Given a string S containing only two special characters ‘*‘ and ‘?‘, and two integers A and B that denotes the count of available 0’s and 1’s. The task is to count the maximum number of characters that can be placed in the place of ‘?‘ such that no two adjacent characters are the same.
Examples:
Input: s = “*???*???*”, A = 4, B = 3
Output: 6
Explanation: The string can be modified to “*010*010*”. Therefore, maximum number of characters that can be placed in the place of ‘?’ are (4 + 2 = 6)
Input: s = “???*??*”, A = 0, B = 5
Output: 3
Explanation: The string can be modified to “*1?1*?1*”. Therefore, maximum number of characters that can be placed in the place of ‘?’ are (0 + 3 = 3)
Approach: The task can be solved by keeping track of contiguous segments of ‘?’s and placing 0’s and 1’s such that after replacing ‘?’s, no two adjacent elements in the resultant string are the same.
Follow the below steps to solve the problem:
- Initialize a vector ‘v’, which will store the lengths of contiguous segments of ?s
- The variable ‘cur’ stores the current count of ?s, as soon as a ‘*’ is encountered, store the value of cur inside v
- Now, start iterating the stored segment lengths of ?s, and greedily assign 0s and 1s in-place of ?s
- Keep track of the maximum number of characters that can be placed in the place of ?s
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to count the maximum number // of characters 'A' and 'B' that can be // placed in position of '?' int maximumChar(string s, int A, int B)
{ // Length of the string
int len = s.size();
// Store the current count of '?'s
int curr = 0;
// Store the lengths of contiguous
// segments of '?'s
vector< int > v;
// Traversing the string
for ( int i = 0; i < len; i++) {
// If character is '?'
// increment curr by 1
if (s[i] == '?' ) {
curr++;
}
// If character is '*'
else {
// If curr is not equal to 0
if (curr != 0) {
v.push_back(curr);
// Re-initialise curr to 0
curr = 0;
}
}
}
// After traversing the string
// if curr is not equal to 0
if (curr != 0) {
v.push_back(curr);
}
// Variable for maximum count
int count = 0;
// Traversing the vector
for ( int i = 0; i < v.size(); i++) {
// Variable to store half of
// each elements in vector
int x = v[i] / 2;
// Variable to store half of
// each elements with its remainder
int y = v[i] / 2 + v[i] % 2;
// If A is greater than B
if (A > B) {
// Swap both
swap(A, B);
}
// Increment count by
// minimum of A and x
count += min(A, x);
// Update A
A -= min(A, x);
// Increment count by
// minimum of B and y
count += min(B, y);
// Update B
B -= min(B, y);
}
// Return count
return count;
} // Driver Code int main()
{ string s = "*???*???*" ;
int A = 4, B = 3;
cout << maximumChar(s, A, B);
return 0;
} |
// Java program for the above approach import java.util.ArrayList;
class GFG {
// Function to count the maximum number
// of characters 'A' and 'B' that can be
// placed in position of '?'
public static int maximumChar(String s, int A, int B)
{
// Length of the string
int len = s.length();
// Store the current count of '?'s
int curr = 0 ;
// Store the lengths of contiguous
// segments of '?'s
ArrayList<Integer> v = new ArrayList<Integer>();
// Traversing the string
for ( int i = 0 ; i < len; i++) {
// If character is '?'
// increment curr by 1
if (s.charAt(i) == '?' ) {
curr++;
}
// If character is '*'
else {
// If curr is not equal to 0
if (curr != 0 ) {
v.add(curr);
// Re-initialise curr to 0
curr = 0 ;
}
}
}
// After traversing the string
// if curr is not equal to 0
if (curr != 0 ) {
v.add(curr);
}
// Variable for maximum count
int count = 0 ;
// Traversing the vector
for ( int i = 0 ; i < v.size(); i++)
{
// Variable to store half of
// each elements in vector
int x = v.get(i) / 2 ;
// Variable to store half of
// each elements with its remainder
int y = v.get(i) / 2 + v.get(i) % 2 ;
// If A is greater than B
if (A > B)
{
// Swap both
int temp = A;
A = B;
B = temp;
}
// Increment count by
// minimum of A and x
count += Math.min(A, x);
// Update A
A -= Math.min(A, x);
// Increment count by
// minimum of B and y
count += Math.min(B, y);
// Update B
B -= Math.min(B, y);
}
// Return count
return count;
}
// Driver Code
public static void main(String args[]) {
String s = "*???*???*" ;
int A = 4 , B = 3 ;
System.out.println(maximumChar(s, A, B));
}
} // This code is contributed by saurabh_jaiswal. |
# python program for the above approach import math
# Function to count the maximum number # of characters 'A' and 'B' that can be # placed in position of '?' def maximumChar(s, A, B):
# Length of the string
le = len (s)
# Store the current count of '?'s
curr = 0
# Store the lengths of contiguous
# segments of '?'s
v = []
# Traversing the string
for i in range ( 0 , le):
