Skip to content
Related Articles

Related Articles

Maximum score possible by removing substrings made up of single distinct character
  • Last Updated : 13 Apr, 2021

Given a binary string S and an array A[], both of size N, the task is to find the maximum score possible by removing substrings of any length, say K, consisting of the same characters, and adding A[K] to the score. 

Examples:

Input: S = “abb”, A = [1, 3, 1]
Output: 4
Explanation: 
Initially, score = 0 and S=”abb” 
Remove the substring {S[1], .. S[2]}, of length 2, and add A[2] to score. Therefore, S modifies to “a”. Score = 3.
Remove the substring {S[0]},  of length 1, and add A[1] to score. Therefore, S modifies to “”. Score = 4.

Input: S = “abb”, A = [2, 3, 1]
Output: 6
Explanation:
Initially, score = 0 and S=”abb”.
Remove the substring {S[2]}, of length 1, and add A[1] to score. Therefore, S modifies to “ab”. Score = 1
Remove the substring {S[1]}, of length 1, and add A[1] to score. Therefore, S modifies to “a”. Score = 4
Remove the substring {S[0]}, of length 1, and add A[1] to score. Therefore, S modifies to “”. Score = 6

Naive Approach: The simplest idea is to solve this problem is to use Recursion. Iterate over the characters of the string. If a substring consisting only of one distinct character is encountered, then proceed with either to continue the search or to remove the substring and recursively call the function for the remaining string. 



Below is the implementation of the above approach:

Java




// Java program for the above approach
import java.util.*;
class GFG
{
   
  // Function to check if the string s consists
  // of a single distinct character or not
  static boolean isUnique(String s)
  {
    HashSet<Character> set = new HashSet<>();
    for (char c : s.toCharArray())
      set.add(c);
    return set.size() == 1;
  }
 
  // Function to calculate the maximum
  // score possible by removing substrings
  static int maxScore(String s, int[] a)
  {
    int n = s.length();
 
    // If string is empty
    if (n == 0)
      return 0;
 
    // If length of string is 1
    if (n == 1)
      return a[0];
 
    // Store the maximum result
    int mx = -1;
     
    // Try to remove all substrings that
    // satisfy the condition and check
    // for resultant string after removal
    for (int i = 0; i < n; i++)
    {
      for (int j = i; j < n; j++)
      {
 
        // Store the substring {s[i], .., s[j]}
        String sub = s.substring(i, j + 1);
 
        // Check if the substring contains
        // only a single distinct character
        if (isUnique(sub))
          mx = Math.max(
          mx,
          a[sub.length() - 1]
          + maxScore(
            s.substring(0, i)
            + s.substring(j + 1),
            a));
      }
    }
     
    // Return the maximum score
    return mx;
  }
   
  // Driver Code
  public static void main(String args[])
  {
    String s = "011";
    int a[] = { 1, 3, 1 };
    System.out.print(maxScore(s, a));
  }
}
 
// This code is contributed by hemanth gadarla.

Python3




# Python program for the above approach
 
# Function to check if the string s consists
# of a single distinct character or not
def isUnique(s):
    return True if len(set(s)) == 1 else False
 
# Function to calculate the maximum
# score possible by removing substrings
def maxScore(s, a):
    n = len(s)
 
    # If string is empty
    if n == 0:
        return 0
 
    # If length of string is 1
    if n == 1:
        return a[0]
 
    # Store the maximum result
    mx = -1
 
    # Try to remove all substrings that
    # satisfy the condition and check
    # for resultant string after removal
    for i in range(n):
        for j in range(i, n):
 
            # Store the substring {s[i], .., s[j]}
            sub = s[i:j + 1]
 
            # Check if the substring contains
            # only a single distinct character
            if isUnique(sub):
                mx = max(mx, a[len(sub)-1]
                         + maxScore(s[:i]+s[j + 1:], a))
 
        # Return the maximum score
    return mx
 
 
# Driver Code
if __name__ == "__main__":
 
    s = "011"
    a = [1, 3, 1]
    print(maxScore(s, a))
Output: 
4

 

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Memoization to store the result of the recursive calls and use Two pointer technique to store the substring consisting only of 1 distinct character.
Follow the steps below to solve the problem:

  • Declare a recursive function that takes the string as input to find the required result.
  • Initialize an array, say dp[] to memoize the results.
    • If the value is already stored in the array dp[], return the result.
    • Otherwise, perform the following steps:
      • Considering the base case if the size of the string is 0, return 0. If it is equal to 1, return A[1].
      • Initialize a variable, say res, to store the result of the current function call.
      • Initialize two pointers, say head and tail, denoting the starting and ending indices of the substring.
      • Generate substrings satisfying the given condition, and for each substring, recursively call the function for the remaining string. Store the maximum score in res.
      • Store the result in the dp[] array and return it.
    • Print the value returned by the function as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Initialize a dictionary to
// store the precomputed results
map<string,int> dp;
 
