Given two strings str1 and str2, the task is to count the minimum number of operations of the following three types on one of the two strings that are required to make str1 and str2 permutations of each other:
- Insert a character into the string.
- Remove a character from the string.
- Replace a character with another character from the string.
Note: All the above operations are of equal cost.
Examples:
Input: str1 = “neveropen”, str2 = “neveropenforcoder”
Output: 4
Explanation: Rearrange the string str2 to “neveropenforcedor”
Replace the value of str1[8] to ‘c’.
Replace the value of str1[10] to ‘d’.
Replace the value of str1[11] to ‘o’.
Replace the value of str1[12] to ‘r’.
Therefore, the required output is 4.Input: str1 = “neveropen”, str2 = “keeg”
Output: 1
Approach: The problem can be solved using Hashing to store the frequency of each character of both the string. Below are the observations to solve the problem:
X = Number of characters which are present in both string, str1 and str2.
N1 – X = Number of characters present only in str1.
N2 – X = Number of characters present only in str2.
Total number of replacement operations = min(N1 – X, N2 – X)
Total number of insert/Remove operations = max(N1 – X, N2 – X) – min(N1 – X, N2 – X).
Therefore, total number of operations = max(N1 – X, N2 – X),
Follow the steps below to solve the problem:
- Initialize two arrays, say freq1[] and freq2[] to store the frequency of all the characters of str1 and str2 respectively.
- Traverse both the strings and store the frequency of each character of both the strings in arrays freq1[] and freq2[] respectively.
- Traverse both the arrays freq1[] and freq2[].
- For every ith character, if freq1[i] exceeds freq2[i], then replace freq1[i] to freq1[i] – freq2[i] and set freq2[i] = 0 and vice-versa.
- Finally, calculate the sum of the arrays freq1[] and freq2[], and print the maximum between them as the answer
Below is the implement the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to minimize the count of// operations to make str1 and str2// permutations of each otherint ctMinEdits(string str1, string str2){ int N1 = str1.length(); int N2 = str2.length(); // Store the frequency of // each character of str1 int freq1[256] = { 0 }; for (int i = 0; i < N1; i++) { freq1[str1[i]]++; } // Store the frequency of // each character of str2 int freq2[256] = { 0 }; for (int i = 0; i < N2; i++) { freq2[str2[i]]++; } // Traverse the freq1[] and freq2[] for (int i = 0; i < 256; i++) { // If frequency of character in // str1 is greater than str2 if (freq1[i] > freq2[i]) { freq1[i] = freq1[i] - freq2[i]; freq2[i] = 0; } // Otherwise else { freq2[i] = freq2[i] - freq1[i]; freq1[i] = 0; } } // Store sum of freq1[] int sum1 = 0; // Store sum of freq2[] int sum2 = 0; for (int i = 0; i < 256; i++) { sum1 += freq1[i]; sum2 += freq2[i]; } return max(sum1, sum2);}// Driver Codeint main(){ string str1 = "neveropen"; string str2 = "neveropenforcoder"; cout << ctMinEdits(str1, str2);} |
Java
// Java program to implement // the above approach import java.util.*;import java.io.*;import java.lang.Math;class GFG{ // Function to minimize the count of // operations to make str1 and str2 // permutations of each other static int ctMinEdits(String str1, String str2) { int N1 = str1.length(); int N2 = str2.length(); // Store the frequency of // each character of str1 int freq1[] = new int[256]; Arrays.fill(freq1, 0); for(int i = 0; i < N1; i++) { freq1[str1.charAt(i)]++; } // Store the frequency of // each character of str2 int freq2[] = new int[256]; Arrays.fill(freq2, 0); for(int i = 0; i < N2; i++) { freq2[str2.charAt(i)]++; } // Traverse the freq1[] and freq2[] for(int i = 0; i < 256; i++) { // If frequency of character in // str1 is greater than str2 if (freq1[i] > freq2[i]) { freq1[i] = freq1[i] - freq2[i]; freq2[i] = 0; } // Otherwise else { freq2[i] = freq2[i] - freq1[i]; freq1[i] = 0; } } // Store sum of freq1[] int sum1 = 0; // Store sum of freq2[] int sum2 = 0; for(int i = 0; i < 256; i++) { sum1 += freq1[i]; sum2 += freq2[i]; } return Math.max(sum1, sum2); } // Driver Codepublic static void main(final String[] args) { String str1 = "neveropen"; String str2 = "neveropenforcoder"; System.out.println(ctMinEdits(str1, str2)); }}// This code is contributed by bikram2001jha |