# If character is '?'
# increment curr by 1
if (s[i] = = '?' ):
curr + = 1
# If character is '*'
else :
# If curr is not equal to 0
if (curr ! = 0 ):
v.append(curr)
# Re-initialise curr to 0
curr = 0
# After traversing the string
# if curr is not equal to 0
if (curr ! = 0 ):
v.append(curr)
# Variable for maximum count
count = 0
# Traversing the vector
for i in range ( 0 , len (v)):
# Variable to store half of
# each elements in vector
x = v[i] / / 2
# Variable to store half of
# each elements with its remainder
y = v[i] / / 2 + v[i] % 2
# If A is greater than B
if (A > B):
# Swap both
temp = A
A = B
B = temp
# Increment count by
# minimum of A and x
count + = min (A, x)
# Update A
A - = min (A, x)
# Increment count by
# minimum of B and y
count + = min (B, y)
# Update B
B - = min (B, y)
# Return count
return count
# Driver Code if __name__ = = "__main__" :
s = "*???*???*"
A = 4
B = 3
print (maximumChar(s, A, B))
# This code is contributed by rakeshsahni
|
// C# program for the above approach using System;
using System.Collections.Generic;
public class GFG {
// Function to count the maximum number
// of characters 'A' and 'B' that can be
// placed in position of '?'
public static int maximumChar(String s, int A, int B)
{
// Length of the string
int len = s.Length;
// Store the current count of '?'s
int curr = 0;
// Store the lengths of contiguous
// segments of '?'s
List< int > v = new List< int >();
// Traversing the string
for ( int i = 0; i < len; i++)
{
// If character is '?'
// increment curr by 1
if (s[i] == '?' ) {
curr++;
}
// If character is '*'
else {
// If curr is not equal to 0
if (curr != 0) {
v.Add(curr);
// Re-initialise curr to 0
curr = 0;
}
}
}
// After traversing the string
// if curr is not equal to 0
if (curr != 0) {
v.Add(curr);
}
// Variable for maximum count
int count = 0;
// Traversing the vector
for ( int i = 0; i < v.Count; i++)
{
// Variable to store half of
// each elements in vector
int x = v[i] / 2;
// Variable to store half of
// each elements with its remainder
int y = v[i] / 2 + v[i] % 2;
// If A is greater than B
if (A > B)
{
// Swap both
int temp = A;
A = B;
B = temp;
}
// Increment count by
// minimum of A and x
count += Math.Min(A, x);
// Update A
A -= Math.Min(A, x);
// Increment count by
// minimum of B and y
count += Math.Min(B, y);
// Update B
B -= Math.Min(B, y);
}
// Return count
return count;
}
// Driver Code
public static void Main(String []args) {
String s = "*???*???*" ;
int A = 4, B = 3;
Console.WriteLine(maximumChar(s, A, B));
}
} // This code is contributed by shikhasingrajput |
<script> // JavaScript Program to implement
// the above approach
// Function to count the maximum number
// of characters 'A' and 'B' that can be
// placed in position of '?'
function maximumChar(s, A, B)
{
// Length of the string
let len = s.length;
// Store the current count of '?'s
let curr = 0;
// Store the lengths of contiguous
// segments of '?'s
let v = [];
// Traversing the string
for (let i = 0; i < len; i++)
{
// If character is '?'
// increment curr by 1
if (s[i] == '?' ) {
curr++;
}
// If character is '*'
else
{
// If curr is not equal to 0
if (curr != 0) {
v.push(curr);
// Re-initialise curr to 0
curr = 0;
}
}
}
// After traversing the string
// if curr is not equal to 0
if (curr != 0) {
v.push(curr);
}
// Variable for maximum count
let count = 0;
// Traversing the vector
for (let i = 0; i < v.length; i++)
{
// Variable to store half of
// each elements in vector
let x = Math.floor(v[i] / 2);
// Variable to store half of
// each elements with its remainder
let y = Math.floor(v[i] / 2) + v[i] % 2;
// If A is greater than B
if (A > B)
{
// Swap both
let temp = A;
A = B;
B = temp;
}
// Increment count by
// minimum of A and x
count = count + Math.min(A, x);
// Update A
A = A - Math.min(A, x);
// Increment count by
// minimum of B and y
count = count + Math.min(B, y);
// Update B
B = B - Math.min(B, y);
}
// Return count
return count;
}
// Driver Code
let s = "*???*???*" ;
let A = 4, B = 3;
document.write(maximumChar(s, A, B));
// This code is contributed by Potta Lokesh
</script>
|
6
Time Complexity: O(N)
Auxiliary Space: O(N)