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Reduce Hamming distance by swapping two characters

Given two strings S and T. Find the positions of two letters to be swapped in S so that the hamming distance between strings S and T is as small as possible. 
Hamming Distance: Hamming distance between two strings S and T of the same length, which is defined as the number of positions in which S and T have different characters 

Examples: 

Input : S = "permanent", T = "pergament"
Output: 4 6

Input : S = "double" T = "bundle"
Output : 4 1
  • Explanation 1: Initially, the hamming distance between S and T is 2(at 4 and 6). After swapping the letters at positions 4 and 6 it becomes “pernament”. Here, the hamming distance is only 1. This is the minimum possible. 
  • Explanation 2: Before swapping: “double” “bundle”, distance = 4 
    After swapping: “boudle” “bundle”, distance = 2

Approach : 

In the given string, hamming distance can be decreased by at most two because, only two characters can be moved. 

  • Case-I: Decrease distance by two is possible, if there are two positions with the same two letters in two strings but that appear in different order(like “bundle” and “double”). 
  • Case-II: Decrease by one is possible by “fixing” the characters that are on wrong positions(like in “permanent” and “pergament”, here n stands in wrong pair with m and there is also unmatched m, which we can fix). 
  • Case-III: If not possible to decrease the hamming distance, print -1.

Implementation: 

  • Case-I: To decrease the distance by two, create a 2-d array dp[i][j](i is s[] – ‘a’ and j is t[] – ‘a’) and assign it to the index of i in string S. If repeated pair is found, print the corresponding indexes. 
  • Case-II: To decrease the distance by one, maintain two arrays A[] and B[] 

Implementation:

C++

// C++ code to decrease
// hamming distance using swap.
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 26
 
// Function to return the
// swapped indexes to get
// minimum hamming distance.
void Swap(string s, string t, int n)
{
    int dp[MAX][MAX];
    memset(dp, -1, sizeof dp);
 
    // Find the initial hamming
    // distance
    int tot = 0;
    for (int i = 0; i < n; i++)
        if (s[i] != t[i])
            tot++;
     
    // Case-I: To decrease distance
    // by two
    for (int i = 0; i < n; i++) {
 
        // ASCII values of present
        // character.
        int a = s[i] - 'a';
        int b = t[i] - 'a';
 
        if (a == b)
            continue;
 
        // If two same letters appear
        // in different positions
        // print their indexes
        if (dp[a][b] != -1) {
            cout << i + 1 << " "
                << dp[a][b] + 1 << endl;
            return;
        }
 
        // Store the index of
        // letters which is
        // in wrong position
        dp[b][a] = i;
    }
 
    // Case:II
    int A[MAX], B[MAX];
    memset(A, -1, sizeof A);
    memset(B, -1, sizeof B);
 
    for (int i = 0; i < n; i++) {
        int a = s[i] - 'a';
        int b = t[i] - 'a';
        if (a == b)
            continue;
 
        // If misplaced letter
        // is found, print its
        // original index and
        // its new index
        if (A[b] != -1) {
            cout << i + 1 << " " <<
                  A[b] + 1 << endl;
            return;
        }
 
        if (B[a] != -1) {
            cout << i + 1 << " " <<
                  B[a] + 1 << endl;
            return;
        }
 
        // Store the index of
        // letters in wrong position
        A[a] = i;
        B[b] = i;
    }
 
    // Case-III
    cout << -1 << endl;
}
 
// Driver code
int main()
{
    string S = "permanent";
    string T = "pergament";
    int n = S.length();
 
    if (S == "" || T == "")
        cout << "Required string is empty.";
    else
        Swap(S, T, n);
 
    return 0;
}

Java

// Java code to decrease
// hamming distance using swap.
import java.util.Arrays;
 
class GFG
{
     
static final int MAX = 26;
 
// Function to return the
// swapped indexes to get
// minimum hamming distance.
static void Swap(String s, String t, int n)
{
    int dp[][] = new int[MAX][MAX];
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < MAX; j++)
        {
            dp[i][j]=-1;
        }
    }
 
    // Find the initial hamming
    // distance
    int tot = 0;
    for (int i = 0; i < n; i++)
        if (s.charAt(i) != t.charAt(i))
            tot++;
     
    // Case-I: To decrease distance
    // by two
    for (int i = 0; i < n; i++)
    {
 
