# Make a lexicographically smallest palindrome with minimal changes

Given a string S. Print the lexicographically smallest string possible. You can make minimal changes to the characters in the string and you can permute the string.

Examples:

Input : S = "aabc" Output : "abba" Input : S = "äabcd" Output : "abcba"

**Explanation 1:**Change the last index “c” to “b”, it becomes “aabb”. Then rearrange the string, it becomes “abba”. This is the lexicographically smallest string.

**Explanation 2:** change “d” to “b” => “aabcb” => “abcba”.

**Approach:**

cnt[a] be the no.of occurences of the character a. Consider odd values of cnt[a], then a palindrome can not contain more than one character a with cnt[a]. Denote characters with odd count cnt[] as follows a_{1}, a_{2}….a_{k}(in alphabetical order). Replace any of the symbols a_{k} with a_{1} and a_{k-1} with a_{2} and so on upto middle of sequence above. Now, there are no more than one odd symbol. Place it in the middle of answer. First half of answer will consist of occurences of symbol a. Second half will contains same symbols in reverse order.

**Example :**

take string S = "aabcd" cnt[a] = 2, cnt[b] = 1, cnt = 1, cnt[d] = 1. Odd characters => b, c, d replace a_{k}('d') with a_{1}('b') we get => b, c, b. cnt[b] = 2, cnt = 1.Place odd character in middle and 'a' and 'b' in first half and also in second half, we get "abcba".

## C++

`// CPP code for changing a string ` `// into lexicographically smallest ` `// pallindromic string ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to create a palindrome ` `int` `Palindrome(string s, ` `int` `n) ` `{ ` ` ` `unordered_map<` `char` `, ` `int` `> cnt; ` ` ` `string R = ` `""` `; ` ` ` ` ` `// Count the occurences of ` ` ` `// every character in the string ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `char` `a = s[i]; ` ` ` `cnt[a]++; ` ` ` `} ` ` ` ` ` `// Create a string of characters ` ` ` `// with odd occurences ` ` ` `for` `(` `char` `i = ` `'a'` `; i <= ` `'z'` `; i++) { ` ` ` `if` `(cnt[i] % 2 != 0) ` ` ` `R += i; ` ` ` `} ` ` ` ` ` `int` `l = R.length(); ` ` ` `int` `j = 0; ` ` ` ` ` `// Change the created string upto ` ` ` `// middle element and update ` ` ` `// count to make sure that only ` ` ` `// one odd character exists. ` ` ` `for` `(` `int` `i = l - 1; i >= l / 2; i--) { ` ` ` ` ` `// decrease the count of ` ` ` `// character updated ` ` ` `cnt[R[i]]--; ` ` ` `R[i] = R[j]; ` ` ` `cnt[R[j]]++; ` ` ` `j++; ` ` ` `} ` ` ` ` ` `// Create three strings to make ` ` ` `// first half second half and ` ` ` `// middle one. ` ` ` `string first, middle, second; ` ` ` ` ` `for` `(` `char` `i = ` `'a'` `; i <= ` `'z'` `; i++) { ` ` ` `if` `(cnt[i] != 0) { ` ` ` ` ` `// characters with even occurrences ` ` ` `if` `(cnt[i] % 2 == 0) { ` ` ` `int` `j = 0; ` ` ` ` ` `// fill the first half. ` ` ` `while` `(j < cnt[i] / 2) { ` ` ` `first += i; ` ` ` `j++; ` ` ` `} ` ` ` `} ` ` ` ` ` `// character with odd occurrence ` ` ` `else` `{ ` ` ` `int` `j = 0; ` ` ` ` ` `// fill the first half with ` ` ` `// half of occurrence except one ` ` ` `while` `(j < (cnt[i] - 1) / 2) { ` ` ` `first += i; ` ` ` `j++; ` ` ` `} ` ` ` ` ` `// For middle element ` ` ` `middle += i; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` ` ` `// create the second half by ` ` ` `// reversing the first half. ` ` ` `second = first; ` ` ` `reverse(second.begin(), second.end()); ` ` ` `string resultant = first + middle + second; ` ` ` `cout << resultant << endl; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `string S = ` `"geeksforgeeks"` `; ` ` ` `int` `n = S.length(); ` ` ` `Palindrome(S, n); ` ` ` `return` `0; ` `} ` |

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Output:

eefgksoskgfee

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