Lexicographically smallest K-length subsequence from a given string
Last Updated :
06 May, 2025
Given a string s of length n, the task is to find the lexicographically smallest k-length subsequence from the string s (where k < n).
Examples:
Input: s = "bbcaab", k = 3
Output: "aab"
Input: s = "aabdaabc", k = 3
Output: "aaa"
[Naive Approach] Generating all Subsequences - O(2^n) time and O(C(n,k)*k+n) space
The idea is to generate all possible subsequences of length k from the input string s, store them in an array, sort them lexicographically, and return the first (smallest) string.
C++
// C++ program to find Lexicographically smallest
// K-length subsequence from a given string
#include <bits/stdc++.h>
using namespace std;
// Recursive function to generate all subsequences of length k
void generateSub(string s, int i, int k, string curr,
vector<string>& result) {
int n = s.length();
// If we've reached end of string and string
// is of length k.
if (i == n) {
if (curr.length() == k) result.push_back(curr);
return;
}
// Include curr character
generateSub(s, i + 1, k, curr + s[i], result);
// Exclude curr character
generateSub(s, i + 1, k, curr, result);
}
string lexSmallest(string s, int k) {
int n = s.length();
vector<string> subsequences;
generateSub(s, 0, k, "", subsequences);
// Sort all subsequences lexicographically
sort(subsequences.begin(), subsequences.end());
// Return the lexicographically smallest subsequence
return subsequences[0];
}
int main() {
string s = "bbcaab";
int k = 3;
cout << lexSmallest(s, k);
return 0;
}
// Java program to find Lexicographically smallest
// K-length subsequence from a given string
import java.util.*;
class GfG {
// Recursive function to generate all subsequences of length k
static void generateSub(String s, int i, int k,
String curr, ArrayList<String> result) {
int n = s.length();
// If we've reached end of string and string
// is of length k.
if (i == n) {
if (curr.length() == k) result.add(curr);
return;
}
// Include curr character
generateSub(s, i + 1, k, curr + s.charAt(i), result);
// Exclude curr character
generateSub(s, i + 1, k, curr, result);
}
static String lexSmallest(String s, int k) {
int n = s.length();
ArrayList<String> subsequences = new ArrayList<>();
generateSub(s, 0, k, "", subsequences);
// Sort all subsequences lexicographically
Collections.sort(subsequences);
// Return the lexicographically smallest subsequence
return subsequences.get(0);
}
public static void main(String[] args) {
String s = "bbcaab";
int k = 3;
System.out.println(lexSmallest(s, k));
}
}
# Python program to find Lexicographically smallest
# K-length subsequence from a given string
# Recursive function to generate all subsequences of length k
def generateSub(s, i, k, curr, result):
n = len(s)
# If we've reached end of string and string
# is of length k.
if i == n:
if len(curr) == k:
result.append(curr)
return
# Include curr character
generateSub(s, i + 1, k, curr + s[i], result)
# Exclude curr character
generateSub(s, i + 1, k, curr, result)
def lexSmallest(s, k):
n = len(s)
subsequences = []
generateSub(s, 0, k, "", subsequences)
# Sort all subsequences lexicographically
subsequences.sort()
# Return the lexicographically smallest subsequence
return subsequences[0]
if __name__ == "__main__":
s = "bbcaab"
k = 3
print(lexSmallest(s, k))
// C# program to find Lexicographically smallest
// K-length subsequence from a given string
using System;
using System.Collections.Generic;
class GfG {
// Recursive function to generate all subsequences of length k
static void generateSub(string s, int i, int k,
string curr, List<string> result) {
int n = s.Length;
// If we've reached end of string and string
// is of length k.
if (i == n) {
if (curr.Length == k) result.Add(curr);
return;
}
// Include curr character
generateSub(s, i + 1, k, curr + s[i], result);
// Exclude curr character
generateSub(s, i + 1, k, curr, result);
}
static string lexSmallest(string s, int k) {
int n = s.Length;
List<string> subsequences = new List<string>();
generateSub(s, 0, k, "", subsequences);
// Sort all subsequences lexicographically
subsequences.Sort();
// Return the lexicographically smallest subsequence
return subsequences[0];
}
static void Main(string[] args) {
string s = "bbcaab";
int k = 3;
Console.WriteLine(lexSmallest(s, k));
}
}
// JavaScript program to find Lexicographically smallest
// K-length subsequence from a given string
// Recursive function to generate all subsequences of length k
function generateSub(s, i, k, curr, result) {
let n = s.length;
// If we've reached end of string and string
// is of length k.
if (i === n) {
if (curr.length === k) result.push(curr);
return;
}
// Include curr character
generateSub(s, i + 1, k, curr + s[i], result);
// Exclude curr character
generateSub(s, i + 1, k, curr, result);
}
function lexSmallest(s, k) {
let n = s.length;
let subsequences = [];
generateSub(s, 0, k, "", subsequences);
// Sort all subsequences lexicographically
subsequences.sort();
// Return the lexicographically smallest subsequence
return subsequences[0];
}
let s = "bbcaab";
let k = 3;
console.log(lexSmallest(s, k));
Time Complexity: O(2^n), at each index we have two choices: include the character or exclude it.