        // ASCII values of present
        // character.
        int a = s.charAt(i)- 'a';
        int b = t.charAt(i) - 'a';
 
        if (a == b)
            continue;
 
        // If two same letters appear
        // in different positions
        // print their indexes
        if (dp[a][b] != -1)
        {
            System.out.println(i + 1 + " " +
                            (dp[a][b] + 1));
            return;
        }
 
        // Store the index of
        // letters which is
        // in wrong position
        dp[b][a] = i;
    }
 
    // Case:II
    int A[] = new int[MAX], B[] = new int[MAX];
    Arrays.fill(A, -1);
    Arrays.fill(B, -1);
 
    for (int i = 0; i < n; i++)
    {
        int a = s.charAt(i)- 'a';
        int b = t.charAt(i) - 'a';
        if (a == b)
            continue;
 
        // If misplaced letter
        // is found, print its
        // original index and
        // its new index
        if (A[b] != -1)
        {
            System.out.println(i + 1 + " " +
                                (A[b] + 1));
            return;
        }
 
        if (B[a] != -1)
        {
            System.out.println(i + 1 + " " +
                                 (B[a] + 1));
            return;
        }
 
        // Store the index of
        // letters in wrong position
        A[a] = i;
        B[b] = i;
    }
 
    // Case-III
    System.out.println(-1);
}
 
// Driver code
public static void main(String[] args)
{
    String S = "permanent";
    String T = "pergament";
    int n = S.length();
 
    if (S == "" || T == "")
        System.out.println("Required string is empty.");
    else
        Swap(S, T, n);
}
}
 
// This code is contributed by Rajput-Ji

Python 3

# Python 3 code to decrease
# hamming distance using swap.
MAX = 26
 
# Function to return the swapped indexes
# to get minimum hamming distance.
def Swap(s, t, n):
 
    dp = [[-1 for x in range(MAX)]
              for y in range(MAX)]
 
    # Find the initial hamming
    # distance
    tot = 0;
    for i in range(n):
        if (s[i] != t[i]):
            tot += 1
     
    # Case-I: To decrease distance
    # by two
    for i in range(n) :
 
        # ASCII values of present
        # character.
        a = ord(s[i]) - ord('a')
        b = ord(t[i]) - ord('a')
 
        if (a == b):
            continue
 
        # If two same letters appear
        # in different positions
        # print their indexes
        if (dp[a][b] != -1) :
            print(i + 1," ", dp[a][b] + 1)
            return
 
        # Store the index of
        # letters which is
        # in wrong position
        dp[b][a] = i
 
    # Case:II
    A = [-1] * MAX
    B = [-1] * MAX
 
    for i in range(n) :
        a = ord(s[i]) - ord('a')
        b = ord(t[i]) - ord('a')
        if (a == b):
            continue
 
        # If misplaced letter is found,
        # print its original index and
        # its new index
        if (A[b] != -1) :
            print( i + 1, A[b] + 1)
            return
 
        if (B[a] != -1) :
            print( i + 1, B[a] + 1)
            return
 
        # Store the index of
        # letters in wrong position
        A[a] = i
        B[b] = i
     
    # Case-III
    print( "-1" )
 
# Driver code
if __name__ == "__main__":
    S = "permanent"
    T = "pergament"
    n = len(S)
 
    if (S == "" or T == ""):
        print("Required string is empty.")
    else:
        Swap(S, T, n)
 
# This code is contributed
# by ChitraNayal

C#

// C# code to decrease
// hamming distance using swap.
using System;
 
class GFG
{
     
static readonly int MAX = 26;
 
    // Function to return the
    // swapped indexes to get
    // minimum hamming distance.
    static void Swap(string s, string t, int n)
    {
        int [,]dp = new int[MAX, MAX];
        for(int i = 0; i < MAX; i++)
        {
            for(int j = 0; j < MAX; j++)
            {
                dp[i, j]=-1;
            }
        }
 
        // Find the initial hamming
        // distance
        int tot = 0;
        for (int i = 0; i < n; i++)
            if (s[i] != t[i])
                tot++;
 
        // Case-I: To decrease distance
        // by two
        for (int i = 0; i < n; i++)
        {
 
