Given a string S consisting of N lowercase characters and character ‘?’ and a positive integer K, the task is to replace each character ‘?’ with some lowercase alphabets such that the given string becomes a period of K. If it is not possible to do so, then print “-1”.
A string is said to be a period of K if and only if the length of the string is a multiple of K and for all possible value of i over the range [0, K) the value S[i + K], S[i + 2*K], S[i + 3*K], …, remains the same.
Examples:
Input: S = “ab??”, K = 2
Output: ababInput: S = “??????”, K = 3
Output: aaaaaa
Naive Approach: The given approach can also be solved by generating all possible combination of strings by replacing each character ‘?’ with any lowercase characters and print that string that have each substring of size K is the same.
Time Complexity: O(26M), where M is the number of ‘?’ in the string S.
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by traversing the string in a way such that the first, second, third, and so on characters are traversed and if all the characters are ‘?’ then replace it with character ‘a’, Otherwise, if there exists only one distinct character at each respective position then replace ‘?’ with that character, Otherwise, the string can’t be modified as per the given criteria and hence, print “-1”. Follow the steps below to solve the problem:
- Iterate a loop over the range [0, K] using the variable i and perform the following steps:
- Initialize a map, say M to store the frequency of characters of substring at positions i.
- Traverse the given string over the range [i, N] using the variable j with an increment of K and store the frequency of the character S[j] by 1 in the map M.
- After completing the above steps perform the following:
- If the size of the map is greater than 2, then print “-1” and break out of the loop.
- Otherwise, if the size of the map is 2, then replace each ‘?’ with that different character.
- Otherwise, replace all the ‘?’ with the character ‘a’.
- After completing the above steps, if the string can be modified, then print the string S as the result string.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include "bits/stdc++.h"using namespace std;// Function to modify the given string// such that string S is a period of Kstring modifyString(string& S, int K){ int N = S.length(); // Iterate over the range [0, K] for (int i = 0; i < K; i++) { // Stores the frequency of the // characters S[i + j*K] map<char, int> M; // Iterate over the string with // increment of K for (int j = i; j < N; j += K) { M[S[j]]++; } // Print "-1" if (M.size() > 2) { return "-1"; } else if (M.size() == 1) { if (M['?'] != 0) { // Replace all characters // of the string with '?' // to 'a' to make it smallest for (int j = i; j < N; j += K) { S[j] = 'a'; } } } // Otherwise else if (M.size() == 2) { char ch; // Find the character other // than '?' for (auto& it : M) { if (it.first != '?') { ch = it.first; } } // Replace all characters // of the string with '?' // to character ch for (int j = i; j < N; j += K) { S[j] = ch; } } // Clear the map M M.clear(); } // Return the modified string return S;}// Driver Codeint main(){ string S = "ab??"; int K = 2; cout << modifyString(S, K); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to modify the given String // such that String S is a period of K static String modifyString(char[] S, int K) { int N = S.length; // Iterate over the range [0, K] for (int i = 0; i < K; i++) { // Stores the frequency of the // characters S[i + j*K] HashMap<Character,Integer> M = new HashMap<>(); // Iterate over the String with // increment of K for (int j = i; j < N; j += K) { if(M.containsKey(S[j])){ M.put(S[j], M.get(S[j])+1); } else{ M.put(S[j], 1); } } // Print "-1" if (M.size() > 2) { return "-1"; } else if (M.size() == 1) { if (M.get('?') != 0) { // Replace all characters // of the String with '?' // to 'a' to make it smallest for (int j = i; j < N; j += K) { S[j] = 'a'; } } } // Otherwise else if (M.size() == 2) { char ch=' '; // Find the character other // than '?' for (Map.Entry<Character,Integer> entry : M.entrySet()) { if (entry.getKey() != '?') { ch = entry.getKey(); } } // Replace all characters // of the String with '?' // to character ch for (int j = i; j < N; j += K) { S[j] = ch; } } // Clear the map M M.clear(); } // Return the modified String return String.valueOf(S); } // Driver Code public static void main(String[] args) { String S = "ab??"; int K = 2; System.out.print(modifyString(S.toCharArray(), K)); }}// This code is contributed by umadevi9616 |