Python3
# Python3 program to implement# the above approach # Function to minimize the count of# operations to make str1 and str2# permutations of each otherdef ctMinEdits(str1, str2): N1 = len(str1) N2 = len(str2) # Store the frequency of # each character of str1 freq1 = [0] * 256 for i in range(N1): freq1[ord(str1[i])] += 1 # Store the frequency of # each character of str2 freq2 = [0] * 256 for i in range(N2): freq2[ord(str2[i])] += 1 # Traverse the freq1[] and freq2[] for i in range(256): # If frequency of character in # str1 is greater than str2 if (freq1[i] > freq2[i]): freq1[i] = freq1[i] - freq2[i] freq2[i] = 0 # Otherwise else: freq2[i] = freq2[i] - freq1[i] freq1[i] = 0 # Store sum of freq1[] sum1 = 0 # Store sum of freq2[] sum2 = 0 for i in range(256): sum1 += freq1[i] sum2 += freq2[i] return max(sum1, sum2)# Driver Codestr1 = "neveropen"str2 = "neveropenforcoder"print(ctMinEdits(str1, str2))# This code is contributed by code_hunt |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to minimize the count of // operations to make str1 and str2 // permutations of each other static int ctMinEdits(string str1, string str2) { int N1 = str1.Length; int N2 = str2.Length; // Store the frequency of // each character of str1 int[] freq1 = new int[256]; freq1[0] = str1[0]; for(int i = 0; i < N1; i++) { freq1[str1[i]]++; } // Store the frequency of // each character of str2 int[] freq2 = new int[256]; freq2[0] = str2[0]; for(int i = 0; i < N2; i++) { freq2[str2[i]]++; } // Traverse the freq1[] and freq2[] for(int i = 0; i < 256; i++) { // If frequency of character in // str1 is greater than str2 if (freq1[i] > freq2[i]) { freq1[i] = freq1[i] - freq2[i]; freq2[i] = 0; } // Otherwise else { freq2[i] = freq2[i] - freq1[i]; freq1[i] = 0; } } // Store sum of freq1[] int sum1 = 0; // Store sum of freq2[] int sum2 = 0; for(int i = 0; i < 256; i++) { sum1 += freq1[i]; sum2 += freq2[i]; } return Math.Max(sum1, sum2); } // Driver Codepublic static void Main() { string str1 = "neveropen"; string str2 = "neveropenforcoder"; Console.WriteLine(ctMinEdits(str1, str2)); }}// This code is contributed by code_hunt |
Javascript
<script>// Javascript program to implement// the above approach// Function to minimize the count of// operations to make str1 and str2// permutations of each otherfunction ctMinEdits(str1, str2){ let N1 = str1.length; let N2 = str2.length; // Store the frequency of // each character of str1 let freq1 = new Array(256).fill(0); for(let i = 0; i < N1; i++) { freq1[str1[i].charCodeAt()]++; } // Store the frequency of // each character of str2 let freq2 = new Array(256).fill(0); for(let i = 0; i < N2; i++) { freq2[str2[i].charCodeAt()]++; } // Traverse the freq1[] and freq2[] for(let i = 0; i < 256; i++) { // If frequency of character in // str1 is greater than str2 if (freq1[i] > freq2[i]) { freq1[i] = freq1[i] - freq2[i]; freq2[i] = 0; } // Otherwise else { freq2[i] = freq2[i] - freq1[i]; freq1[i] = 0; } } // Store sum of freq1[] let sum1 = 0; // Store sum of freq2[] let sum2 = 0; for(let i = 0; i < 256; i++) { sum1 += freq1[i]; sum2 += freq2[i]; } return Math.max(sum1, sum2);}// Driver Code let str1 = "neveropen"; let str2 = "neveropenforcoder"; document.write(ctMinEdits(str1.split(''), str2.split(''))); </script> |
Output
4
Time Complexity: O(N)
Auxiliary Space: O(1)
Another method using Hashing: Unordered_map():
It made the code simpler and also reduce space requirement.
