io.*;import java.util.*;class GFG { // Function to display longest names // contained in the array public static ArrayList<String> solve(String arr[], int N) { // Edge Case if (N == 0) { ArrayList<String> temp = new ArrayList<String>(); return temp; } // Initialize Max int Max = arr[0].length(); // Create an arraylist res ArrayList<String> res = new ArrayList<String>(); // Insert first element in res res.add(arr[0]); // Traverse the array for (int i = 1; i < N; i++) { // If string with greater length // is found if (arr[i].length() > Max) { Max = arr[i].length(); res.clear(); res.add(arr[i]); } // If string with current max length else if (arr[i].length() == Max) { res.add(arr[i]); } } // Return the final answer return res; } // Driver Code public static void main(String[] args) { String arr[] = { "neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool" }; int N = arr.length; // Function call ArrayList<String> v = solve(arr, N); // Printing the answer for (String i : v) { System.out.print(i + " "); } System.out.println(); }}// This code is contributed by Rohit Pradhan |
Python3
# Python code for the above approach# Function to display longest names# contained in the arraydef solve(arr, N): # Edge Case if (N == 0): return [] # Initialize Max Max = len(arr[0]) # Create an array res res = [] # Insert first element in res res.append(arr[0]) # Traverse the array for i in range(1,N): # If string with greater length # is found if (len(arr[i]) > Max): Max = len(arr[i]) res.clear() res.append(arr[i]); # If string with current max length elif(len(arr[i]) == Max): res.append(arr[i]) # Return the final answer return res# Driver Codeif __name__ == "__main__": arr = ["neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"] # Value of N N = len(arr) # Function call v = solve(arr, N) # Printing the answer for i in v: print(i,end=" ") # This code is contributed by Abhishek Thakur. |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class GFG{ // Function to display longest names // contained in the array public static List<string> solve(string[] arr, int N) { // Edge Case if (N == 0) { List<string> temp = new List<string>(); return temp; } // Initialize Max int Max = arr[0].Length; // Create an List res List<string> res = new List<string>(); // Insert first element in res res.Add(arr[0]); // Traverse the array for (int i = 1; i < N; i++) { // If string with greater length // is found if (arr[i].Length > Max) { Max = arr[i].Length; res.Clear(); res.Add(arr[i]); } // If string with current max length else if (arr[i].Length == Max) { res.Add(arr[i]); } } // Return the final answer return res; }// Driver Codepublic static void Main(){ string[] arr = { "neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool" }; int N = arr.Length; // Function call List<string> v = solve(arr, N); // Printing the answer foreach (string i in v) { Console.Write(i + " "); } Console.WriteLine();}}// This code is contributed by code_hunt. |
Javascript
<script>// JS code for the above approach// Function to display longest names// contained in the arrayfunction solve(arr,N){ // Edge Case if (N == 0) return []; // Initialize Max let Max = arr[0].length; // Create an array res res = []; // Insert first element in res res.push(arr[0]); // Traverse the array for (let i = 1; i < N; i++) { // If string with greater length // is found if (arr[i].length > Max) { Max = arr[i].length; res = []; res.push(arr[i]); } // If string with current max length else if (arr[i].length == Max) { res.push(arr[i]); } } // Return the final answer return res;}// Driver Codelet arr = [ "neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool" ];let N = arr.length;// Function calllet v = solve(arr, N);// Printing the answerconsole.log(v);// This code is contributed by akashish__</script> |
Output
neveropen StackOverFlow
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), for storing the names in the res array.
Another Approach: Hashing
We can make a hash-map of key-value pair where key will be length of string and value will be the string themself. This allows us to quickly access the longest names by retrieving the group with the maximum length.
Follow the steps to implement the above idea:
- Create a hash map to store names grouped by their lengths, and a variable maxLen to store the length of longest string.
- Iterate through each name in the input array.
- Calculate the length of the current name.
- Update the hash map with the current name added to its corresponding length group.
- Update maxLen if the current length is greater.
- Retrieve all the names from the hash map for the length maxLen.
