Convert given string to another by minimum replacements of subsequences by its smallest character

Last Updated : 07 Aug, 2023

Given two strings A and B, the task is to count the minimum number of operations required to convert the string A to B. In one operation, select a subsequence from string A and convert every character of that subsequence to the smallest character present in it. If it is not possible to transform, then print “-1”.

Examples:

Input: A = “abcab”, B = “aabab”
Output: 2
Explanation:
Operation 1: Replacing characters from indices {2, 1} by the smallest character from those indices(i.e. ‘b’), transforms A to “abbab”.
Operation 2: Replacing characters from indices {1, 0}, by the smallest character from those indices(i.e. ‘a’), transforms A to “aabab”.
Therefore, the count of operations required to convert string A to B is 2.

Input: A = “aaa”, B = “aab”
Output: -1
Explanation:
There is no possible way to convert A to B as string A doesn’t contain ‘b’.

Approach: The approach is based on the idea that if any character at index i of string A is less than the character at index i of string B, then it is impossible to change A to B because changing a character to a character smaller than itself is not allowed.

Below is the implementation of the above approach:

C++

// C++ program for the above approach 
 
#include <bits/stdc++.h> 
using namespace std; 
 
// Function to return the minimum number 
// of operation 
void transformString(string str1, 
                    string str2) 
{ 
    // Storing data 
    int N = str1.length(); 
 
    vector<int> convChar[26]; 
    vector<int> str1array[26]; 
 
    // Initialize both arrays 
    for (int i = 0; i < 26; i++) { 
        vector<int> v; 
        convChar[i] = v; 
        str1array[i] = v; 
    } 
 
    // Stores the index of character 
    map<int, char> convertMap; 
 
    // Filling str1array, convChar 
    // and hashmap convertMap. 
    for (int i = 0; i < N; i++) { 
        str1array[str1[i] - 'a'].push_back(i); 
    } 
 
    for (int i = 0; i < N; i++) { 
 
        // Not possible to convert 
        if (str1[i] < str2[i]) { 
            cout << -1 << endl; 
            return; 
        } 
        else if (str1[i] == str2[i]) 
            continue; 
        else { 
            convChar[str2[i] - 'a'].push_back(i); 
            convertMap[i] = str2[i]; 
        } 
    } 
 
    // Calculate result 
    // Initializing return values 
    int ret = 0; 
    vector<vector<int> > retv; 
 
    // Iterating the character from 
    // the end 
    for (int i = 25; i >= 0; i--) { 
 
        vector<int> v = convChar[i]; 
 
        if (v.size() == 0) 
            continue; 
 
        // Increment the number of 
        // operations 
        ret++; 
        vector<int> v1 = str1array[i]; 
 
        // Not possible to convert 
        if (v1.size() == 0) { 
            cout << -1 << endl; 
            return; 
        } 
 
        // to check whether the final 
        // element has been added 
        // in set S or not. 
        bool isScompleted = false; 
 
        for (int j = 0; j < v1.size(); j++) { 
 
            // Check if v1[j] is present 
            // in hashmap or not 
            if (convertMap.find(v1[j]) 
                != convertMap.end()) { 
 
                char a = convertMap[v1[j]]; 
 
                // Already converted then 
                // then continue 
                if (a > i + 'a') 
                    continue; 
                else { 
                    v.push_back(v1[j]); 
                    isScompleted = true; 
                    retv.push_back(v); 
                    break; 
                } 
            } 
            else { 
                v.push_back(v1[j]); 
                isScompleted = true; 
                retv.push_back(v); 
                break; 
            } 
        } 
 
        // Not possible to convert 
        if (!isScompleted) { 
            cout << -1 << endl; 
            return; 
        } 
    } 
 
    // Print the result 
    cout << ret << endl; 
} 
 
// Driver Code 
int main() 
{ 
    // Given strings 
    string A = "abcab"; 
    string B = "aabab"; 
 
    // Function Call 
    transformString(A, B); 
    return 0; 
} 

Java

// Java program for the above approach 
import java.util.*; 
import java.lang.*; 
 
class GFG{ 
 
// Function to return the minimum number 
// of operation 
static void transformString(String str1, 
                            String str2) 
{ 
    // Storing data 
    int N = str1.length(); 
    ArrayList< 
    ArrayList<Integer>> convChar = new ArrayList<>(); 
    ArrayList< 
    ArrayList<Integer>> str1array = new ArrayList<>(); 
 
    // Initialize both arrays 
    for(int i = 0; i < 26; i++) 
    { 
        convChar.add(new ArrayList<>()); 
        str1array.add(new ArrayList<>()); 
    } 
 
    // Stores the index of character 
    Map<Integer, 
        Character> convertMap = new HashMap<>(); 
 
