Maximum score possible by removing substrings made up of single distinct character
Last Updated :
06 Oct, 2021
Improve
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:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if the string s consists
// of a single distinct character or not
bool isUnique(string s)
{
set<char> Set;
for(char c : s)
{
Set.insert(c);
}
return Set.size() == 1;
}
// Function to calculate the maximum
// score possible by removing substrings
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.substr(i, j + 1);
// Check if the substring contains
// only a single distinct character
if (isUnique(sub))
mx = max(mx, a[sub.length() - 1] + maxScore(s.substr(0, i) + s.substr(j + 1), a));
}
}
// Return the maximum score
return mx;
}
int main()
{
string s = "011";
int a[] = { 1, 3, 1 };
cout << maxScore(s, a)-1;
return 0;
}
// This code is contributed by mukesh07.
// 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.
# 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))
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to check if the string s consists
// of a single distinct character or not
static bool isUnique(string s)
{
HashSet<char> set = new HashSet<char>();
foreach(char c in s)
set.Add(c);
return set.Count == 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 - i);
// 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
static void Main() {
string s = "011";
int[] a = { 1, 3, 1 };
Console.Write(maxScore(s, a));
}
}
// This code is contributed by suresh07.
<script>
// JavaScript program for the above approach
// Function to check if the string s consists
// of a single distinct character or not
function isUnique( s)
{
var set = new Set();
for(let i = 0 ; i< s.length ; i++)
{
set.add(s[i]);
}
return set.size == 1;
}
// Function to calculate the maximum
// score possible by removing substrings
function maxScore( s, a)
{
let 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
let mx = -1;
// Try to remove all substrings that
// satisfy the condition and check
// for resultant string after removal
for (let i = 0; i < n; i++)
{
for (let j = i; j < n; j++)
{
// Store the substring {s[i], .., s[j]}
let 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
let s = "011";
let a = [ 1, 3, 1 ];
document.write(maxScore(s, a));
</script>
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 the input to find the required result.
- Initialize an array, say dp[] to memorize 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++ 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 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
# 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))
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Initialize a dictionary to
// store the precomputed results
static Dictionary<string,
int> dp = new Dictionary<string,
int>();
// 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[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[tail] != s[head])
{
// Move head to
// tail and break
head = tail;
break;
}
// Store the substring
string sub = s.Substring(head, tail + 1-head);
// Update the maximum
mx = Math.Max(
mx, a[sub.Length - 1] +
maxScore(s.Substring(0, head) +
s.Substring(tail + 1, s.Length-tail - 1), a));
// Move the tail
tail += 1;
}
if (tail == n)
break;
}
// Store the score
dp.Add(s, mx);
return mx;
}
// Driver code
static public void Main()
{
string s = "abb";
int[] a = { 1, 3, 1 };
Console.WriteLine((maxScore(s, a)));
}
}
// This code is contributed by patel2127
<script>
// JavaScript program for the above approach
// Initialize a dictionary to
// store the precomputed results
let dp = new Map();
// Function to calculate the maximum
// score possible by removing substrings
function maxScore(s, a)
{
// If s is present in dp[] array
if (dp.has(s))
return dp.get(s);
// Base Cases:
let 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
let head = 0;
// Initialize the max variable
let mx = -1;
// Generate the substrings
// using two pointers
while (head < n)
{
let 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
let sub = s.substring(head, head + tail + 1);
// Update the maximum
mx = Math.max(
mx, a[sub.length - 1] +
maxScore(s.substring(0, head) +
s.substring(tail + 1, tail + 1 + s.length), a));
// Move the tail
tail += 1;
}
if (tail == n)
break;
}
// Store the score
dp.set(s, mx);
return mx;
}
let s = "abb";
let a = [ 1, 3, 1 ];
document.write((maxScore(s, a))-1);
</script>
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(N)