            // ASCII values of present
            // character.
            int a = s[i] - 'a';
            int b = t[i] - 'a';
 
            if (a == b)
                continue;
 
            // If two same letters appear
            // in different positions
            // print their indexes
            if (dp[a,b] != -1)
            {
                Console.WriteLine(i + 1 + " " +
                                (dp[a, b] + 1));
                return;
            }
 
            // Store the index of
            // letters which is
            // in wrong position
            dp[b,a] = i;
        }
 
        // Case:II
        int []A = new int[MAX]; int []B = new int[MAX];
        for (int i = 0; i < MAX; i++)
        {
            A[i]=-1;
            B[i]=-1;
        }
 
        for (int i = 0; i < n; i++)
        {
            int a = s[i] - 'a';
            int b = t[i] - 'a';
            if (a == b)
                continue;
 
            // If misplaced letter
            // is found, print its
            // original index and
            // its new index
            if (A[b] != -1)
            {
                Console.WriteLine(i + 1 + " " +
                                    (A[b] + 1));
                return;
            }
 
            if (B[a] != -1)
            {
                Console.WriteLine(i + 1 + " " +
                                    (B[a] + 1));
                return;
            }
 
            // Store the index of
            // letters in wrong position
            A[a] = i;
            B[b] = i;
        }
 
        // Case-III
        Console.WriteLine(-1);
    }
 
    // Driver code
    public static int Main()
    {
        string S = "permanent";
        string T = "pergament";
        int n = S.Length;
 
        if (S == "" || T == "")
            Console.WriteLine("Required string is empty.");
        else
            Swap(S, T, n);
 
        return 0;
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript code to decrease
// hamming distance using swap.
     
    let MAX = 26;
     
// Function to return the
// swapped indexes to get
// minimum hamming distance. 
function Swap(s,t,n)
{
       let dp=new Array(MAX);
    for(let i = 0; i < MAX; i++)
    {
        dp[i]=new Array(MAX);
        for(let j = 0; j < MAX; j++)
        {
            dp[i][j]=-1;
        }
    }
    // Find the initial hamming
    // distance
    let tot = 0;
    for (let i = 0; i < n; i++)
        if (s[i] != t[i])
            tot++;
       
    // Case-I: To decrease distance
    // by two
    for (let i = 0; i < n; i++)
    {
   
        // ASCII values of present
        // character.
        let a = s[i].charCodeAt(0)- 'a'.charCodeAt(0);
        let b = t[i].charCodeAt(0) - 'a'.charCodeAt(0);
   
        if (a == b)
            continue;
   
        // If two same letters appear
        // in different positions
        // print their indexes
        if (dp[a][b] != -1)
        {
            document.write(i + 1 + " " +
                            (dp[a][b] + 1)+"<br>");
            return;
        }
   
        // Store the index of
        // letters which is
        // in wrong position
        dp[b][a] = i;
    }
   
    // Case:II
    let A = new Array(MAX), B = new Array(MAX);
    for(let i=0;i<MAX;i++)
    {
        A[i]=-1;
        B[i]=-1;
    }
   
    for (let i = 0; i < n; i++)
    {
        let a = s[i].charCodeAt(0)- 'a'.charCodeAt(0);
        let b = t[i].charCodeAt(0) - 'a'.charCodeAt(0);
        if (a == b)
            continue;
   
        // If misplaced letter
        // is found, print its
        // original index and
        // its new index
        if (A[b] != -1)
        {
            document.write(i + 1 + " " +
                                (A[b] + 1)+"<br>");
            return;
        }
   
        if (B[a] != -1)
        {
            document.write(i + 1 + " " +
                                 (B[a] + 1)+"<br>");
            return;
        }
   
        // Store the index of
        // letters in wrong position
        A[a] = i;
        B[b] = i;
    }
   
    // Case-III
    document.write(-1);
}
 
// Driver code
let S = "permanent";
let T = "pergament";
let n = S.length;
if (S == "" || T == "")
    document.write("Required string is empty.");
else
    Swap(S, T, n);
 
     
    // This code is contributed by avanitrachhadiya2155
     
</script>

Output

6 4

Time Complexity: O(n), where n is the length of the given strings.
Auxiliary Space: O(MAX * MAX), where MAX is a predefined value. Here, MAX = 26


 


Reffered: https://www.geeksforgeeks.org

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Category:
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