Make largest palindrome by changing at most K-digits

Given a string containing all digits, we need to convert this string to a palindrome by changing at most K digits. If many solutions are possible then print lexicographically largest one.

Examples:

```Input   : str = “43435”
k = 3
Output  : "93939"
Lexicographically largest palindrome
after 3 changes is "93939"

Input :  str = “43435”
k = 1
Output : “53435”
Lexicographically largest palindrome
after 3 changes is “53435”

Input  : str = “12345”
k = 1
Output : "Not Possible"
It is not possible to make str palindrome
after 1 change.
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can solve this problem using two pointers method. We start from left and right and if both digits are not equal then we replace the smaller value with larger value and decrease k by 1. We stop when the left and right pointers cross each other, after they stop if value of k is negative, then it is not possible to make string palindrome using k changes. If k is positive, then we can further maximize the string by looping once again in the same manner from left and right and converting both the digits to 9 and decreasing k by 2. If k value remains to 1 and string length is odd then we make the middle character as 9 to maximize whole value.

```// C++ program to get largest palindrome changing
// atmost K digits
#include <bits/stdc++.h>
using namespace std;

// Returns maximum possible palindrome using k changes
string maximumPalinUsingKChanges(string str, int k)
{
string palin = str;

// Iinitialize l and r by leftmost and
// rightmost ends
int l = 0;
int r = str.length() - 1;

//  first try to make string palindrome
while (l < r)
{
// Replace left and right character by
// maximum of both
if (str[l] != str[r])
{
palin[l] = palin[r] = max(str[l], str[r]);
k--;
}
l++;
r--;
}

// If k is negative then we can't make
// string palindrome
if (k < 0)
return "Not possible";

l = 0;
r = str.length() - 1;

while (l <= r)
{
// At mid character, if K>0 then change
// it to 9
if (l == r)
{
if (k > 0)
palin[l] = '9';
}

// If character at lth (same as rth) is
// less than 9
if (palin[l] < '9')
{
/* If none of them is changed in the
previous loop then subtract 2 from K
and convert both to 9 */
if (k >= 2 && palin[l] == str[l] &&
palin[r] == str[r])
{
k -= 2;
palin[l] = palin[r] = '9';
}

/*  If one of them is changed in the previous
loop then subtract 1 from K (1 more is
subtracted already) and make them 9  */
else if (k >= 1 && (palin[l] != str[l] ||
palin[r] != str[r]))
{
k--;
palin[l] = palin[r] = '9';
}
}
l++;
r--;
}

return palin;
}

//  Driver code to test above methods
int main()
{
string str = "43435";
int k = 3;
cout << maximumPalinUsingKChanges(str, k);
return 0;
}
```

Output:

```93939
```

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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