Given a string S and integers P and Q, which denotes the cost of removal of substrings “ab” and “ba” respectively from S, the task is to find the maximum cost of removing all occurrences of substrings “ab” and “ba”.
Examples:
Input: S = “cbbaabbaab”, P = 6, Q = 4
Output: 22
Explanation:
Removing substring “ab” from “cbbaabbaab”, the string obtained is “cbbabaab”. Cost = 6.
Removing substring “ab” from “cbbabaab”, the string obtained is “cbbaab”. Cost = 6.
Removing substring “ba” from “cbbaab”, the string obtained is “cbab”. Cost = 4.
Removing substring “ab” from “cbab“, the string obtained is “cb”. Cost = 6.
Total cost = 6 + 6 + 4 + 6 = 22Input: S = “bbaanaybbabd”, P = 3, Q = 5
Output: 15
Explanation:
Removing substring “ba” from “bbaanaybbabd”, the string obtained is “banaybbabd”. Cost = 5.
Removing substring “ba”, the string obtained is “banaybbabd”, the string obtained is “naybbabd”. Cost = 5.
Removing substring “ba” from “naybbabd”, the string obtained is “naybbd”. Cost = 5.
Total cost = 5 + 5 + 5 = 15
Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:
-
Traverse the string and remove one type of substring. This can be done using greedy approach as:
- If P >= Q, remove all occurrences of “ab” substring and then remove all occurrences of “ba” substring.
- Otherwise, remove all occurrences of “ba” substring and then remove all occurrences of “ab” substring.
- Stack Data Structure can be used
- Initialize our higher cost and lower cost string as “ab” or “ba” according to the value of P and Q as character arrays maxstr[] and minstr[] of size 2 and initialize maximum cost and minimum cost as maxp and minp respectively.
- Initialize variable, say cost, to store maximum cost
-
Traverse the string and perform the following steps:
- If stack is not empty and top of stack and the current character forms maxstr[], then pop the stack top and add maxp to cost.
- Otherwise, add the current character to the stack.
- Traverse the remaining string.
- If stack is not empty and top of stack and the current character forms the minstr[], then pop the stack top and add minp to cost.
- Otherwise, add the current character to the stack.
- Print cost as the maximum cost.
Below is the implementation of the above approach:
// C++ program for the above approach: #include <bits/stdc++.h> using namespace std;
// Function to find the maximum cost of // removing substrings "ab" and "ba" from S int MaxCollection(string S, int P, int Q)
{ // MaxStr is the substring char
// array with larger cost
char maxstr[2];
string x = (P >= Q ? "ab" : "ba" );
strcpy (maxstr, x.c_str());
// MinStr is the substring char
// array with smaller cost;
char minstr[2];
x = (P >= Q ? "ba" : "ab" );
strcpy (minstr, x.c_str());
// Denotes larger point
int maxp = max(P, Q);
// Denotes smaller point
int minp = min(P, Q);
// Stores cost scored
int cost = 0;
// Removing all occurrences of
// maxstr from the S
// Stack to keep track of characters
stack< char > stack1;
char s[S.length()];
strcpy (s, S.c_str());
// Traverse the string
for ( auto &ch : s) {
// If the substring is maxstr
if (!stack1.empty()
&& (stack1.top() == maxstr[0]
&& ch == maxstr[1])) {
// Pop from the stack
stack1.pop();
// Add maxp to cost
cost += maxp;
}
// Push the character to the stack
else {
stack1.push(ch);
}
}
// Remaining string after removing maxstr
string sb = "" ;
// Find remaining string
while (stack1.size() > 0)
{
sb = sb + stack1.top();
stack1.pop();
}
// Reversing the string
// retrieved from the stack
reverse(sb.begin(), sb.end());
// Removing all occurrences of minstr
for ( auto &ch : sb) {
// If the substring is minstr
if (!stack1.empty()
&& (stack1.top() == minstr[0]
&& ch == minstr[1])) {
// Pop from the stack
