Given a string and array of strings, find whether the array contains a string with one character difference from the given string. Array may contain strings of different lengths.
Examples:
Input : str = "banana"
arr[] = {"bana", "apple", "banaba",
bonanzo", "banamf"}
Output :True
Explanation:-There is only a one character difference
between banana and banaba
Input : str = "banana"
arr[] = {"bana", "apple", "banabb", bonanzo",
"banamf"}
Output : False
We traverse through given string and check for every string in arr. Follow the two steps as given below for every string contained in arr:-
- Check whether the string contained in arr is of the same length as the target string.
- If yes, then check if there is only one character mismatch, if yes then return true else return false.
Implementation:
C++
// C++ program to find if given string is present// with one mismatch.#include <bits/stdc++.h>using namespace std;bool check(vector<string> list, string s){ int n = (int)list.size(); // If the array is empty if (n == 0) return false; for (int i = 0; i < n; i++) { // If sizes are same if (list[i].size() != s.size()) continue; bool diff = false; for (int j = 0; j < (int)list[i].size(); j++) { if (list[i][j] != s[j]) { // If first mismatch if (!diff) diff = true; // Second mismatch else { diff = false; break; } } } if (diff) return true; } return false;}// Driver codeint main(){ vector<string> s; s.push_back("bana"); s.push_back("apple"); s.push_back("banacb"); s.push_back("bonanza"); s.push_back("banamf"); cout << check(s, "banana"); return 0;} |
Java
import java.util.*;// Java program to find if// given string is present// with one mismatch.class GFG{ static boolean check(Vector<String> list, String s) { int n = (int) list.size(); // If the array is empty if (n == 0) { return false; } for (int i = 0; i < n; i++) { // If sizes are same if (list.get(i).length() != s.length()) { continue; } boolean diff = false; for (int j = 0; j < (int) list.get(i).length(); j++) { if (list.get(i).charAt(j) != s.charAt(j)) { // If first mismatch if (!diff) { diff = true; } // Second mismatch else { diff = false; break; } } } if (diff) { return true; } } return false; } // Driver code public static void main(String[] args) { Vector<String> s = new Vector<>(); s.add("bana"); s.add("apple"); s.add("banacb"); s.add("bonanza"); s.add("banamf"); System.out.println(check(s, "banana") == true ? 1 : 0); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python 3 program to find if given # string is present with one mismatch.def check(list, s): n = len(list) # If the array is empty if (n == 0): return False for i in range(0, n, 1): # If sizes are same if (len(list[i]) != len(s)): continue diff = False for j in range(0, len(list[i]), 1): if (list[i][j] != s[j]): # If first mismatch if (diff == False): diff = True # Second mismatch else: diff = False break if (diff): return True return False# Driver codeif __name__ == '__main__': s = [] s.append("bana") s.append("apple") s.append("banacb") s.append("bonanza") s.append("banamf") print(int(check(s, "banana")))# This code is contributed by# Sahil_shelangia |
C#
// C# program to find if// given string is present// with one mismatch.using System; using System.Collections.Generic;public class GFG{ static bool check(List<String> list, String s) { int n = (int) list.Count; // If the array is empty if (n == 0) { return false; } for (int i = 0; i < n; i++) { // If sizes are same if (list[i].Length != s.Length) { continue; } bool diff = false; for (int j = 0; j < (int) list[i].Length; j++) { if (list[i][j] != s[j]) { // If first mismatch if (!diff) { diff = true; } // Second mismatch else { diff = false; break; } } } if (diff) { return true; } } return false; } // Driver code public static void Main(String[] args) { List<String> s = new List<String>(); s.Add("bana"); s.Add("apple"); s.Add("banacb"); s.Add("bonanza"); s.Add("banamf"); Console.WriteLine(check(s, "banana") == true ? 1 : 0); }}// This code has been contributed by 29AjayKumar |
Javascript
<script> // Javascript program to find if // given string is present // with one mismatch. function check(list, s) { let n = list.length; // If the array is empty if (n == 0) { return false; } for (let i = 0; i < n; i++) { // If sizes are same if (list[i].length != s.length) { continue; } let diff = false; for (let j = 0; j < list[i].length; j++) { if (list[i][j] != s[j]) { // If first mismatch if (!diff) { diff = true; } // Second mismatch else { diff = false; break; } } } if (diff) { return true; } } return false; } let s = []; s.push("bana"); s.push("apple"); s.push("banacb"); s.push("bonanza"); s.push("banamf"); document.write(check(s, "banana") == true ? 1 : 0); </script> |
Output
0
Time complexity: O(n2)
Auxiliary space: O(1)
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Approach#2: Using sorting
One more approach is to sort the array and then loop through each string in the array. For each string, compare it with the given string character by character and count the number of mismatches. If the count of mismatches is equal to 1, then return True.
