# Modify string by inserting characters such that every K-length substring consists of unique characters only

• Difficulty Level : Expert
• Last Updated : 27 May, 2021

Given string S of size N consisting of K distinct characters and (N – K) ‘?’s, the task is to replace all ‘?’ with existing characters from the string such that every substring of size K has consisted of unique characters only. If it is not possible to do so, then print “-1”.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: S = “????abcd”, K = 4
Output: abcdabcd
Explanation:
Replacing the 4 ‘?’s with “abcd” modifies string S to “abcdabcd”, which satisfies the given condition.

Input: S = “?a?b?c”, K = 3
Output: bacbac
Explanation:
Replacing S[0] with ‘b’, S[2] with ‘c’ and S[4] with ‘a’ modifies string S to “bacbac”, which satisfies the given condition.

Approach: The idea is based on the observation that in the final resultant string, each character must appear after exactly K places, like the (K + 1)th character must be the same as 1st, (K + 2)th character must be the same as 2nd, and so on.

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to replace all '?'``// characters in a string such``// that the given conditions are satisfied``void` `fillString(string s, ``int` `k)``{``    ``unordered_map<``int``, ``char``> mp;` `    ``// Traverse the string to Map the``    ``// characters with respective positions``    ``for` `(``int` `i = 0; i < s.size(); i++) {``        ``if` `(s[i] != ``'?'``) {``            ``mp[i % k] = s[i];``        ``}``    ``}` `    ``// Traverse the string again and``    ``// replace all unknown characters``    ``for` `(``int` `i = 0; i < s.size(); i++) {` `        ``// If i % k is not found in``        ``// the Map M, then return -1``        ``if` `(mp.find(i % k) == mp.end()) {` `            ``cout << -1;``            ``return``;``        ``}` `        ``// Update S[i]``        ``s[i] = mp[i % k];``    ``}` `    ``// Print the string S``    ``cout << s;``}` `// Driver Code``int` `main()``{``    ``string S = ``"????abcd"``;``    ``int` `K = 4;``    ``fillString(S, K);` `    ``return` `0;``}`

## Java

 `// Java Program to implement``// the above approach``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``// Function to replace all '?'``    ``// characters in a string such``    ``// that the given conditions are satisfied``    ``static` `void` `fillString(String str, ``int` `k)``    ``{` `        ``char` `s[] = str.toCharArray();` `        ``HashMap mp = ``new` `HashMap<>();` `        ``// Traverse the string to Map the``        ``// characters with respective positions``        ``for` `(``int` `i = ``0``; i < s.length; i++) {``            ``if` `(s[i] != ``'?'``) {``                ``mp.put(i % k, s[i]);``            ``}``        ``}` `        ``// Traverse the string again and``        ``// replace all unknown characters``        ``for` `(``int` `i = ``0``; i < s.length; i++) {` `            ``// If i % k is not found in``            ``// the Map M, then return -1``            ``if` `(!mp.containsKey(i % k)) {``                ``System.out.println(-``1``);``                ``return``;``            ``}` `            ``// Update S[i]``            ``s[i] = mp.getOrDefault(i % k, s[i]);``        ``}` `        ``// Print the string S``        ``System.out.println(``new` `String(s));``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``String S = ``"????abcd"``;``        ``int` `K = ``4``;``        ``fillString(S, K);``    ``}``}` `// This code is contributed by Kingash.`

## Python3

 `# Python 3 program for the above approach` `# Function to replace all '?'``# characters in a string such``# that the given conditions are satisfied``def` `fillString(s, k):``    ``mp ``=` `{}` `    ``# Traverse the string to Map the``    ``# characters with respective positions``    ``for` `i ``in` `range``(``len``(s)):``        ``if` `(s[i] !``=` `'?'``):``            ``mp[i ``%` `k] ``=` `s[i]` `    ``# Traverse the string again and``    ``# replace all unknown characters``    ``s ``=` `list``(s)``    ``for` `i ``in` `range``(``len``(s)):``      ` `        ``# If i % k is not found in``        ``# the Map M, then return -1``        ``if` `((i ``%` `k) ``not` `in` `mp):``            ``print``(``-``1``)``            ``return` `        ``# Update S[i]``        ``s[i] ``=` `mp[i ``%` `k]` `    ``# Print the string S``    ``s ``=`   `''.join(s)``    ``print``(s)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``S ``=` `"????abcd"``    ``K ``=` `4``    ``fillString(S, K)` `    ``# This code is contributed by bgangwar59.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `  ``// Function to replace all '?'``  ``// characters in a string such``  ``// that the given conditions are satisfied``  ``static` `void` `fillString(``string` `str, ``int` `k)``  ``{` `    ``char``[] s = str.ToCharArray();` `    ``Dictionary<``int``,``    ``int``> mp = ``new` `Dictionary<``int``,``    ``int``>();` `    ``// Traverse the string to Map the``    ``// characters with respective positions``    ``for` `(``int` `i = 0; i < s.Length; i++) {``      ``if` `(s[i] != ``'?'``) {``        ``mp[i % k] = s[i];``      ``}``    ``}` `    ``// Traverse the string again and``    ``// replace all unknown characters``    ``for` `(``int` `i = 0; i < s.Length; i++) {` `      ``// If i % k is not found in``      ``// the Map M, then return -1``      ``if` `(!mp.ContainsKey(i % k)) {``        ``Console.WriteLine(-1);``        ``return``;``      ``}` `      ``// Update S[i]``      ``s[i] = (``char``)mp[i % k];` `    ``}` `    ``// Print the string S``    ``Console.WriteLine(``new` `string``(s));``  ``}` `  ``// Driver code``  ``static` `void` `Main()``  ``{``    ``string` `S = ``"????abcd"``;``    ``int` `K = 4;``    ``fillString(S, K);``  ``}``}` `// This code is contributed by susmitakundugoaldanga.`

## Javascript

 ``
Output:
`abcdabcd`

Time Complexity: O(N)
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up