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# Maximize the Binary String value by replacing A[i]th elements from start or end

Given a string S of size M consisting of only zeroes (and hence representing the integer 0). Also, given an array A[] of size N whose, each element is an integer in the range [1, M]. The task is to maximize the integer represented by the string by performing the following operation N times :

• In ith (1 â‰¤ i â‰¤ N) operation, replace either A[i]th term or (M+1-A[i])th term with 1.
• Characters at the same position can be changed more than once.

Examples :

Input: N = 4, M = 5, S = “00000”, A = {1, 1, 3, 1}
Output: “10101”
Explanation: In 1st operation, the element at 1st position (0-th index)
is transformed into 1.
In 2nd operation, since element at 1st position is already 1,
so make element at 5th position equal to 1.
In 3rd operation, make element at 3rd position equal to 1.
In 4th operation, we can make either element at 1st
or 5th position equal to 1. Both of them are already 1.

Input : N=1, M=5, S=”00000″, A={2}
Output : “01000”

Approach :

The problem can be solved easily by a greedy approach. The catch here is that to maximize the integer represented by the string, we must try to convert the 0s to 1s which are on as left as possible i.e., nearest to the most significant bit.

The following steps can be taken to solve this problem:

• Iterate through the elements of A[].
• For convenience, make A[i] = min(A[i], M+1-A[i]).
• To handle the 0-indexing of the array, M-1-A[i] would be written instead of M+1-A[i].
• If the A[i]th character of S is not 1, we will replace it.
• Else, we’ll replace the (M+1-A[i])th character.
• Return the final string as the required answer.

Following is the code based on the above approach :

## C++

```// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to maximize the integer represented by
// the string by performing the given operations
string maxInteger(int N, int M, string S, int A[])
{
int i = 0;

// Loop to perform N operations
while (N--) {

// To handle 0-indexing
A[i]--;

// Initializing A[i] as the minimum
// of A[i] and M-1-A[i]
A[i] = min(A[i], M - 1 - A[i]);

// If element at the A[i] position of S
// is not 1 make it 1
if (S[A[i]] != '1') {
S[A[i]] = '1';
}

// Else make the element at M-1-A[i]th position 1
else {
S[M - 1 - A[i]] = '1';
}
i++;
}

// Returning maximized string after N operations
return S;
}

// Driver Code
int main()
{
int N = 4, M = 5;
string S = "00000";
int A[4] = { 1, 1, 3, 1 };

// Function call
string ans = maxInteger(N, M, S, A);
cout << ans << endl;
return 0;
}```

## Java

```/*package whatever //do not write package name here */
import java.io.*;
class GFG {

// Function to maximize the integer represented by
// the string by performing the given operations
static String maxInteger(int N, int M, String S, int A[])
{
int i = 0;
char str[] = S.toCharArray();

// Loop to perform N operations
while (N-->0) {

// To handle 0-indexing
A[i]--;

// Initializing A[i] as the minimum
// of A[i] and M-1-A[i]
A[i] = Math.min(A[i], M - 1 - A[i]);

// If element at the A[i] position of S
// is not 1 make it 1
if (str[A[i]] != '1') {
str[A[i]] = '1';
}

// Else make the element at M-1-A[i]th position 1
else {
str[M - 1 - A[i]] = '1';
}
i++;
}

// Returning maximized string after N operations
return new String(str);
}

public static void main (String[] args) {

int N = 4, M = 5;
String S = "00000";
int A[] = { 1, 1, 3, 1 };

// Function call
String ans = maxInteger(N, M, S, A);
System.out.println(ans);
}
}

// This code is contributed by aadityapburujwale```

## Python3

```# python code to implement the approach

# Function to maximize the integer represented by
# the string by performing the given operations

def maxInteger(N, M, S, A):

i = 0

# Loop to perform N operations
while (N):

# To handle 0-indexing
A[i] -= 1

# Initializing A[i] as the minimum
# of A[i] and M-1-A[i]
A[i] = min(A[i], M - 1 - A[i])

# If element at the A[i] position of S
# is not 1 make it 1
if (S[A[i]] != '1'):
S[A[i]] = '1'

# Else make the element at M-1-A[i]th position 1
else:
S[M - 1 - A[i]] = '1'

i += 1
N -= 1

# Returning maximized string after N operations
return "".join(S)

# Driver Code
if __name__ == "__main__":

N = 4
M = 5
S = "00000"
A = [1, 1, 3, 1]

# Function call
ans = maxInteger(N, M, list(S), A)
print(ans)

# This code is contributed by rakeshsahni
```

## C#

```// Include namespace system
using System;

public class GFG
{
// Function to maximize the integer represented by
// the string by performing the given operations
public static String maxInteger(int N, int M, String S, int[] A)
{
var i = 0;
char[] str = S.ToCharArray();

// Loop to perform N operations
while (N-- > 0)
{
// To handle 0-indexing
A[i]--;

// Initializing A[i] as the minimum
// of A[i] and M-1-A[i]
A[i] = Math.Min(A[i],M - 1 - A[i]);

// If element at the A[i] position of S
// is not 1 make it 1
if (str[A[i]] != '1')
{
str[A[i]] = '1';
}
else
{
str[M - 1 - A[i]] = '1';
}
i++;
}

// Returning maximized string after N operations
return new  String(str);
}
public static void Main(String[] args)
{
var N = 4;
var M = 5;
var S = "00000";
int[] A = {1, 1, 3, 1};

// Function call
var ans = GFG.maxInteger(N, M, S, A);
Console.WriteLine(ans);
}
}

// This code is contributed by aadityapburujwale.```

## Javascript

```<script>
// Javascript program for above approach

// Function to maximize the integer represented by
// the string by performing the given operations
function maxInteger(N, M, S, A)
{
let i = 0;
let str = S.split('');

// Loop to perform N operations
while (N-->0) {

// To handle 0-indexing
A[i]--;

// Initializing A[i] as the minimum
// of A[i] and M-1-A[i]
A[i] = Math.min(A[i], M - 1 - A[i]);

// If element at the A[i] position of S
// is not 1 make it 1
if (str[A[i]] != '1') {
str[A[i]] = '1';
}

// Else make the element at M-1-A[i]th position 1
else {
str[M - 1 - A[i]] = '1';
}
i++;
}

// Returning maximized string after N operations
return new String(str);
}

// Driver Code
let N = 4, M = 5;
let S = "00000";
let A = [ 1, 1, 3, 1 ];

// Function call
let ans = maxInteger(N, M, S, A);
document.write(ans);

// This code is contributed by code_hunt.
</script>```
Output

`10101`

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

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