Below is the implementation:
C++
#include <bits/stdc++.h>using namespace std;vector<string> findLongestNames(const vector<string>& arr) { int maxLen = 0; unordered_map<int, vector<string>> lengthMap; // Hash map to store names grouped by their lengths // Iterate through each name in the array for (const string& name : arr) { int len = name.length(); // Calculate the length of the current name // Update the hash map with the current name added to its corresponding length group lengthMap[len].push_back(name); // Update maxLen if the current length is greater if (len > maxLen) { maxLen = len; } } // Retrieve all the names from the hash map for the length maxLen return lengthMap[maxLen];}// Driver Codeint main() { vector<string> arr = {"neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"}; vector<string> longestNames = findLongestNames(arr); for (const string& name : longestNames) { cout << name << " "; } cout << endl; return 0;}// This code is contributed by Veerendra_Singh_Rajpoot |
Java
// Java code to Display the Longest Name Using Hashingimport java.util.ArrayList;import java.util.HashMap;import java.util.List;import java.util.Map;public class Main { // Function to display longest names public static List<String> findLongestNames(List<String> arr) { int maxLen = 0; Map<Integer, List<String> > lengthMap = new HashMap<>(); // Hash map to store names // grouped by their lengths // Iterate through each name in the list for (String it : arr) { int len = it.length(); // Calculate the length // of the current name // Update the hash map with the current name // added to its corresponding length group List<String> namesList = lengthMap.getOrDefault( len, new ArrayList<>()); namesList.add(it); lengthMap.put(len, namesList); // Update maxLen if the current length is // greater if (len > maxLen) { maxLen = len; } } // Retrieve all the names from the hash map for the // length maxLen return lengthMap.get(maxLen); } // Driver Code public static void main(String[] args) { List<String> arr = List.of("neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"); List<String> longestNames = findLongestNames(arr); for (String name : longestNames) { System.out.print(name + " "); } System.out.println(); }}// This code is contributed by Veerendra_Singh_Rajpoot |
Python3
def findLongestNames(arr): # Hash map to store names grouped by their lengths maxLen = 0 lengthMap = {} # Iterate through each name in the array for name in arr: # Calculate the length of the current name length = len(name) # Update the hash map with the current name added to its corresponding length group if length in lengthMap: lengthMap[length].append(name) else: lengthMap[length] = [name] # Update maxLen if the current length is greater if length > maxLen: maxLen = length # Retrieve all the names from the hash map for the length maxLen return lengthMap[maxLen]# Test casearr = ["neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"]longestNames = findLongestNames(arr)for name in longestNames: print(name, end=" ")print() |
C#
using System;using System.Collections.Generic;class Gfg { static List<string> findLongestNames(List<string> arr) { int maxLen = 0; Dictionary<int, List<string>> lengthMap = new Dictionary<int, List<string>>(); // Dictionary to store names grouped by their lengths // Iterate through each name in the array foreach (string name in arr) { int len = name.Length; // Calculate the length of the current name // Update the dictionary with the current name added to its corresponding length group if (!lengthMap.ContainsKey(len)) { lengthMap[len] = new List<string>(); } lengthMap[len].Add(name); // Update maxLen if the current length is greater if (len > maxLen) { maxLen = len; } } // Retrieve all the names from the dictionary for the length maxLen return lengthMap[maxLen]; } static void Main(string[] args) { List<string> arr = new List<string> { "neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool" }; List<string> longestNames = findLongestNames(arr); foreach (string name in longestNames) { Console.Write(name + " "); } Console.WriteLine(); }} |
Javascript
function findLongestNames(arr) { let maxLen = 0; // Hash map to store names grouped by their lengths let lengthMap = {}; // Iterate through each name in the array for (let name of arr) { // Calculate the length of the current name let length = name.length; // Update the hash map with the current name added to its corresponding length group if (length in lengthMap) { lengthMap[length].push(name); } else { lengthMap[length] = [name]; } // Update maxLen if the current length is greater if (length > maxLen) { maxLen = length; } } // Retrieve all the names from the hash map for the length maxLen return lengthMap[maxLen];}// Test caselet arr = ["neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"];let longestNames = findLongestNames(arr);for (let name of longestNames) { console.log(name + " ");}console.log("\n"); |
Output
neveropen StackOverFlow
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), for hash map
Sorting Approach:
Sort the array of names in descending order of length. Then, iterate through the sorted array and print all names with the same length as the first name in the sorted array (which will be the longest).
Below is the implementation:
C++
#include <iostream>#include <vector>#include <string>#include <algorithm>using namespace std;bool compareByLength(const string& a, const string& b) { return a.length() > b.length();}vector<string> findLongestNames(const vector<string>& arr) { // Create a copy of the input array to avoid modifying the original vector<string> sortedArr = arr; // Sort the array in descending order of length sort(sortedArr.begin(), sortedArr.end(), compareByLength); // Find the length of the first (longest) name int maxLength = sortedArr[0].length(); // Initialize a vector to store the longest names vector<string> longestNames; // Iterate through the sorted array and add names with the same length as the first name for (const string& name : sortedArr) { if (name.length() == maxLength) { longestNames.push_back(name); } else { break; // Names are sorted by length, so we can stop when a shorter name is encountered } } return longestNames;}//Driver codeint main() { // Input array of names vector<string> arr = {"neveropen", "FreeCodeCamp", "StackOverFlow", "MyCodeSchool"}; // Call the function to find the longest names vector<string> longestNames = findLongestNames(arr); // Print the longest names cout << "Longest Names: "; for (const string& name : longestNames) { cout << name << " "; } cout << endl; return 0;}//This Code is contributed by Veerendra_Singh_Rajpoot |
Output
Longest Names: neveropen StackOverFlow
Time Complexity: O(N log N)
Space Complexity: O(1)
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