Count occurrences of a sub-string with one variable character

Last Updated : 24 Mar, 2023

Given two strings a and b, and an integer k which is the index in b at which the character can be changed to any other character, the task is to check if b is a sub-string in a and print out how many times b occurs in a in total after replacing the b[k] with every possible lowercase character of English alphabet.

Examples:

Input: a = "geeks", b = "ee", k = 1
Output: 1 Replace b[1] with 'k' and "ek" is a sub-string in "geeks" "ee" is also a sub-string in "geeks" Hence the total count is 2
Input: a = "dogdog", b = "dop", k = 2
Output: 2 Replace b[2] with 'g', "dog" is a sub-string in "dogdog" which appears twice.

Approach: Make a list of all possible versions of the string b by iterating through all the lowercase letters and replacing the k^th i.e. b[k] character in b with the current character. Then count the number of occurrence of the new string b in the original string a and store it in a variable count. After all the lowercase characters are used, print the count.

Below is the implementation of the above approach:

C++

// c++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// function to implement count
int count_(string a, string var)
{
    int c = 0;
    int l = var.length();
    for (int i = 0; i < a.length() - l; i++)
        if (a.substr(i, l) == var)
            c++;
    return c;
}
// Function to return the count of occurrences
int countOccurrence(string a, string b, int k)
{

    // Generate all possible substrings to
    // be searched
    vector<string> x;
    for (int i = 0; i < 26; i++) {
        string temp = b;
        temp[k] = 'a' + i;
        x.push_back(temp);
    }
    // Now search every substring 'a' and
    // increment count
    int count = 0;
    for (auto var : x) {
        if (a.find(var) != string::npos)
            count += count_(a, var);
    }
    return count;
}
int main()
{
    string a = "geeks", b = "ee";
    int k = 1;
    cout << countOccurrence(a, b, k) << endl;
    return 0;
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)

Java

// Java program for the above approach

import java.util.*;

public class Main {
    // function to implement count
    public static int count_(String a, String var) {
        int c = 0;
        int l = var.length();
        for (int i = 0; i < a.length() - l; i++) {
            if (a.substring(i, i + l).equals(var)) {
                c++;
            }
        }
        return c;
    }

    // Function to return the count of occurrences
    public static int countOccurrence(String a, String b, int k) {
        // Generate all possible substrings to be searched
        List<String> x = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            char[] temp = b.toCharArray();
            temp[k] = (char)('a' + i);
            x.add(String.valueOf(temp));
        }

        // Now search every substring 'a' and increment count
        int count = 0;
        for (String var : x) {
            if (a.indexOf(var) != -1) {
                count += count_(a, var);
            }
        }
        return count;
    }

    public static void main(String[] args) {
        String a = "geeks";
        String b = "ee";
        int k = 1;
        System.out.println(countOccurrence(a, b, k));
    }
}
// This code is contributed by Prince Kumar

Python3

# Python3 implementation of the approach
import string

# Function to return the count of occurrences
def countOccurrence(a, b, k):    

    # Generate all possible substrings to
    # be searched 
    x = []
    for i in range(26):
        x.append(b[0:k] + string.ascii_lowercase[i] + b[k + 1:])

    # Now search every substring 'a' and
    # increment count   
    count = 0
    for var in x:
        if var in a:
            count += a.count(var)
            
    return count

# Driver code
a, b = "geeks", "ee"
k = 1
print(countOccurrence(a, b, k))

JavaScript

// JavaScript implementation of the approach

// Function to return the count of occurrences
function countOccurrence(a, b, k) {

let x = [];

// Generate all possible substrings to be searched
for (let i = 0; i < 26; i++) {
x.push(b.substring(0, k) + String.fromCharCode(97 + i) + b.substring(k + 1));
}

// Now search every substring 'a' and increment count
let count = 0;

for (let var1 of x) {
if (a.includes(var1)) {
count += a.split(var1).length - 1;
}
}

return count;
}

// Driver code
let a = "geeks";
let b = "ee";
let k = 1;
console.log(countOccurrence(a, b, k));

// c# implementation of above approach
using System;
using System.Collections.Generic;

class GFG {
    // function to implement count
    static int count_(string a, string var)
    {
        int c = 0;
        int l = var.Length;
        for (int i = 0; i < a.Length - l; i++)
            if (a.Substring(i, l) == var)
                c++;
        return c;
    }

    // Function to return the count of occurrences
    static int countOccurrence(string a, string b, int k)
    {

        // Generate all possible substrings to
        // be searched
        List<string> x = new List<string>();
        for (int i = 0; i < 26; i++) {
            string temp = b;
            temp = temp.Remove(k, 1).Insert(
                k, Convert.ToString((char)('a' + i)));
            x.Add(temp);
        }
        // Now search every substring 'a' and
        // increment count
        int count = 0;
        foreach(var var in x)
        {
            if (a.Contains(var))
                count += count_(a, var);
        }
        return count;
    }

    static void Main()
    {
        string a = "geeks", b = "ee";
        int k = 1;
        Console.WriteLine(countOccurrence(a, b, k));
    }
}

Output:

Count occurrences of a sub-string with one variable character

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