1. Create an unordered map mp to store the frequency of each character in both strings.
2. Loop over the str1 and update character frequency in the map.
3. Loop over the str2 and update characters frequency in the map.
4. Calculate the absolute difference in frequency for each character in the map.
5. Now store the sum of frequency differences in the two variables f1 and f2 for each strings.
6. Now, return the maximum of f1 and f2.
C++
// C++ code for above approach#include <bits/stdc++.h>using namespace std;// function to find the minimum no. of edits required.int CntMinEdit(string str1, string str2){ // counter variable int ans = 0; // unordered map to store the frequency of each character in both strings unordered_map<char,pair<int,int>>mp; // loop to store frequency of str1 in the map for(int i = 0; i<str1.length();i++) mp[str1[i]].first++; // loop to store frequency of str2 in the map for(int i = 0; i<str2.length();i++) mp[str2[i]].second++; // Variables to store the sum of frequency difference of each strings int f1 = 0, f2 = 0; // loop to calculate the difference in frequency for each character for(auto it:mp) { if(it.second.first>it.second.second) f1+=it.second.first-it.second.second; else f2 += it.second.second-it.second.first; } // return max of f1 and f2 return max(f2,f1);}// Driver codeint main(){ // Two strings taken string str1 = "neveropen"; string str2 = "neveropenforcoder"; // printing the output cout << CntMinEdit(str1, str2)<<endl; //Example 2 str1 = "neveropen"; str2 = "keeg"; cout << CntMinEdit(str1, str2)<<endl;} |
Java
// Java implementation of above approachimport java.util.*;public class Main { // function to find the minimum no. of edits required. public static int cntMinEdit(String str1, String str2) { int ans = 0; // HashMap to store the frequency HashMap<Character, int[]> mp = new HashMap<>(); // loop to store frequency of str1 in the map for(int i = 0; i<str1.length(); i++) { char c = str1.charAt(i); if(mp.containsKey(c)) { mp.get(c)[0]++; } else { mp.put(c, new int[]{1, 0}); } } // loop to store frequency of str2 in the map for(int i = 0; i<str2.length(); i++) { char c = str2.charAt(i); if(mp.containsKey(c)) { mp.get(c)[1]++; } else { mp.put(c, new int[]{0, 1}); } } // Variables to store the sum of frequency difference of each strings int f1 = 0, f2 = 0; // loop to calculate the difference in frequency for each character for(Map.Entry<Character, int[]> entry : mp.entrySet()) { if(entry.getValue()[0]>entry.getValue()[1]) f1+=(entry.getValue()[0] - entry.getValue()[1]); else f2+=(entry.getValue()[1] - entry.getValue()[0]); } // divide the answer by 2 since each change is counted 2 times. return Math.max(f1, f2); } // Driver code public static void main(String[] args) { // Two strings taken String str1 = "neveropen"; String str2 = "neveropenforcoder"; // Output System.out.println(cntMinEdit(str1, str2)); // Example 2 str1 = "neveropen"; str2 = "keeg"; // Output System.out.println(cntMinEdit(str1, str2)); }} |
Python