Auxiliary Space: O(C(n,k)*k + n), where C(n,k) equals the number of subsequences of length k, which can be created from a string of length n.
[Expected Approach] Using Stack - O(n) time and O(n) space
We use a stack that maintain the result so far. We process each character of the string from left to right, always maintaining the lexicographically smallest subsequence of the appropriate length seen so far. Whenever we encounter a smaller character than top of the stack, we remove larger characters to ensure the smallest possible subsequence, also ensuring we'll still have enough remaining characters to form a subsequence of length k.
Step by step approach:
- Process each character from left to right, maintaining a stack of the best possible subsequence.
- Pop characters from the stack when a smaller one is found and we have enough remaining characters.
- Add the current character to the stack only if we haven't collected k characters yet.
- Build the final result by emptying the stack and reversing the characters.
C++
// C++ program to find Lexicographically smallest
// K-length subsequence from a given string
#include <bits/stdc++.h>
using namespace std;
string lexSmallest(string s, int k) {
int n = s.length();
stack<char> st;
for (int i=0; i<n; i++) {
// While stack is not empty, and top
// of stack is less than current char
// while ensuring there are enough
// characters left to make subsequence
// of length k.
while (!st.empty() && st.top()>s[i] && st.size()-1+n-i >= k) {
st.pop();
}
// If size of stack is less than k.
if (st.size() < k) st.push(s[i]);
}
string res = "";
while (!st.empty()) {
res += st.top();
st.pop();
}
// Reverse the string to get the
// final resultant string
reverse(res.begin(), res.end());
return res;
}
int main() {
string s = "bbcaab";
int k = 3;
cout << lexSmallest(s, k);
return 0;
}
// Java program to find Lexicographically smallest
// K-length subsequence from a given string
import java.util.*;
class GfG {
static String lexSmallest(String s, int k) {
int n = s.length();
Stack<Character> st = new Stack<>();
for (int i = 0; i < n; i++) {
// While stack is not empty, and top
// of stack is less than current char
// while ensuring there are enough
// characters left to make subsequence
// of length k.
while (!st.isEmpty() && st.peek() > s.charAt(i)
&& st.size() - 1 + n - i >= k) {
st.pop();
}
// If size of stack is less than k.
if (st.size() < k) st.push(s.charAt(i));
}
StringBuilder res = new StringBuilder();
while (!st.isEmpty()) {
res.append(st.pop());
}
// Reverse the string to get the
// final resultant string
res.reverse();
return res.toString();
}
public static void main(String[] args) {
String s = "bbcaab";
int k = 3;
System.out.println(lexSmallest(s, k));
}
}
# Python program to find Lexicographically smallest
# K-length subsequence from a given string
def lexSmallest(s, k):
n = len(s)
st = []
for i in range(n):
# While stack is not empty, and top
# of stack is less than current char
# while ensuring there are enough
# characters left to make subsequence
# of length k.
while st and st[-1] > s[i] and len(st) - 1 + n - i >= k:
st.pop()
# If size of stack is less than k.
if len(st) < k:
st.append(s[i])
res = ""
while st:
res += st.pop()
# Reverse the string to get the
# final resultant string
res = res[::-1]
return res
if __name__ == "__main__":
s = "bbcaab"
k = 3
print(lexSmallest(s, k))
// C# program to find Lexicographically smallest
// K-length subsequence from a given string
using System;
using System.Collections.Generic;
using System.Text;
class GfG {
static string lexSmallest(string s, int k) {
int n = s.Length;
Stack<char> st = new Stack<char>();
for (int i = 0; i < n; i++) {
// While stack is not empty, and top
// of stack is less than current char
// while ensuring there are enough
// characters left to make subsequence
// of length k.
while (st.Count > 0 && st.Peek() > s[i] &&
st.Count - 1 + n - i >= k) {
st.Pop();
}
// If size of stack is less than k.
if (st.Count < k) st.Push(s[i]);
}
StringBuilder res = new StringBuilder();
while (st.Count > 0) {
res.Append(st.Pop());
}
// Reverse the string to get the
// final resultant string
char[] arr = res.ToString().ToCharArray();
Array.Reverse(arr);
return new string(arr);
}
static void Main(string[] args) {
string s = "bbcaab";
int k = 3;
Console.WriteLine(lexSmallest(s, k));
}
}
// JavaScript program to find Lexicographically smallest
// K-length subsequence from a given string
function lexSmallest(s, k) {
let n = s.length;
let st = [];
for (let i = 0; i < n; i++) {
// While stack is not empty, and top
// of stack is less than current char
// while ensuring there are enough
// characters left to make subsequence
// of length k.
while (st.length > 0 && st[st.length - 1] > s[i]
&& st.length - 1 + n - i >= k) {
st.pop();
}
// If size of stack is less than k.
if (st.length < k) st.push(s[i]);
}
let res = "";
while (st.length > 0) {
res += st.pop();
}
// Reverse the string to get the
// final resultant string
res = res.split("").reverse().join("");
return res;
}
let s = "bbcaab";
let k = 3;
console.log(lexSmallest(s, k));
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