Algorithm
1. Sort the array
2. Loop through each string in the array
3. Initialize a variable ‘count’ to 0
4. Loop through each character in the string
5. If the character in the given string at index ‘i’ is not equal to the character in the current string at index ‘i’, increment ‘count’ by 1
6. If ‘count’ becomes more than 1 or if the length of the string in the array is less than the length of the given string minus 1, continue to the next string
7. If the length of the string in the array is greater than the length of the given string plus 1, break out of the loop
8. If ‘count’ is equal to 1 and the length of the string in the array is either equal to the length of the given string or the length of the given string minus 1, return True
9. If the loop completes without finding a match, return False
C++
#include <bits/stdc++.h>using namespace std;// Function to check if there is one mismatch between str and any string in arrbool has_one_mismatch(string str, vector<string> arr) { sort(arr.begin(), arr.end()); for (string s : arr) { int count = 0; int i = 0; while (i < s.length() && count <= 1) { if (str[i] != s[i]) { count += 1; } i += 1; } if (count > 1 || s.length() < str.length()-1) { continue; } if (s.length() > str.length()+1) { break; } if (count == 1 && (s.length() == str.length() || s.length() == str.length()-1)) { return true; } } return false;}// Driver codeint main() { string str = "banana"; vector<string> arr = {"bana", "apple", "banaba", "bonanzo", "banamf"}; if(has_one_mismatch(str, arr)) cout<<"True"; else cout<<"False";} |
Java
import java.util.*;public class Main { // Function to check if there is one mismatch between // str and any string in arr public static boolean has_one_mismatch(String str, ArrayList<String> arr) { Collections.sort(arr); for (String s : arr) { int count = 0; int i = 0; while (i < s.length() && count <= 1) { if (str.charAt(i) != s.charAt(i)) { count += 1; } i += 1; } if (count > 1 || s.length() < str.length() - 1) { continue; } if (s.length() > str.length() + 1) { break; } if (count == 1 && (s.length() == str.length() || s.length() == str.length() - 1)) { return true; } } return false; } // driver code public static void main(String[] args) { String str = "banana"; ArrayList<String> arr = new ArrayList<String>( Arrays.asList("bana", "apple", "banaba", "bonanzo", "banamf")); if (has_one_mismatch(str, arr)) System.out.println("True"); else System.out.println("False"); }} |
Python3
def has_one_mismatch(str, arr): arr.sort() for s in arr: count = 0 i = 0 while i < len(s) and count <= 1: if str[i] != s[i]: count += 1 i += 1 if count > 1 or len(s) < len(str)-1: continue if len(s) > len(str)+1: break if count == 1 and (len(s) == len(str) or len(s) == len(str)-1): return True return Falsestr = "banana"arr= ["bana", "apple", "banaba", "bonanzo", "banamf"]print(has_one_mismatch(str, arr)) |
C#
using System;using System.Collections.Generic;using System.Linq;class Program { // Function to check if there is one mismatch between // str and any string in arr static bool HasOneMismatch(string str, List<string> arr) { arr.Sort(); foreach(string s in arr) { int count = 0; int i = 0; while (i < s.Length && count <= 1) { if (str[i] != s[i]) { count += 1; } i += 1; } if (count > 1 || s.Length < str.Length - 1) { continue; } if (s.Length > str.Length + 1) { break; } if (count == 1 && (s.Length == str.Length || s.Length == str.Length - 1)) { return true; } } return false; } // driver code static void Main(string[] args) { string str = "banana"; List<string> arr = new List<string>{ "bana", "apple", "banaba", "bonanzo", "banamf" }; if (HasOneMismatch(str, arr)) Console.WriteLine("True"); else Console.WriteLine("False"); }} |
Javascript
// JavaScript code to check if a string has one mismatchfunction has_one_mismatch(str, arr) { arr.sort(); for (let s of arr) { let count = 0; let i = 0; while (i < s.length && count <= 1) { if (str[i] != s[i]) { count += 1; } i += 1; } if (count > 1 || s.length < str.length - 1) { continue; } if (s.length > str.length + 1) { break; } if (count == 1 && (s.length == str.length || s.length == str.length - 1)) { return true; } } return false;}let str = "banana";let arr = ["bana", "apple", "banaba", "bonanzo", "banamf"];console.log(has_one_mismatch(str, arr)); |
Output
True
Time Complexity: O(nmlog(m)), where n is the length of the array and m is the length of the longest string in the array or the given string
Auxiliary Space: O(1)
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