Count ways to select two N sized Substrings differing by one letter

Given a string str, the task is to find the number of ways two non-overlapping substrings of length N can be selected out of a string that differs by just one letter.

Examples:

Input: str = “abcd”, N = 1
Output: 6
Explanation: Possible non overlapping substrings pairs are (“a”, “b”), (“b”, “c”), (“c”, “d”), (“a”, “c”), (“a”, “d”) and (“b”, “d”).

Input: str = “abcb”, N = 2
Output: 1
Explanation: The only possible substring pair is (“ab”, “cb”).

Naive Approach: To solve the problem follow the below idea:

The approach is to find the number of ways two substrings of a given string, “str”, of length “n” can differ by just one letter. The code first initializes a variable “count” to store the number of ways and a variable “len” to store the length of the given string. Then, the code uses two nested for loops to iterate through all possible substrings of length “n” within the given string. For each pair of substrings found, the code compares each letter of the two substrings. If the two substrings differ by just one letter, the code increases the count by 1. Finally, the code returns the count as the result.

Follow the steps of the code:

• Declare a variable “count” and initialize it with 0. This variable will be used to store the number of ways the substrings differ by just one letter.
• Use a for loop to iterate through all substrings of length “N” in the input string “str”. The loop starts from index 0 and goes up to the (length of the string – N)
  • Within the for loop, use the substring method to store the first substring in a variable “s1”.
• Use another for loop to iterate through all substrings of length “N” starting from the (i + 1)^th position.
  • Within the second for loop, use the substring method to store the second substring in a variable “s2”.
• Declare a variable “diff” and initialize it with 0. This variable will be used to store the number of different letters between the two substrings.
• Use another for loop to iterate through the letters of the two substrings.
  • Within the third for loop, compare each letter of the two substrings. 
  • If they are different, increment the “diff” variable by 1.
• Check if the value of “diff” is equal to 1, if so increase the “count” variable by 1.
• Exit the second for loop
• Exit the first for loop
• Return the value of “count”

Below is the implementation for the above approach:

C++14

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

int countWays(string str, int n)
{
    int len = str.length();

    // To store the number of ways
    int count = 0;

    // Iterate through all the
    // substrings of length n
    for (int i = 0; i <= len - n; i++) {

        // To store the first substring
        string s1 = str.substr(i, n);

        // Iterate through all the
        // substrings of length n starting
        // from (i + 1)th position
        for (int j = i + 1; j <= len - n; j++) {

            // To store the second substring
            string s2 = str.substr(j, n);

            // Check if the two substrings
            // differ by just one letter
            int diff = 0;
            for (int k = 0; k < n; k++)
                if (s1[k] != s2[k])
                    diff++;

            // If the two substrings differ
            // by just one letter, increase
            // the count
            if (diff == 1)
                count++;
        }
    }

    return count;
}

// Driver code
int main()
{
    string str = "abcb";
    int N = 2;

    // Function Call
    cout << countWays(str, N);
    return 0;
}

Java

// Java code for the above approach:
import java.util.*;

class GFG {
    public static int countWays(String str, int n)
    {
        int len = str.length();

        // To store the number of ways
        int count = 0;

        // Iterate through all the
        // substrings of length n
        for (int i = 0; i <= len - n; i++) {

            // To store the first substring
            String s1 = str.substring(i, i + n);

            // Iterate through all the
            // substrings of length n starting
            // from (i + 1)th position
            for (int j = i + 1; j <= len - n; j++) {

                // To store the second substring
                String s2 = str.substring(j, j + n);

                // Check if the two substrings
                // differ by just one letter
                int diff = 0;
                for (int k = 0; k < n; k++) {
                    if (s1.charAt(k) != s2.charAt(k)) {
                        diff++;
                    }
                }

                // If the two substrings differ
                // by just one letter, increase
                // the count
                if (diff == 1) {
                    count++;
                }
            }
        }
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        String str = "abcb";
        int N = 2;

        // Function Call
        System.out.println(countWays(str, N));
    }
}
// This Code is Contributed by Prasad Kandekar(prasad264)

Python

# Java code for the above approach
def countWays(str, n):

    len_ = len(str)

    # To store the number of ways
    count = 0

    # Iterate through all the
    # substrings of length n
    for i in range(len_ - n + 1):

        # To store the first substring
        s1 = str[i:i+n]

        # Iterate through all the
        # substrings of length n starting
        # from (i + 1)th position
        for j in range(i + 1, len_ - n + 1):

              # To store the second substring
            s2 = str[j:j+n]

            # Check if the two substrings
            # differ by just one letter
            diff = 0
            for k in range(n):
                if s1[k] != s2[k]:
                    diff += 1

            # If the two substrings differ
            # by just one letter, increase
            # the count
            if diff == 1:
                count += 1
    return count

# Driver Code
str_ = "abcb"
N = 2

# Function Call
print(countWays(str_, N))

# This Code is Contributed by Prasad Kandekar(prasad264)

C#

// C# code for the above approach:
using System;

public class GFG {
    static int CountWays(string str, int n)
    {
        int len = str.Length;

          // To store the number of ways
        int count = 0;

          // Iterate through all the
        // substrings of length n
        for (int i = 0; i <= len - n; i++) {

              // To store the first substring
            string s1 = str.Substring(i, n);

              // Iterate through all the
            // substrings of length n starting
            // from (i + 1)th position
            for (int j = i + 1; j <= len - n; j++) {

                  // To store the second substring
                string s2 = str.Substring(j, n);

                  // Check if the two substrings
                // differ by just one letter
                int diff = 0;
                for (int k = 0; k < n; k++) {
                    if (s1[k] != s2[k])
                        diff++;
                }
                if (diff == 1)
                    count++;
            }
        }

        return count;
    }

      // Driver code
    static void Main(string[] args)
    {
        string str = "abcb";
        int N = 2;

          // Function Call
        Console.WriteLine(CountWays(str, N));
    }
}
// This Code is Contributed by Prasad Kandekar(prasad264)

Javascript

// Javascript code for the above approach:
function countWays( str,  n)
{
    let len = str.length;

    // To store the number of ways
    let count = 0;

    // Iterate through all the
    // substrings of length n
    for (let i = 0; i <= len - n; i++) {

        // To store the first substring
        let s1 = str.substr(i);

        // Iterate through all the
        // substrings of length n starting
        // from (i + 1)th position
        for (let j = i + 1; j <= len - n; j++) {

            // To store the second substring
            let s2 = str.substr(j);

            // Check if the two substrings
            // differ by just one letter
            let diff = 0;
            for (let k = 0; k < n; k++)
                if (s1[k] != s2[k])
                    diff++;

            // If the two substrings differ
            // by just one letter, increase
            // the count
            if (diff == 1)
                count++;
        }
    }

    return count;
}

// Driver code
let str = "abcb";
let N = 2;

// Function Call
console.log(countWays(str, N));

  // this code is contributed by poojaagarwal2.

Output

1

Time Complexity: O((len – N)*(len – N) * N), where N is the required substring length and len is the length of string str.
Auxiliary Space: O(1)