    // Filling str1array, convChar 
    // and hashmap convertMap. 
    for(int i = 0; i < N; i++) 
    { 
        str1array.get(str1.charAt(i) - 'a').add(i); 
    } 
 
    for(int i = 0; i < N; i++) 
    { 
         
        // Not possible to convert 
        if (str1.charAt(i) < str2.charAt(i)) 
        { 
            System.out.println(-1); 
            return; 
        } 
        else if (str1.charAt(i) == str2.charAt(i)) 
            continue; 
        else
        { 
            convChar.get(str2.charAt(i) - 'a').add(i); 
            convertMap.put(i,str2.charAt(i)); 
        } 
    } 
 
    // Calculate result 
    // Initializing return values 
    int ret = 0; 
    ArrayList< 
    ArrayList<Integer>> retv = new ArrayList<>(); 
 
    // Iterating the character from 
    // the end 
    for(int i = 25; i >= 0; i--) 
    { 
        ArrayList<Integer> v = convChar.get(i); 
 
        if (v.size() == 0) 
            continue; 
 
        // Increment the number of 
        // operations 
        ret++; 
        ArrayList<Integer> v1 = str1array.get(i); 
 
        // Not possible to convert 
        if (v1.size() == 0) 
        { 
            System.out.println(-1); 
            return; 
        } 
 
        // To check whether the final 
        // element has been added 
        // in set S or not. 
        boolean isScompleted = false; 
 
        for(int j = 0; j < v1.size(); j++) 
        { 
             
            // Check if v1[j] is present 
            // in hashmap or not 
            if (convertMap.containsKey(v1.get(j))) 
            { 
                char a = convertMap.get(v1.get(j)); 
 
                // Already converted then 
                // then continue 
                if (a > i + 'a') 
                    continue; 
                else
                { 
                    v.add(v1.get(j)); 
                    isScompleted = true; 
                    retv.add(v); 
                    break; 
                } 
            } 
            else
            { 
                v.add(v1.get(j)); 
                isScompleted = true; 
                retv.add(v); 
                break; 
            } 
        } 
 
        // Not possible to convert 
        if (!isScompleted) 
        { 
            System.out.println(-1); 
            return; 
        } 
    } 
 
    // Print the result 
    System.out.println(ret); 
} 
 
// Driver Code 
public static void main (String[] args) 
{ 
     
    // Given strings 
    String A = "abcab"; 
    String B = "aabab"; 
     
    // Function call 
    transformString(A, B); 
} 
} 
 
// This code is contributed by offbeat 

Python3

# Python3 program for the above approach
 
# Function to return the minimum number
# of operation
def transformString(str1, str2):
     
    # Storing data
    N = len(str1)
    convChar = []
    str1array = []
     
    # Initialize both arrays
    for i in range(26):
        convChar.append([])
        str1array.append([])
         
    # Stores the index of character
    convertMap = {}
     
    # Filling str1array, convChar 
    # and hashmap convertMap.
    for i in range(N):
        str1array[ord(str1[i]) -
                  ord('a')].append(i)
         
    for i in range(N):
         
        # Not possible to convert
        if (str1[i] < str2[i]):
            print(-1)
            return
        elif (str1[i] == str2[i]):
            continue
        else:
            convChar[ord(str2[i]) -
                     ord('a')].append(i)
            convertMap[i] = str2[i]
     
    # Calculate result 
    # Initializing return values 
    ret = 0
    retv = []
     
    # Iterating the character from 
    # the end
    for i in range(25, -1, -1):
        v = convChar[i]
         
        if (len(v) == 0):
            continue
         
        # Increment the number of 
        # operations 
        ret += 1;
        v1 = str1array[i]
         
        # Not possible to convert
        if (len(v1) == 0):
            print(-1)
            return
         
        # To check whether the final 
        # element has been added 
        # in set S or not.
        isScompleted = False
         
        for j in range(len(v1)):
             
            # Check if v1[j] is present 
            # in hashmap or not 
            if (v1[j] in convertMap):
                a = v1[j]
                 
                # Already converted then 
                # then continue 
                if (a > i + ord('a')):
                    continue
                else:
                    v.append(v1[j])
                    isScompleted = True
                    retv.append(v)
                    break
            else:
                v.append(v1[j])
                isScompleted = True
                retv.append(v)
                break
         
        # Not possible to convert 
        if (isScompleted == False):
            print(-1)
            return
     
    # Print the result
    print(ret)
             
# Driver Code
A = "abcab"
B = "aabab"
 
# Function call
transformString(A, B)
 
# This code is contributed by dadi madhav

C#

// C# program for the above approach  
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to return the minimum number  
// of operation  
static void transformString(string str1, string str2)  
{ 
     
    // Storing data  
    int N = str1.Length;  
    List<List<int>> convChar = new List<List<int>>();  
    List<List<int>> str1array = new List<List<int>>();  
   
    // Initialize both arrays  
    for(int i = 0; i < 26; i++)  
    {  
        convChar.Add(new List<int>());  
        str1array.Add(new List<int>());  
    }  
   
    // Stores the index of character  
    Dictionary<int, 
               char> convertMap = new Dictionary<int, 
                                                 char>();  
   
