Maximum adjacent index difference for a subsequence T present in given string S

Given two strings S and T of lengths n and m respectively. Find the maximum adjacent index difference for the indices where string T is present in string S as a subsequence.

• Cost is defined as the difference between indices of subsequence T
• The maximum cost of a sequence is defined as max(a_i+1−a_i) for 1 ≤ i < m.
• It is guaranteed that there is at least subsequence as T in S.

Examples:

Input: S = “abbbc”, T = “abc” 
Output: 3
Explanation: One of the subsequences is {1, 4, 5} denoting the sequence “abc” same as T, thus the maximum cost is 4 – 1 = 3 .

Input: S = “aaaaa”, T = “aa” 
Output: 4
Explanation: One of the subsequences is {1, 5} and the max cost is 5 – 1 = 4 .

Approach: For some subsequence of the string S, let a_i denote the position of the character T_i in the string S. For fixed i, we can find a subsequence that maximizes a_i+1−a_i.

Follow the below steps to solve the problem:

1. Let left_i and right_i be the minimum and maximum possible value of a_i among all valid a’s respectively. Now, it is easy to see that the maximum possible value of a_i+1 − a_i is equal to right_i+1 − left_i.
2. To calculate left_i we just need to find the first element after left_i−1 which is equal to T_i , right_i can be found in the same way by iterating in the reverse direction.
3. So, after finding left and right, the answer is max(right_i+1−left_i) for 1 ≤ i < m.

Below is the implementation of the above approach.

C++

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

// Function to find the maximum
// cost of a subsequence
int maxCost(string s, string t, int n, int m)
{
    // mxCost stores the maximum
    // cost of a subsequence
    int mxCost = 0;

    // left[i] and right[i] store
    // the minimum and maximum
    // value of ai among all
    // valid a's respectively
    int left[m], right[m];

    int j = 0;
    for (int i = 0; i < n; i++) {
        if (j >= m)
            break;

        if (s[i] == t[j]) {
            left[j] = i;
            j++;
        }
    }

    j = m - 1;
    for (int i = n - 1; i >= 0; i--) {
        if (j < 0)
            break;

        if (s[i] == t[j]) {
            right[j] = i;
            j--;
        }
    }

    for (int i = 0; i < m - 1; i++) {
        mxCost = max(mxCost,
                     right[i + 1] - left[i]);
    }

    return mxCost;
}

// Driver function
int main()
{
    string S = "abbbc", T = "abc";
    int n = S.length(), m = T.length();

    // Function Call
    cout << maxCost(S, T, n, m);
    return 0;
}

Java

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

class GFG{

  // Function to find the maximum
  // cost of a subsequence
  static int maxCost(String s, String t, int n, int m)
  {
    // mxCost stores the maximum
    // cost of a subsequence
    int mxCost = 0;

    // left[i] and right[i] store
    // the minimum and maximum
    // value of ai among all
    // valid a's respectively
    int []left = new int[m];
    int []right = new int[m];

    int j = 0;
    for (int i = 0; i < n; i++) {
      if (j >= m)
        break;

      if (s.charAt(i) == t.charAt(j)) {
        left[j] = i;
        j++;
      }
    }

    j = m - 1;
    for (int i = n - 1; i >= 0; i--) {
      if (j < 0)
        break;

      if (s.charAt(i) == t.charAt(j)) {
        right[j] = i;
        j--;
      }
    }

    for (int i = 0; i < m - 1; i++) {
      mxCost = Math.max(mxCost,
                        right[i + 1] - left[i]);
    }

    return mxCost;
  }

  // Driver function
  public static void main(String[] args)
  {
    String S = "abbbc", T = "abc";
    int n = S.length(), m = T.length();

    // Function Call
    System.out.print(maxCost(S, T, n, m));
  }
}

// This code is contributed by 29AjayKumar

Python

# Python program for the above approach

# Function to find the maximum
# cost of a subsequence
def maxCost(s, t, n, m):

    # mxCost stores the maximum
    # cost of a subsequence
    mxCost = 0

    # left[i] and right[i] store
    # the minimum and maximum
    # value of ai among all
    # valid a's respectively
    left = []
    right = []

    j = 0
    for i in range(0, n):
        if (j >= m):
            break

        if (s[i] == t[j]):
            left.append(i)
            j += 1

    j = m - 1
    for i in reversed(range(n)):
        if (j < 0):
            break

        if (s[i] == t[j]):
            right.append(i)
            j -= 1

    for i in range(0, m - 1):
        mxCost = max(mxCost, right[i + 1] - left[i])

    return mxCost

# Driver function
S = "abbbc"
T = "abc"
n = len(S)
m = len(T)

print(maxCost(S, T, n, m))

# This code is contributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;
class GFG{

  // Function to find the maximum
  // cost of a subsequence
  static int maxCost(String s, String t, int n, int m)
  {

    // mxCost stores the maximum
    // cost of a subsequence
    int mxCost = 0;

    // left[i] and right[i] store
    // the minimum and maximum
    // value of ai among all
    // valid a's respectively
    int []left = new int[m];
    int []right = new int[m];

    int j = 0;
    for (int i = 0; i < n; i++) {
      if (j >= m)
        break;

      if (s[i] == t[j]) {
        left[j] = i;
        j++;
      }
    }

    j = m - 1;
    for (int i = n - 1; i >= 0; i--) {
      if (j < 0)
        break;

      if (s[i] == t[j]) {
        right[j] = i;
        j--;
      }
    }

    for (int i = 0; i < m - 1; i++) {
      mxCost = Math.Max(mxCost,  right[i + 1] - left[i]);
    }

    return mxCost;
  }

  // Driver function
  public static void Main()
  {
    String S = "abbbc", T = "abc";
    int n = S.Length, m = T.Length;

    // Function Call
    Console.Write(maxCost(S, T, n, m));
  }
}

// This code is contributed by gfgking

Javascript

<script>
    // JavaScript program for the above approach

    // Function to find the maximum
    // cost of a subsequence
    const maxCost = (s, t, n, m) => {

        // mxCost stores the maximum
        // cost of a subsequence
        let mxCost = 0;

        // left[i] and right[i] store
        // the minimum and maximum
        // value of ai among all
        // valid a's respectively
        let left = new Array(m).fill(0), right = new Array(m).fill(0);

        let j = 0;
        for (let i = 0; i < n; i++) {
            if (j >= m)
                break;

            if (s[i] == t[j]) {
                left[j] = i;
                j++;
            }
        }

        j = m - 1;
        for (let i = n - 1; i >= 0; i--) {
            if (j < 0)
                break;

            if (s[i] == t[j]) {
                right[j] = i;
                j--;
            }
        }

        for (let i = 0; i < m - 1; i++) {
            mxCost = Math.max(mxCost,
                right[i + 1] - left[i]);
        }

        return mxCost;
    }

    // Driver function

    let S = "abbbc", T = "abc";
    let n = S.length, m = T.length;

    // Function Call
    document.write(maxCost(S, T, n, m));

// This code is contributed by rakeshsahni

</script>

Output

3

Time Complexity: O(n + m)
Auxiliary Space: O(n)