Python3
# python 3 program for the above approach# Function to modify the given string# such that string S is a period of Kdef modifyString(S,K): N = len(S) S = list(S) # Iterate over the range [0, K] for i in range(K): # Stores the frequency of the # characters S[i + j*K] M = {} # Iterate over the string with # increment of K for j in range(i,N,K): if S[j] in M: M[S[j]] += 1 else: M[S[j]] = 1 # Print "-1" if (len(M) > 2): return "-1" elif (len(M) == 1): if (M['?'] != 0): # Replace all characters # of the string with '?' # to 'a' to make it smallest for j in range(i,N,K): S[j] = 'a' # Otherwise elif(len(M) == 2): ch = '' # Find the character other # than '?' for key,value in M.items(): if (key != '?'): ch = key # Replace all characters # of the string with '?' # to character ch for j in range(i,N,K): S[j] = ch # Clear the map M M.clear() S = ''.join(S) # Return the modified string return S# Driver Codeif __name__ == '__main__': S = "ab??" K = 2 print(modifyString(S, K)) # This code is contributed by ipg2016107. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG { // Function to modify the given String // such that String S is a period of K static String modifyString(char[] S, int K) { int N = S.Length; // Iterate over the range [0, K] for (int i = 0; i < K; i++) { // Stores the frequency of the // characters S[i + j*K] Dictionary<char, int> M = new Dictionary<char, int>(); // Iterate over the String with // increment of K for (int j = i; j < N; j += K) { if (M.ContainsKey(S[j])) { M.Add(S[j], M[S[j]] + 1); } else { M.Add(S[j], 1); } } // Print "-1" if (M.Count > 2) { return "-1"; } else if (M.Count == 1) { if (M['?'] != 0) { // Replace all characters // of the String with '?' // to 'a' to make it smallest for (int j = i; j < N; j += K) { S[j] = 'a'; } } } // Otherwise else if (M.Count == 2) { char ch = ' '; // Find the character other // than '?' foreach (KeyValuePair<char, int> entry in M) { if (entry.Key != '?') { ch = entry.Key; } } // Replace all characters // of the String with '?' // to character ch for (int j = i; j < N; j += K) { S[j] = ch; } } // Clear the map M M.Clear(); } // Return the modified String return String.Join("",S); } // Driver Code public static void Main(String[] args) { String S = "ab??"; int K = 2; Console.Write(modifyString(S.ToCharArray(), K)); }}// This code is contributed by umadevi9616 |
Javascript
<script>// JavaScript program for the above approach// Function to modify the given string// such that string S is a period of Kfunction modifyString(S, K){ let N = S.length; // Iterate over the range [0, K] for(let i = 0; i < K; i++) { // Stores the frequency of the // characters S[i + j*K] let M = new Map(); // Iterate over the string with // increment of K for(let j = i; j < N; j += K) { if (M.has(S[j])) { M.set(M.get(S[j]), M.get(S[j]) + 1); } else { M.set(S[j], 1); } } // Print "-1" if (M.size > 2) { return "-1"; } else if (M.size == 1) { if (M.has('?')) { // Replace all characters // of the string with '?' // to 'a' to make it smallest for(let j = i; j < N; j += K) { S = S.replace(S[j], 'a'); } } } // Otherwise else if (M.size == 2) { let ch; // Find the character other // than '?' for(let it of M.keys()) { if (it != '?') { ch = it; } } // Replace all characters // of the string with '?' // to character ch for(let j = i; j < N; j += K) { S = S.replace(S[j], ch); } } // Clear the map M M.clear(); } // Return the modified string return S;}// Driver Codelet S = "ab??";let K = 2;document.write(modifyString(S, K));// This code is contributed by Potta Lokesh</script> |
Output:
abab
Time Complexity: O(N)
Auxiliary Space: O(N)
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