stack1.pop();
// Add minp to the cost
cost += minp;
}
// Otherwise
else {
stack1.push(ch);
}
}
// Return the maximum cost
return cost;
} int main()
{ // Input String
string S = "cbbaabbaab" ;
// Costs
int P = 6;
int Q = 4;
cout << MaxCollection(S, P, Q);
return 0;
} // This code is contributed by decode2207. |
// Java program for the above approach: import java.util.*;
class GFG {
// Function to find the maximum cost of
// removing substrings "ab" and "ba" from S
public static int MaxCollection(
String S, int P, int Q)
{
// MaxStr is the substring char
// array with larger cost
char maxstr[]
= (P >= Q ? "ab" : "ba" ).toCharArray();
// MinStr is the substring char
// array with smaller cost;
char minstr[]
= (P >= Q ? "ba" : "ab" ).toCharArray();
// Denotes larger point
int maxp = Math.max(P, Q);
// Denotes smaller point
int minp = Math.min(P, Q);
// Stores cost scored
int cost = 0 ;
// Removing all occurrences of
// maxstr from the S
// Stack to keep track of characters
Stack<Character> stack1 = new Stack<>();
char [] s = S.toCharArray();
// Traverse the string
for ( char ch : s) {
// If the substring is maxstr
if (!stack1.isEmpty()
&& (stack1.peek() == maxstr[ 0 ]
&& ch == maxstr[ 1 ])) {
// Pop from the stack
stack1.pop();
// Add maxp to cost
cost += maxp;
}
// Push the character to the stack
else {
stack1.push(ch);
}
}
// Remaining string after removing maxstr
StringBuilder sb = new StringBuilder();
// Find remaining string
while (!stack1.isEmpty())
sb.append(stack1.pop());
// Reversing the string
// retrieved from the stack
sb = sb.reverse();
String remstr = sb.toString();
// Removing all occurrences of minstr
for ( char ch : remstr.toCharArray()) {
// If the substring is minstr
if (!stack1.isEmpty()
&& (stack1.peek() == minstr[ 0 ]
&& ch == minstr[ 1 ])) {
// Pop from the stack
stack1.pop();
// Add minp to the cost
cost += minp;
}
// Otherwise
else {
stack1.push(ch);
}
}
// Return the maximum cost
return cost;
}
// Driver Code
public static void main(String[] args)
{
// Input String
String S = "cbbaabbaab" ;
// Costs
int P = 6 ;
int Q = 4 ;
System.out.println(MaxCollection(S, P, Q));
}
} |
# Python3 program for the above approach: # Function to find the maximum cost of # removing substrings "ab" and "ba" from S def MaxCollection(S, P, Q):
# MaxStr is the substring char
# array with larger cost
maxstr = [i for i in ( "ab" if P > = Q else "ba" )]
# MinStr is the substring char
# array with smaller cost;
minstr = [i for i in ( "ba" if P > = Q else "ab" )]
# Denotes larger point
maxp = max (P, Q)
# Denotes smaller point
minp = min (P, Q)
# Stores cost scored
cost = 0
# Removing all occurrences of
# maxstr from the S
# Stack to keep track of characters
stack1 = []
s = [i for i in S]
# Traverse the string
for ch in s:
# If the substring is maxstr
if ( len (stack1)> 0 and (stack1[ - 1 ] = = maxstr[ 0 ] and ch = = maxstr[ 1 ])):
# Pop from the stack
del stack1[ - 1 ]
# Add maxp to cost
cost + = maxp
# Push the character to the stack
else :
stack1.append(ch)
# Remaining string after removing maxstr
sb = ""
# Find remaining string
while ( len (stack1) > 0 ):
sb + = stack1[ - 1 ]
del stack1[ - 1 ]
# Reversing the string
# retrieved from the stack
sb = sb[:: - 1 ]
remstr = sb
# Removing all occurrences of minstr
for ch in remstr:
# If the substring is minstr
if ( len (stack1) > 0 and (stack1[ - 1 ] = = minstr[ 0 ] and ch = = minstr[ 1 ])):
# Pop from the stack
del stack1[ - 1 ]
# Add minp to the cost
cost + = minp
# Otherwise
else :
stack1.append(ch)
# Return the maximum cost
return cost
# Driver Code if __name__ = = '__main__' :
# Input String
S = "cbbaabbaab"
# Costs
P = 6 ;
Q = 4 ;
print (MaxCollection(S, P, Q));
