# Lexicographically smallest string formed by appending a character from first K characters of a string | Set 2

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2022

Given a string str consisting of lowercase alphabets and an integer K, you can perform the following operations on str

1. Initialize an empty string X = “”.
2. Take any character from the first K characters of str and append it to X.
3. Remove the chosen character from str.
4. Repeat the above steps while there are characters left in str.

The task is to generate X such that it is lexicographically smallest possible then print the generated string. Examples:

Input: str = “geek”, K = 2 Output: eegk Operation 1: str = “gek”, X = “e” Operation 2: str = “gk”, X = “ee” Operation 3: str = “k”, X = “eeg” Operation 4: str = “”, X = “eegk” Input: str = “geeksforgeeks”, K = 5 Output: eefggeekkorss

Approach: In order to get the lexicographically smallest string, we need to take the minimum character from the first K characters every time we choose a character from str. To do that, we can put the first K characters in a priority_queue (min-heap) and then choose the smallest character and append it to X. Then, push the next character in str to the priority queue and repeat the process until there are characters left to process. Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the lexicographically``// smallest required string``string getSmallestStr(string S, ``int` `K)``{` `    ``// Initially empty string``    ``string X = ``""``;` `    ``// min heap of characters``    ``priority_queue<``char``, vector<``char``>, greater<``char``> > pq;` `    ``// Length of the string``    ``int` `i, n = S.length();` `    ``// K cannot be greater than``    ``// the size of the string``    ``K = min(K, n);` `    ``// First push the first K characters``    ``// into the priority_queue``    ``for` `(i = 0; i < K; i++)``        ``pq.push(S[i]);` `    ``// While there are characters to append``    ``while` `(!pq.empty()) {` `        ``// Append the top of priority_queue to X``        ``X += pq.top();` `        ``// Remove the top element``        ``pq.pop();` `        ``// Push only if i is less than``        ``// the size of string``        ``if` `(i < S.length())``            ``pq.push(S[i]);` `        ``i++;``    ``}` `    ``// Return the generated string``    ``return` `X;``}` `// Driver code``int` `main()``{``    ``string S = ``"geeksforgeeks"``;``    ``int` `K = 5;` `    ``cout << getSmallestStr(S, K);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.PriorityQueue;` `class` `GFG``{` `    ``// Function to return the lexicographically``    ``// smallest required string``    ``static` `String getSmallestStr(String S, ``int` `K)``    ``{` `        ``// Initially empty string``        ``String X = ``""``;` `        ``// min heap of characters``        ``PriorityQueue pq = ``new` `PriorityQueue<>();` `        ``// Length of the string``        ``int` `i, n = S.length();` `        ``// K cannot be greater than``        ``// the size of the string``        ``K = Math.min(K, n);` `        ``// First push the first K characters``        ``// into the priority_queue``        ``for` `(i = ``0``; i < K; i++)``            ``pq.add(S.charAt(i));` `        ``// While there are characters to append``        ``while` `(!pq.isEmpty())``        ``{` `            ``// Append the top of priority_queue to X``            ``X += pq.peek();` `            ``// Remove the top element``            ``pq.remove();` `            ``// Push only if i is less than``            ``// the size of string``            ``if` `(i < S.length())``                ``pq.add(S.charAt(i));``                ` `            ``i++;``        ``}` `        ``// Return the generated string``        ``return` `X;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String S = ``"geeksforgeeks"``;``        ``int` `K = ``5``;``        ``System.out.println(getSmallestStr(S, K));``    ``}``}` `// This code is contributed by``// sanjeev2552`

Output:

`eefggeekkorss`

Time Complexity: O(nlogn) where n is the length of the string.
Auxiliary Space: O(K), as extra space of size K is used to build priorityQueue

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