用最小的变化制作一个字典最小的回文

给定一个字符串S。打印可能的字典最小字符串。您可以对字符串中的字符进行最少的更改，并且可以置换字符串。

例子：

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

Input : S = "äabcd"
Output : "abcba"

解释1：将最后一个索引“c”改为“b”，变为“aabb”。然后重新排列字符串，它变成了“abba”。这是字典上最小的字符串。
解释2：将“d”改为“b”=>“aabcb”=>“abcba”。

方法：
cnt[a] 是字符a 的出现次数。考虑 cnt[a] 的奇数值，则回文不能包含多个带有 cnt[a] 的字符a。表示奇数 cnt[] 的字符如下 a ₁ , a ₂ .... _ak (按字母顺序)。将任何符号 a _k替换为₁ ，将_k-1替换为₂ ，依此类推，直到上述序列的中间。现在，奇数符号不超过一个。把它放在答案的中间。答案的前半部分将包括$cnt[a]/2$符号 a 的出现。下半部分将包含相反顺序的相同符号。
例子 ：

take string S =  "aabcd"
  cnt[a] = 2, cnt[b] = 1, cnt = 1, cnt[d] = 1.
 Odd characters => b, c, d
  replace ak('d') with a1('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 
using namespace std;
  
// Function to create a palindrome
int Palindrome(string s, int n)
{
    unordered_map cnt;
    string R = "";
  
    // Count the occurrences 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 occurrences
    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;
}

Python3

# Python code for changing a string
# into lexicographically smallest
# pallindromic string
  
# Function to create a palindrome
def palindrome(s: str, n: int) -> int:
    cnt = dict()
    R = []
  
    # Count the occurrences of
    # every character in the string
    for i in range(n):
        a = s[i]
        if a in cnt:
            cnt[a] += 1
        else:
            cnt[a] = 1
  
    # Create a string of characters
    # with odd occurrences
    i = 'a'
    while i <= 'z':
        if i in cnt and cnt[i] % 2 != 0:
            R += i
        i = chr(ord(i) + 1)
  
    l = len(R)
    j = 0
  
    # Change the created string upto
    # middle element and update
    # count to make sure that only
    # one odd character exists.
    for i in range(l - 1, (l // 2) - 1, -1):
  
        # decrease the count of
        # character updated
        if R[i] in cnt:
            cnt[R[i]] -= 1
        else:
            cnt[R[i]] = -1
  
        R[i] = R[j]
  
        if R[j] in cnt:
            cnt[R[j]] += 1
        else:
            cnt[R[j]] = 1
        j += 1
  
    # Create three strings to make
    # first half second half and
    # middle one.
    first, middle, second = "", "", ""
  
    i = 'a'
    while i <= 'z':
        if i in cnt:
  
            # characters with even occurrences
            if cnt[i] % 2 == 0:
                j = 0
  
                # fill the first half.
                while j < cnt[i] // 2:
                    first += i
                    j += 1
  
            # character with odd occurrence
            else:
                j = 0
  
                # fill the first half with
                # half of occurrence except one
                while j < (cnt[i] - 1) // 2:
                    first += i
                    j += 1
  
                # For middle element
                middle += i
        i = chr(ord(i) + 1)
  
    # create the second half by
    # reversing the first half.
    second = first
    second = ''.join(reversed(second))
    resultant = first + middle + second
    print(resultant)
  
# Driver Code
if __name__ == "__main__":
  
    S = "geeksforgeeks"
    n = len(S)
    palindrome(S, n)
  
# This code is contributed by
# sanjeev2552

输出：

eefgksoskgfee