// Function to calculate the maximum
// score possible by removing substrings
int maxScore(string s, vector<int> a)
{
 
  // If s is present in dp[] array
  if (dp.find(s) != dp.end())
    return dp[s];
 
  // Base Cases:
  int n = s.size();
 
  // If length of string is 0
  if (n == 0)
    return 0;
 
  // If length of string is 1
  if (n == 1)
    return a[0];
 
  // Put head pointer at start
  int head = 0;
 
  // Initialize the max variable
  int mx = -1;
 
  // Generate the substrings
  // using two pointers
  while (head < n)
  {
    int tail = head;
    while (tail < n)
    {
 
      // If s[head] and s[tail]
      // are different
      if (s[tail] != s[head])
      {
 
        // Move head to
        // tail and break
        head = tail;
        break;
      }
 
      // Store the substring
      string sub = s.substr(head, tail + 1);
 
      // Update the maximum
      mx = max(mx, a[sub.size() - 1] +
               maxScore(s.substr(0, head) +
                        s.substr(tail + 1,s.size()), a));
 
      // Move the tail
      tail += 1;
    }
    if (tail == n)
      break;
  }
 
  // Store the score
  dp[s] = mx;
  return mx;
}
 
// Driver Code
int main()
{
  string s = "abb";
  vector<int> a = {1, 3, 1};
  cout<<(maxScore(s, a)-1);
}
 
// This code is contributed by mohit kumar 29.

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Initialize a dictionary to
// store the precomputed results
static Map<String, Integer> dp = new HashMap<>();
 
// Function to calculate the maximum
// score possible by removing substrings
static int maxScore(String s, int[] a)
{
     
    // If s is present in dp[] array
    if (dp.containsKey(s))
        return dp.get(s);
 
    // Base Cases:
    int n = s.length();
 
    // If length of string is 0
    if (n == 0)
        return 0;
 
    // If length of string is 1
    if (n == 1)
        return a[0];
 
    // Put head pointer at start
    int head = 0;
 
    // Initialize the max variable
    int mx = -1;
 
    // Generate the substrings
    // using two pointers
    while (head < n)
    {
        int tail = head;
        while (tail < n)
        {
             
            // If s[head] and s[tail]
            // are different
            if (s.charAt(tail) != s.charAt(head))
            {
                 
                // Move head to
                // tail and break
                head = tail;
                break;
            }
 
            // Store the substring
            String sub = s.substring(head, tail + 1);
 
            // Update the maximum
            mx = Math.max(
                mx, a[sub.length() - 1] +
                maxScore(s.substring(0, head) +
                s.substring(tail + 1, s.length()), a));
 
            // Move the tail
            tail += 1;
        }
        if (tail == n)
            break;
    }
 
    // Store the score
    dp.put(s, mx);
    return mx;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "abb";
    int[] a = { 1, 3, 1 };
     
    System.out.println((maxScore(s, a)));
}
}
 
// This code is contributed by offbeat

Python3




# Python program for the above approach
 
# Initialize a dictionary to
# store the precomputed results
dp = dict()
 
# Function to calculate the maximum
# score possible by removing substrings
def maxScore(s, a):
 
    # If s is present in dp[] array
    if s in dp:
        return dp[s]
 
    # Base Cases:
    n = len(s)
     
    # If length of string is 0
    if n == 0:
        return 0
       
    # If length of string is 1
    if n == 1:
        return a[0]
 
    # Put head pointer at start
    head = 0
 
    # Initialize the max variable
    mx = -1
 
    # Generate the substrings
    # using two pointers
    while head < n:
        tail = head
        while tail < n:
             
            # If s[head] and s[tail]
            # are different
            if s[tail] != s[head]:
               
                  # Move head to
                # tail and break
                head = tail
                break
             
            # Store the substring
            sub = s[head:tail + 1]
 
            # Update the maximum
            mx = max(mx, a[len(sub)-1]
                     + maxScore(s[:head] + s[tail + 1:], a))
 
            # Move the tail
            tail += 1
        if tail == n:
            break
 
    # Store the score
    dp[s] = mx
    return mx
 
 
# Driver Code
if __name__ == "__main__":
   
    s = "abb"
    a = [1, 3, 1]
 
    print(maxScore(s, a))
Output: 
4

 

Time Complexity: O(N)
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up
Recommended Articles
Page :