Horje
Find longest substring of S1 that matches in S2 with a given cost

Given two strings S1 and S2 of length n. Additionally, two positive integers, target and C. The task is to determine the maximum length of a contiguous substring in S1 such that, by changing any character in the substring of S1 to another character, the resulting substring matches the corresponding segment in S2, and the total cost of these changes is at most target, where the cost of transformation for each character is C units.

Note: If there are multiple valid substrings with the same maximum length, provide any one of them as the answer.

Examples:

Input: S1 = “abcd”, S2 = “bcdf”, C = 1, target = 3
Output: 1 3
Explanation: Substring of S1 from index 1 to 3 is “bcd” that can change to “cdf”. The cost of changing would be 3. So the maximum length is 3.

Input: S1 = “adcd”, S2 = “cdef”, C = 2, target = 5
Output: 1 3

Approach:

The idea is to use Binary Search on Answer method to find what can be the maximum possible length. For each potential length (from the midpoint of the current search range [start, end]), use a sliding window approach to check all substrings of that length (i.e, mid) in S1. Calculates the transformation cost for each substring and checks if it’s less than or equal to the target. If a valid substring is found, the starting and ending positions are updated, and the search continues with larger lengths. If no valid substring is found for a certain length, the search continues with smaller lengths.

Steps-by-step approach:

  • Initialize a variable start with the minimum valid length substring possible i.e., 0.
  • Initialize a variable end with the maximum valid length substring possible i.e., the size of the string itself.
  • Initialize a variable result for storing the maximum length of valid substring
  • While the start is less than the end, do the following:
    • Calculate the mid = (start + end) /2
    • Check if mid is the possible length for a valid substring
      • Assign the result to mid
      • Shift start to mid + 1.
    • Otherwise, shift end to mid.
  • If any valid substring is possible then print the starting index and the ending index of the valid substring.

Below is the impelementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;
 
// Initilise S1 variable startPos and endPos
// for storing the starting and ending
// position of valid substring
// respectively.
int startPos = -1, endPos = -1;
 
bool isValid(int len, string& S1, string& S2, int C, int target)
{
    int i = 0, j = 0, n = S1.size(), cost = 0;
 
    // Initilise S1 variable flag to false.
    // It will keep track of any valid
    // substring of length len is
    // possible or not
    bool flag = false;
 
    while (j < n) {
 
        // Include the cost required to
        // convert S1[j] to S2[j]
        cost += ((S1[j] != S2[j]) ? C : 0);
 
        // If window size is equal to len
        // then shift the window.
        if (j - i + 1 == len) {
 
            // Check if any valid substring
            // then keep the indices of
            // valid substring in startPos
            // and endPos
            if (cost <= target) {
                flag = true;
                startPos = i;
                endPos = j;
            }
 
            // Remove the calculation of
            // ith indices as our window
            // we have to slide our window
            cost -= ((S1[j] != S2[j]) ? C : 0);
            i++;
        }
 
        j++;
    }
 
    // Return the flag
    return flag;
}
 
void validSubstring(string S1, string S2, int C, int target)
{
    startPos = -1, endPos = -1;
 
    // Initilise S1 variable start with
    // minimum valid length
    // substring possible.
    int start = 0;
 
    // Initilise S1 variable end with
    // maximum valid length substring
    // possible.
    int end = S1.size();
    int n = S1.size();
 
    // Initilise S1 variable result for
    // storing the maximum length of
    // valid substring
    int result = 0;
 
    // Do while start is less
    // than equal to end
    while (start < end) {
 
        // Calculate the mid
        int mid = (start + end) / 2;
 
        // Check if mid is possible length
        // for valid substring
        if (isValid(mid, S1, S2, C, target)) {
 
            // Assign result to mid
            result = mid;
 
            // Shift start to mid + 1
            start = mid + 1;
        }
 
        // Otherwise shift end to mid - 1
        else {
            end = mid;
        }
    }
 
    if (startPos == -1)
        cout << "Not possible\n";
    else
        cout << startPos << " " << endPos << "\n";
}
 
// Driver code
int main()
{
    // First test case
    string S1 = "abcd", S2 = "bcdf";
    int C = 1, target = 5;
    validSubstring(S1, S2, C, target);
     
    // Second test case
    S1 = "abc", S2 = "cdf";
    C = 10, target = 3;
    validSubstring(S1, S2, C, target);
 
    // Third test case
    S1 = "abcb", S2 = "cdfb";
    C = 2, target = 2;
    validSubstring(S1, S2, C, target);
 
    return 0;
}

Java

import java.util.*;
 
public class Main {
 
    // Initialize variables to store the starting and ending positions of a valid substring.
    static int startPos = -1, endPos = -1;
 