    // Filling str1array, convChar  
    // and hashmap convertMap.  
    for(int i = 0; i < N; i++)  
    {  
        str1array[str1[i] - 'a'].Add(i);  
    }  
   
    for(int i = 0; i < N; i++)  
    {  
         
        // Not possible to convert  
        if (str1[i] < str2[i])  
        {  
            Console.WriteLine(-1);  
            return;  
        }  
        else if (str1[i] == str2[i])  
            continue;  
        else
        {  
            convChar[str2[i] - 'a'].Add(i);  
            convertMap[i] = str2[i];  
        }  
    }  
   
    // Calculate result  
    // Initializing return values  
    int ret = 0;  
    List<List<int>> retv = new List<List<int>>();  
   
    // Iterating the character from  
    // the end  
    for(int i = 25; i >= 0; i--)  
    {  
        List<int> v = convChar[i];  
   
        if (v.Count == 0)  
            continue;  
   
        // Increment the number of  
        // operations  
        ret++;  
        List<int> v1 = str1array[i];  
   
        // Not possible to convert  
        if (v1.Count == 0)  
        {  
            Console.WriteLine(-1);  
            return;  
        }  
   
        // To check whether the final  
        // element has been added  
        // in set S or not.  
        bool isScompleted = false;  
   
        for(int j = 0; j < v1.Count; j++)  
        {  
             
            // Check if v1[j] is present  
            // in hashmap or not  
            if (convertMap.ContainsKey(v1[j]))  
            {  
                char a = convertMap[v1[j]];  
   
                // Already converted then  
                // then continue  
                if (a > i + 'a')  
                    continue;  
                else
                {  
                    v.Add(v1[j]);  
                    isScompleted = true;  
                    retv.Add(v);  
                    break;  
                }  
            }  
            else
            {  
                v.Add(v1[j]);  
                isScompleted = true;  
                retv.Add(v);  
                break;  
            }  
        }  
   
        // Not possible to convert  
        if (!isScompleted)  
        {  
            Console.WriteLine(-1);  
            return;  
        }  
    }  
   
    // Print the result  
    Console.WriteLine(ret);  
}
 
// Driver code
static void Main() 
{
    // Given strings  
    string A = "abcab";  
    string B = "aabab";  
       
    // Function call  
    transformString(A, B);
}
}
 
// This code is contributed by divyesh072019

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to return the minimum number
// of operation
function transformString(str1, str2)
{
     
    // Storing data
    let N = str1.length
    let convChar = []
    let str1array = []
     
    // Initialize both arrays
    for(let i = 0; i < 26; i++)
    {
        convChar.push([])
        str1array.push([])
    }
         
    // Stores the index of character
    let convertMap = new Map()
     
    // Filling str1array, convChar 
    // and hashmap convertMap.
    for(let i = 0; i < N; i++)
    {
        str1array[str1.charCodeAt(i) - 'a'.charCodeAt(0)].push(i)
    }
         
    for(let i = 0; i < N; i++)
    {
         
        // Not possible to convert
        if (str1.charCodeAt(i) < str2.charCodeAt(i))
        {
            document.write(-1)
            return
        }
        else if (str1[i] == str2[i])
            continue
        else
            convChar[str2.charCodeAt(i) - 'a'.charCodeAt(0)].push(i)
            convertMap[i] = str2[i]
    }
     
    // Calculate result 
    // Initializing return values 
    let ret = 0
    let retv = []
     
    // Iterating the character from 
    // the end
    for(let i = 25; i >= 0; i--)
    {
        let v = convChar[i]
         
        if (v.length == 0)
            continue
         
        // Increment the number of 
        // operations 
        ret += 1;
        v1 = str1array[i]
         
        // Not possible to convert
        if (v1.length == 0){
            document.write(-1)
            return
        }
         
        // To check whether the final 
        // element has been added 
        // in set S or not.
        isScompleted = false
         
        for(let j = 0; j < v1.length; j++)
        {
             
            // Check if v1[j] is present 
            // in hashmap or not 
            if (convertMap.has(v1[j])){
                let a = v1[j]
                 
                // Already converted then 
                // then continue 
                if (a > i + 'a'.charCodeAt(0))
                    continue
                else{
                    v.push(v1[j])
                    isScompleted = true
                    retv.append(v)
                    break
                }
            }
            else{
                v.push(v1[j])
                isScompleted = true
                retv.push(v)
                break
            }
        }
         
        // Not possible to convert 
        if (isScompleted == false){
            document.write(-1)
            return
        }
    }
     
    // Print the result
    document.write(ret)
}
             
// Driver Code
let A = "abcab"
let B = "aabab"
 
// Function call
transformString(A, B)
 
// This code is contributed by shinjanpatra
 
</script>

Output

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

Check if a palindromic matrix can be formed from the given array elements

manoj_26

Improve

Article Tags :

Practice Tags :

Convert given string to another by minimum replacements of subsequences by its smallest character

C++

Java

Python3

C#

Javascript

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