# python code for above approachdef CntMinEdit(str1, str2): # vairable to store the answer ans = 0 # map to store the frequency mp = {} # Loop to store frequency of str1 in the dictionary for char in str1: if char in mp: mp[char][0] += 1 else: mp[char] = [1, 0] # Loop to store frequency of str2 in the dictionary for char in str2: if char in mp: mp[char][1] += 1 else: mp[char] = [0, 1] # Variables to store the sum of frequency difference of each string f1, f2 = 0, 0 # Loop to calculate the difference in frequency for each character for char, freq in mp.items(): if freq[0] > freq[1]: f1 += freq[0] - freq[1] else: f2 += freq[1] - freq[0] # Return the max of f1 and f2 return max(f1, f2)# Driver codeif __name__ == "__main__": # Two strings taken str1 = "neveropen" str2 = "neveropenforcoder" # Printing the output print(CntMinEdit(str1, str2)) # Example 2 str1 = "neveropen" str2 = "keeg" print(CntMinEdit(str1, str2)) |
C#
using System;using System.Collections.Generic;class Program{ // function to find the minimum no. of edits required. static int CntMinEdit(string str1, string str2) { // dictionary to store the frequency of each character in both strings Dictionary<char, Tuple<int, int>> mp = new Dictionary<char, Tuple<int, int>>(); // loop to store frequency of str1 in the dictionary foreach (char c in str1) { if (!mp.ContainsKey(c)) mp = new Tuple<int, int>(0, 0); mp = Tuple.Create(mp.Item1 + 1, mp.Item2); } // loop to store frequency of str2 in the dictionary foreach (char c in str2) { if (!mp.ContainsKey(c)) mp = new Tuple<int, int>(0, 0); mp = Tuple.Create(mp.Item1, mp.Item2 + 1); } // Variables to store the sum of frequency difference of each strings int f1 = 0, f2 = 0; // loop to calculate the difference in frequency for each character foreach (var kvp in mp) { if (kvp.Value.Item1 > kvp.Value.Item2) f1 += kvp.Value.Item1 - kvp.Value.Item2; else f2 += kvp.Value.Item2 - kvp.Value.Item1; } // return max of f1 and f2 return Math.Max(f2, f1); } // Driver code static void Main(string[] args) { // Two strings taken string str1 = "neveropen"; string str2 = "neveropenforcoder"; // printing the output Console.WriteLine(CntMinEdit(str1, str2)); // Example 2 str1 = "neveropen"; str2 = "keeg"; Console.WriteLine(CntMinEdit(str1, str2)); }} |
Javascript
// Function to find the minimum number of edits required.function countMinEdit(str1, str2) { // Counter variable let ans = 0; // Map to store the frequency of each character in both strings const mp = new Map(); // Loop to store frequency of str1 in the map for (let i = 0; i < str1.length; i++) { if (!mp.has(str1[i])) { mp.set(str1[i], { first: 1, second: 0 }); } else { mp.get(str1[i]).first++; } } // Loop to store frequency of str2 in the map for (let i = 0; i < str2.length; i++) { if (!mp.has(str2[i])) { mp.set(str2[i], { first: 0, second: 1 }); } else { mp.get(str2[i]).second++; } } // Variables to store the sum of frequency difference of each string let f1 = 0, f2 = 0; // Loop to calculate the difference in frequency for each character for (const [key, value] of mp.entries()) { if (value.first > value.second) { f1 += value.first - value.second; } else { f2 += value.second - value.first; } } // Return the maximum of f1 and f2 return Math.max(f2, f1);}// Driver code// Example 1let str1 = "neveropen";let str2 = "neveropenforcoder";console.log(countMinEdit(str1, str2));// Example 2str1 = "neveropen";str2 = "keeg";console.log(countMinEdit(str1, str2)); |
Output
4 1
Time Complexity: O(N)
Auxiliary Space: O(N)
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