    // Check if a substring of a given length is a valid substring with cost less than or equal to the target.
    static boolean isValid(int len, String S1, String S2, int C, int target) {
        int i = 0, j = 0, n = S1.length(), cost = 0;
        boolean flag = false;
 
        // Iterate through the strings to check validity of the substring.
        while (j < n) {
            // Include the cost required to convert S1[j] to S2[j].
            cost += (S1.charAt(j) != S2.charAt(j)) ? C : 0;
 
            // If the window size is equal to len, then shift the window.
            if (j - i + 1 == len) {
                // Check if any valid substring and keep the indices of the valid substring in startPos and endPos.
                if (cost <= target) {
                    flag = true;
                    startPos = i;
                    endPos = j;
                }
 
                // Remove the calculation of ith indices as the window is sliding.
                cost -= (S1.charAt(i) != S2.charAt(i)) ? C : 0;
                i++;
            }
 
            j++;
        }
 
        // Return the flag indicating if a valid substring was found.
        return flag;
    }
 
    // Find the maximum length of a valid substring and its starting and ending positions.
    static void validSubstring(String S1, String S2, int C, int target) {
        startPos = -1;
        endPos = -1;
        int start = 0;
        int end = S1.length();
        int n = S1.length();
        int result = 0;
 
        // Perform binary search to find the maximum length of a valid substring.
        while (start < end) {
            int mid = (start + end) / 2;
 
            // Check if mid is a possible length for a valid substring.
            if (isValid(mid, S1, S2, C, target)) {
                // Assign result to mid and shift start to mid + 1.
                result = mid;
                start = mid + 1;
            } else {
                // Otherwise, shift end to mid - 1.
                end = mid;
            }
        }
 
        // If no valid substring is found, print "Not possible"; otherwise, print the starting and ending positions.
        if (startPos == -1)
            System.out.println("Not possible");
        else
            System.out.println(startPos + " " + endPos);
    }
 
    // Driver code
    public static void main(String[] args) {
        // First test case
        String S1 = "abcd", S2 = "bcdf";
        int C = 1, target = 5;
        validSubstring(S1, S2, C, target);
 
        // Second test case
        S1 = "abc"; S2 = "cdf";
        C = 10; target = 3;
        validSubstring(S1, S2, C, target);
 
        // Third test case
        S1 = "abcb"; S2 = "cdfb";
        C = 2; target = 2;
        validSubstring(S1, S2, C, target);
    }
}
 
// This code is contributed by shivamgupta0987654321

C#

using System;
 
class Program {
    // Initialize startPos and endPos to store the starting
    // and ending position of the valid substring
    static int startPos = -1, endPos = -1;
 
    // Function to check if a valid substring of length
    // 'len' exists
    static bool IsValid(int len, string S1, string S2,
                        int C, int target)
    {
        int i = 0, j = 0, n = S1.Length, cost = 0;
        bool flag = false;
 
        while (j < n) {
            // Include the cost required to convert S1[j] to
            // S2[j]
            cost += (S1[j] != S2[j]) ? C : 0;
 
            // If window size is equal to len, then check if
            // it's a valid substring
            if (j - i + 1 == len) {
                if (cost <= target) {
                    flag = true;
                    startPos = i;
                    endPos = j;
                }
 
                // Remove the cost of the ith index as the
                // window slides
                cost -= (S1[i] != S2[i]) ? C : 0;
                i++;
            }
 
            j++;
        }
 
        return flag;
    }
 
    // Function to find the maximum length of a valid
    // substring
    static void ValidSubstring(string S1, string S2, int C,
                               int target)
    {
        startPos = -1;
        endPos = -1;
        int start = 0;
        int end = S1.Length;
 
        while (start < end) {
            int mid = (start + end) / 2;
 
            // Check if mid is a possible length for a valid
            // substring
            if (IsValid(mid, S1, S2, C, target)) {
                // Shift start to mid + 1
                start = mid + 1;
            }
            else {
                // Otherwise shift end to mid
                end = mid;
            }
        }
 
        if (startPos == -1)
            Console.WriteLine("Not possible");
        else
            Console.WriteLine(startPos + " " + endPos);
    }
 
    // Driver code
    static void Main(string[] args)
    {
        // First test case
        string S1 = "abcd", S2 = "bcdf";
        int C = 1, target = 5;
        ValidSubstring(S1, S2, C, target);
 
        // Second test case
        S1 = "abc";
        S2 = "cdf";
        C = 10;
        target = 3;
        ValidSubstring(S1, S2, C, target);
 
        // Third test case
        S1 = "abcb";
        S2 = "cdfb";
        C = 2;
        target = 2;
        ValidSubstring(S1, S2, C, target);
    }
}

