Given some text lines in one string, each line is separated by ‘\n’ character. Print the last N lines. If the number of lines is less than N, then print all lines.
An approach for this problem has been already discussed in Set-1 where only 10 lines were printed. In this post, another approach is discussed for printing the last N lines.
Algorithm:
- Split the string around ‘\n’ using strtok.
- Store the individual strings in a vector.
- Print the last N strings from vector.
Below is the implementation of the above approach:
C++
// C++ program to print the last N lines#include <bits/stdc++.h>using namespace std;void PrintLast(string s, int t){ // Vector to store individual strings. vector<string> v; // Get a pointer to string. char* str = &s[0]; // Split the string around '\n'. char* token = strtok(str, "\n"); // Save all strings in the vector. while (token) { v.push_back(token); token = strtok(NULL, "\n"); } if (v.empty()) { cout << "ERROR: string doesn't contain '\\n' character\n"; return; } // If the string has t lines if (v.size() >= t) { for (int i = v.size() - t; i < v.size(); i++) cout << v[i] << endl; } else { for (int i = 0; i < v.size(); i++) cout << v[i] << endl; }}// Driver Codeint main(){ string s1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7\nstr8\nstr9" "\nstr10\nstr11\nstr12\nstr13\nstr14\nstr15\nstr16\nstr17" "\nstr18\nstr19\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25"; int n = 10; PrintLast(s1, n); return 0;} |
Java
// Java program to print the last N linesimport java.util.*;class GFG { static void printLast(String s, int t) { // Vector to store individual strings. // Save all strings in the vector. String[] v = s.split("\n"); if (v.length == 0) { System.out.println("ERROR: string doesn't " + "contain '\\n' character"); return; } // If the string has t lines if (v.length >= t) { for (int i = v.length - t; i < v.length; i++) { System.out.println(v[i]); } } else { for (int i = 0; i < v.length; i++) { System.out.println(v[i]); } } } // Driver Code public static void main(String[] args) { String s1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7" + "\nstr8\nstr9\nstr10\nstr11\nstr12\nstr13" + "\nstr14\nstr15\nstr16\nstr17\nstr18\nstr19" + "\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25"; int n = 10; printLast(s1, n); }}// This code is contributed by// sanjeev2552 |
Python3
# Python3 program to print the last N lines def PrintLast(s, t): # Vector to store individual strings. v = s.split('\n') if len(v) == 0: print("ERROR: string doesn't ", "contain '\\n' character\n") return # If the string has t lines elif len(v) >= t: for i in range(len(v) - t, len(v)): print(v[i]) else: for i in range(0, len(v)): print(v[i]) # Driver Code if __name__ == "__main__": s1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6" + \ "\nstr7\nstr8\nstr9\nstr10\nstr11" + \ "\nstr12\nstr13\nstr14\nstr15\nstr16" + \ "\nstr17\nstr18\nstr19\nstr20\nstr21" + \ "\nstr22\nstr23\nstr24\nstr25" n = 10 PrintLast(s1, n) # This code is contributed by Rituraj Jain |
C#
// C# program to print the last N linesusing System;class GFG { static void printLast(String s, int t) { // List to store individual strings. // Save all strings in the vector. String[] v = s.Split('\n'); if (v.Length == 0) { Console.WriteLine("ERROR: string doesn't " + "contain '\\n' character"); return; } // If the string has t lines if (v.Length >= t) { for (int i = v.Length - t; i < v.Length; i++) { Console.WriteLine(v[i]); } } else { for (int i = 0; i < v.Length; i++) { Console.WriteLine(v[i]); } } } // Driver Code public static void Main(String[] args) { String s1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7" + "\nstr8\nstr9\nstr10\nstr11\nstr12\nstr13" + "\nstr14\nstr15\nstr16\nstr17\nstr18\nstr19" + "\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25"; int n = 10; printLast(s1, n); }}// This code is contributed by Rajput-Ji |
Javascript
function PrintLast( s, t){ const v = s.split("\n"); if (v.length==0) { console.log( "ERROR: string doesn't contain '\\n' character\n"); return; } // If the string has t lines if (v.length >= t) { for (let i = v.length - t; i < v.length; i++) console.log(v[i]); } else { for (let i = 0; i < v.length; i++) console.log(v[i]); }} let s1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7\nstr8\nstr9\nstr10\nstr11\nstr12\nstr13\nstr14\nstr15\nstr16\nstr17\nstr18\nstr19\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25"; let n = 10; PrintLast(s1, n); // This code is contributed by garg28harsh. |
Output
str16 str17 str18 str19 str20 str21 str22 str23 str24 str25
Time Complexity: O(n)
Auxiliary Space: O(n)
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