Given a string S of length N, the task is to find the lexicographically smallest subsequence of length (N – 1), i.e. by removing a single character from the given string.
Examples:
Input: S = “neveropen”
Output: “eeksforneveropen”
Explanation: Lexicographically smallest subsequence possible is “eeksforneveropen”.Input: S = “zxvsjas”
Output: “xvsjas”
Explanation: Lexicographically smallest subsequence possible is “xvsjas”.
Naive Approach: The simplest approach is to generate all possible subsequences of length (N – 1) from the given string and store all subsequences in an array. Now, sort the array and print the string at 0th position for the smallest lexicographically subsequence.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the lexicographically// smallest subsequence of length N-1void firstSubsequence(string s){ vector<string> allsubseq; string k; // Generate all subsequence of // length N-1 for (int i = 0; i < s.length(); i++) { // Store main value of string str k = s; // Erasing element at position i k.erase(i, 1); allsubseq.push_back(k); } // Sort the vector sort(allsubseq.begin(), allsubseq.end()); // Print first element of vector cout << allsubseq[0];}// Driver Codeint main(){ // Given string S string S = "neveropen"; // Function Call firstSubsequence(S); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the lexicographically// smallest subsequence of length N-1static void firstSubsequence(String s){ Vector<String> allsubseq = new Vector<>(); // Generate all subsequence of // length N-1 for(int i = 0; i < s.length(); i++) { String k = ""; // Store main value of String str for(int j = 0; j < s.length(); j++) { if (i != j) { k += s.charAt(j); } } allsubseq.add(k); } // Sort the vector Collections.sort(allsubseq); // Print first element of vector System.out.print(allsubseq.get(0));}// Driver Codepublic static void main(String[] args){ // Given String S String S = "neveropen"; // Function Call firstSubsequence(S);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approach# Function to find the lexicographically# smallest subsequence of length N-1def firstSubsequence(s): allsubseq = [] k = [] # Generate all subsequence of # length N-1 for i in range(len(s)): # Store main value of string str k = [i for i in s] # Erasing element at position i del k[i] allsubseq.append("".join(k)) # Sort the vector allsubseq = sorted(allsubseq) # Print first element of vector print(allsubseq[0])# Driver Codeif __name__ == '__main__': # Given string S S = "neveropen" # Function Call firstSubsequence(S)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the lexicographically// smallest subsequence of length N-1static void firstSubsequence(string s){ List<string> allsubseq = new List<string>(); // Generate all subsequence of // length N-1 for(int i = 0; i < s.Length; i++) { string k = ""; // Store main value of string str for(int j = 0; j < s.Length; j++) { if (i != j) { k += s[j]; } } allsubseq.Add(k); } // Sort the vector allsubseq.Sort(); // Print first element of vector Console.WriteLine(allsubseq[0]);}// Driver Codepublic static void Main(){ // Given string S string S = "neveropen"; // Function Call firstSubsequence(S);}}// This code is contributed by ipg2016107 |
Javascript
<script>// Javascript program for the above approach// Function to find the lexicographically// smallest subsequence of length N-1function firstSubsequence(s){ let allsubseq = []; // Generate all subsequence of // length N-1 for(let i = 0; i < s.length; i++) { let k = ""; // Store main value of String str for(let j = 0; j < s.length; j++) { if (i != j) { k += s[j]; } } allsubseq.push(k); } // Sort the vector (allsubseq).sort(); // Print first element of vector document.write(allsubseq[0]);}// Driver Code// Given String Slet S = "neveropen";// Function CallfirstSubsequence(S);// This code is contributed by patel2127</script> |
Output:
eeksforneveropen