Javascript

// Initialize startPos and endPos for storing the starting and ending position of a valid substring respectively
let startPos = -1, endPos = -1;
 
// Function to check if a substring of length 'len' is a valid substring
function isValid(len, S1, S2, C, target) {
    let i = 0, j = 0, n = S1.length, cost = 0;
    let flag = false;
 
    while (j < n) {
        // Include the cost required to convert S1[j] to S2[j]
        cost += (S1[j] !== S2[j] ? C : 0);
 
        // If the window size is equal to len, check if it's a valid substring
        if (j - i + 1 === len) {
            if (cost <= target) {
                flag = true;
                startPos = i;
                endPos = j;
            }
 
            // Remove the calculation of ith indices as we slide the window
            cost -= (S1[j] !== S2[j] ? C : 0);
            i++;
        }
 
        j++;
    }
 
    return flag;
}
 
// Function to find the maximum length of a valid substring
function validSubstring(S1, S2, C, target) {
    startPos = -1;
    endPos = -1;
    let start = 0;
    let end = S1.length;
    let result = 0;
 
    while (start < end) {
        let mid = Math.floor((start + end) / 2);
 
        // Check if mid is a possible length for a valid substring
        if (isValid(mid, S1, S2, C, target)) {
            result = mid;
            start = mid + 1;
        } else {
            end = mid;
        }
    }
 
    // Output the result
    if (startPos === -1)
        console.log("Not possible");
    else
        console.log(startPos + " " + endPos);
}
 
// First test case
let S1 = "abcd", S2 = "bcdf";
let C = 1, target = 5;
validSubstring(S1, S2, C, target);
 
// Second test case
S1 = "abc", S2 = "cdf";
C = 10, target = 3;
validSubstring(S1, S2, C, target);
 
// Third test case
S1 = "abcb", S2 = "cdfb";
C = 2, target = 2;
validSubstring(S1, S2, C, target);

Python3

def is_valid(length, S1, S2, C, target):
    i, j, n, cost = 0, 0, len(S1), 0
    flag = False
    start_pos, end_pos = -1, -1  # Initialize start_pos and end_pos
 
    while j < n:
        # Include the cost required to convert S1[j] to S2[j]
        cost += (C if S1[j] != S2[j] else 0)
 
        # If window size is equal to length, then shift the window
        if j - i + 1 == length:
            # Check if any valid substring, then keep the indices of valid substring in start_pos and end_pos
            if cost <= target:
                flag = True
                start_pos, end_pos = i, j
 
            # Remove the calculation of ith indices as our window, we have to slide our window
            cost -= (C if S1[j] != S2[j] else 0)
            i += 1
 
        j += 1
 
    return flag, start_pos, end_pos
 
 
def valid_substring(S1, S2, C, target):
    start_pos, end_pos = -1, -1
    start, end, n = 0, len(S1), len(S1)
    result = 0
 
    while start < end:
        mid = (start + end) // 2
 
        # Check if mid is a possible length for a valid substring
        is_valid_flag, i, j = is_valid(mid, S1, S2, C, target)
 
        if is_valid_flag:
            # Assign result to mid, shift start to mid + 1
            result = mid
            start = mid + 1
            start_pos, end_pos = i, j
        else:
            # Otherwise, shift end to mid
            end = mid
 
    if start_pos == -1:
        print("Not possible")
    else:
        print(start_pos, end_pos)
 
 
# First test case
S1, S2 = "abcd", "bcdf"
C, target = 1, 5
valid_substring(S1, S2, C, target)
 
# Second test case
S1, S2 = "abc", "cdf"
C, target = 10, 3
valid_substring(S1, S2, C, target)
 
# Third test case
S1, S2 = "abcb", "cdfb"
C, target = 2, 2
valid_substring(S1, S2, C, target)

Output

1 3
Not possible
2 3

Time Complexity: O(N* log(N))
Auxiliary Space: O(1)


 


Reffered: https://www.geeksforgeeks.org

DSA

Related
Find the winner by incrementing co-ordinates till Euclidean distance &lt;= D Find the winner by incrementing co-ordinates till Euclidean distance &lt;= D
Modify shortest distance between source and destination to K Modify shortest distance between source and destination to K
Minimum number of keypresses to type the given string Minimum number of keypresses to type the given string
Maximize unique elements in set after K deletions from each array Maximize unique elements in set after K deletions from each array
Which is the best YouTube channel to Learn DSA? Which is the best YouTube channel to Learn DSA?

Type:
 Geek
Category:
 Coding
Sub Category:
 Tutorial
Uploaded by:
 Admin
Views:
 12



 

Horje