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Largest Palindromic Number LeetCode Solution

Problem – Largest Palindromic Number

You are given a string `num` consisting of digits only.

Return the largest palindromic integer (in the form of a string) that can be formed using digits taken from `num`. It should not contain leading zeroes.

Notes:

• You do not need to use all the digits of `num`, but you must use at least one digit.
• The digits can be reordered.

Example 1:

``````Input: num = "444947137"
Output: "7449447"
Explanation:
Use the digits "4449477" from "444947137" to form the palindromic integer "7449447".
It can be shown that "7449447" is the largest palindromic integer that can be formed.``````

Example 2:

``````Input: num = "00009"
Output: "9"
Explanation:
It can be shown that "9" is the largest palindromic integer that can be formed.
Note that the integer returned should not contain leading zeroes.``````

Constraints:

• `1 <= num.length <= 105`
• `num` consists of digits.

Largest Palindromic Number LeetCode Solution in Python

``````    def largestPalindromic(self, num: str) -> str:
count = Counter(num)
res = ''.join(count[i] // 2 * i for i in '9876543210').lstrip('0')
mid = max(count[i] % 2 * i for i in count)
return (res + mid + res[::-1]) or '0'``````

Largest Palindromic Number LeetCode Solution in C++

``````class Solution {
public:
string largestPalindromic(string num) {
vector<int> cnt(10);
for (char c: num) {
cnt[c - '0']++;
}

string lp, rp; // left and right partition
for (int i: num) {
for (int j = 9; j >= 0; j--)  {
if (cnt[j] > 1 && (j > 0 || lp.size())) {
lp += '0' + j;
rp += '0' + j;
cnt[j] -= 2;
break;
}
}
}
for (int i = 9; i >= 0; i--) {
if (cnt[i]) {
lp += '0' + i; break;
}
}
reverse(begin(rp), end(rp));
return lp + rp;
}
};``````

Largest Palindromic Number LeetCode Solution in Java

``````class Solution {
public String largestPalindromic(String num) {
int[] n = new int[10];
for (char c : num.toCharArray()) {
n[c - '0']++;
}
boolean only = false;
//seperate the result into 3 parts, the first half and second half, also the center (if exist)
StringBuilder sbPre = new StringBuilder();
StringBuilder sbPost = new StringBuilder();
StringBuilder sbOnly = new StringBuilder();
//check from 9 to 1, to make it the largest
for (int i = 9; i >= 0; i--) {
//check for center
if (n[i] % 2 != 0 && !only) {
only = true;
sbOnly.append(Integer.toString(i));
}
//handle special case of all '0' or '0' is the first digit(invalid)
if (i == 0 && sbPre.length() == 0) {
if (n[i] > 0 && sbOnly.length() == 0) sbOnly.append(Integer.toString(i));
return sbOnly.toString();
}
for (int j = 0; j < n[i] / 2; j++) {
sbPre.append(Integer.toString(i));
sbPost.insert(0, Integer.toString(i));
}
}
sbPre.append(sbOnly);
sbPre.append(sbPost);
return sbPre.toString();
}
}``````
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