Time Complexity: O(N *N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to iterate over the string and check if the ith character is greater that (i + 1)th character, then simply remove the ith character and print the remaining string. Otherwise, remove the last element and print the desired subsequence.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the lexicographically// smallest subsequence of length N-1void firstSubsequence(string s){ // Store index of character // to be deleted int isMax = -1; // Traverse the string for (int i = 0; i < s.length() - 1; i++) { // If ith character > (i + 1)th // character then store it if (s[i] > s[i + 1]) { isMax = i; break; } } // If any character found in non // alphabetical order then remove it if (isMax >= 0) { s.erase(isMax, 1); } // Otherwise remove last character else { s.erase(s.length() - 1, 1); } // Print the resultant subsequence cout << s;}// Driver Codeint main(){ // Given string S string S = "neveropen"; // Function Call firstSubsequence(S); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the lexicographically// smallest subsequence of length N-1static void firstSubsequence(String s){ // Store index of character // to be deleted int isMax = -1; // Traverse the String for(int i = 0; i < s.length() - 1; i++) { // If ith character > (i + 1)th // character then store it if (s.charAt(i) > s.charAt(i + 1)) { isMax = i; break; } } // If any character found in non // alphabetical order then remove it if (isMax >= 0) { s = s.substring(0, isMax) + s.substring(isMax + 1); // s.rerase(isMax, 1); } // Otherwise remove last character else { //s.erase(s.length() - 1, 1); s = s.substring(0, s.length() - 1); } // Print the resultant subsequence System.out.print(s);}// Driver Codepublic static void main(String[] args){ // Given String S String S = "neveropen"; // Function Call firstSubsequence(S);}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach# Function to find the lexicographically# smallest subsequence of length N-1def firstSubsequence(s): # Store index of character # to be deleted isMax = -1 # Traverse the String for i in range(len(s)): # If ith character > (i + 1)th # character then store it if (s[i] > s[i + 1]): isMax = i break # If any character found in non # alphabetical order then remove it if (isMax >= 0): s = s[0 : isMax] + s[isMax + 1 : len(s)] # s.rerase(isMax, 1); # Otherwise remove last character else: # s.erase(s.length() - 1, 1); s = s[0: s.length() - 1] # Print the resultant subsequence print(s)# Driver Codeif __name__ == '__main__': # Given String S S = "neveropen" # Function Call firstSubsequence(S)# This code is contributed by Princi Singh |
C#
// C# program for the above approachusing System;class GFG{// Function to find the lexicographically// smallest subsequence of length N-1static void firstSubsequence(String s){ // Store index of character // to be deleted int isMax = -1; // Traverse the String for(int i = 0; i < s.Length - 1; i++) { // If ith character > (i + 1)th // character then store it if (s[i] > s[i + 1]) { isMax = i; break; } } // If any character found in non // alphabetical order then remove it if (isMax >= 0) { s = s.Substring(0, isMax) + s.Substring(isMax + 1); // s.rerase(isMax, 1); } // Otherwise remove last character else { //s.erase(s.Length - 1, 1); s = s.Substring(0, s.Length - 1); } // Print the resultant subsequence Console.Write(s);}// Driver Codepublic static void Main(String[] args){ // Given String S String S = "neveropen"; // Function Call firstSubsequence(S);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Function to find the lexicographically// smallest subsequence of length N-1function firstSubsequence(s){ // Store index of character // to be deleted let isMax = -1; // Traverse the String for(let i = 0; i < s.length - 1; i++) { // If ith character > (i + 1)th // character then store it if (s[i] > s[i + 1]) { isMax = i; break; } } // If any character found in non // alphabetical order then remove it if (isMax >= 0) { s = s.substring(0, isMax) + s.substring(isMax + 1); // s.rerase(isMax, 1); } // Otherwise remove last character else { //s.erase(s.length() - 1, 1); s = s.substring(0, s.length - 1); } // Print the resultant subsequence document.write(s);}// Driver Code// Given String Slet S = "neveropen";// Function CallfirstSubsequence(S);// This code is contributed by unknown2108</script> |
Output:
eeksforneveropen
Time Complexity: O(N)
Auxiliary Space: O(1)
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