# This code is contributed by mohit kumar 29. |
// C# program for the above approach: using System;
using System.Collections;
class GFG {
// Function to find the maximum cost of
// removing substrings "ab" and "ba" from S
static int MaxCollection( string S, int P, int Q)
{
// MaxStr is the substring char
// array with larger cost
char [] maxstr = (P >= Q ? "ab" : "ba" ).ToCharArray();
// MinStr is the substring char
// array with smaller cost;
char [] minstr = (P >= Q ? "ba" : "ab" ).ToCharArray();
// Denotes larger point
int maxp = Math.Max(P, Q);
// Denotes smaller point
int minp = Math.Min(P, Q);
// Stores cost scored
int cost = 0;
// Removing all occurrences of
// maxstr from the S
// Stack to keep track of characters
Stack stack1 = new Stack();
char [] s = S.ToCharArray();
// Traverse the string
foreach ( char ch in s) {
// If the substring is maxstr
if (stack1.Count > 0 && (( char )stack1.Peek() == maxstr[0] && ch == maxstr[1])) {
// Pop from the stack
stack1.Pop();
// Add maxp to cost
cost += maxp;
}
// Push the character to the stack
else {
stack1.Push(ch);
}
}
// Remaining string after removing maxstr
string sb = "" ;
// Find remaining string
while (stack1.Count > 0)
{
sb = sb + stack1.Peek();
stack1.Pop();
}
// Reversing the string
// retrieved from the stack
char [] chars = sb.ToCharArray();
Array.Reverse(chars);
string remstr = new string (chars);
// Removing all occurrences of minstr
foreach ( char ch in remstr.ToCharArray()) {
// If the substring is minstr
if (stack1.Count > 0 && (( char )stack1.Peek() == minstr[0] && ch == minstr[1])) {
// Pop from the stack
stack1.Pop();
// Add minp to the cost
cost += minp;
}
// Otherwise
else {
stack1.Push(ch);
}
}
// Return the maximum cost
return cost;
}
static void Main()
{
// Input String
string S = "cbbaabbaab" ;
// Costs
int P = 6;
int Q = 4;
Console.Write(MaxCollection(S, P, Q));
}
} // This code is contributed by mukesh07. |
<script> // JavaScript program for the above approach: // Function to find the maximum cost of // removing substrings "ab" and "ba" from S
function MaxCollection(S,P,Q)
{ // MaxStr is the substring char
// array with larger cost
let maxstr
= (P >= Q ? "ab" : "ba" ).split( "" );
// MinStr is the substring char
// array with smaller cost;
let minstr
= (P >= Q ? "ba" : "ab" ).split( "" );
// Denotes larger point
let maxp = Math.max(P, Q);
// Denotes smaller point
let minp = Math.min(P, Q);
// Stores cost scored
let cost = 0;
// Removing all occurrences of
// maxstr from the S
// Stack to keep track of characters
let stack1 = [];
let s = S.split( "" );
// Traverse the string
for (let ch=0;ch< s.length;ch++) {
// If the substring is maxstr
if (stack1.length!=0
&& (stack1[stack1.length-1] == maxstr[0]
&& s[ch] == maxstr[1])) {
// Pop from the stack
stack1.pop();
// Add maxp to cost
cost += maxp;
}
// Push the character to the stack
else {
stack1.push(s[ch]);
}
}
// Remaining string after removing maxstr
let sb = [];
// Find remaining string
while (stack1.length!=0)
sb.push(stack1.pop());
// Reversing the string
// retrieved from the stack
sb = sb.reverse();
let remstr = sb.join( "" );
// Removing all occurrences of minstr
for (let ch =0;ch<remstr.length;ch++) {
// If the substring is minstr
if (stack1.length!=0
&& (stack1[stack1.length-1] == minstr[0]
&& remstr[ch] == minstr[1])) {
// Pop from the stack
stack1.pop();
// Add minp to the cost
cost += minp;
}
// Otherwise
else {
stack1.push(remstr[ch]);
}
}
// Return the maximum cost
return cost;
} // Driver Code // Input String let S = "cbbaabbaab" ;
// Costs
let P = 6;
let Q = 4;
document.write(MaxCollection(S, P, Q));
// This code is contributed by patel2127 </script> |
22
Time Complexity : O(N)
Auxiliary Space : O(N), since N extra space has been taken.