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2384. Largest Palindromic Number

Description

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.

Solutions

Solution 1

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class Solution:
    def largestPalindromic(self, num: str) -> str:
        cnt = Counter(num)
        ans = ''
        for i in range(9, -1, -1):
            v = str(i)
            if cnt[v] % 2:
                ans = v
                cnt[v] -= 1
                break
        for i in range(10):
            v = str(i)
            if cnt[v]:
                cnt[v] //= 2
                s = cnt[v] * v
                ans = s + ans + s
        return ans.strip('0') or '0'

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class Solution {
    public String largestPalindromic(String num) {
        int[] cnt = new int[10];
        for (char c : num.toCharArray()) {
            ++cnt[c - '0'];
        }
        String mid = "";
        for (int i = 9; i >= 0; --i) {
            if (cnt[i] % 2 == 1) {
                mid += i;
                --cnt[i];
                break;
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < 10; ++i) {
            if (cnt[i] > 0) {
                cnt[i] >>= 1;
                sb.append(("" + i).repeat(cnt[i]));
            }
        }
        while (sb.length() > 0 && sb.charAt(sb.length() - 1) == '0') {
            sb.deleteCharAt(sb.length() - 1);
        }
        String t = sb.toString();
        String ans = sb.reverse().toString() + mid + t;
        return "".equals(ans) ? "0" : ans;
    }
}

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class Solution {
public:
    string largestPalindromic(string num) {
        vector<int> cnt(10);
        for (char c : num) ++cnt[c - '0'];
        string mid = "";
        for (int i = 9; ~i; --i) {
            if (cnt[i] % 2) {
                mid += (i + '0');
                --cnt[i];
                break;
            }
        }
        string t = "";
        for (int i = 0; i < 10; ++i) {
            if (cnt[i]) {
                cnt[i] >>= 1;
                while (cnt[i]--) {
                    t += (i + '0');
                }
            }
        }
        while (t.size() && t.back() == '0') {
            t.pop_back();
        }
        string ans = t;
        reverse(ans.begin(), ans.end());
        ans += mid + t;
        return ans == "" ? "0" : ans;
    }
};

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func largestPalindromic(num string) string {
    cnt := make([]int, 10)
    for _, c := range num {
        cnt[c-'0']++
    }
    ans := ""
    for i := 9; i >= 0; i-- {
        if cnt[i]%2 == 1 {
            ans = strconv.Itoa(i)
            cnt[i]--
            break
        }
    }
    for i := 0; i < 10; i++ {
        if cnt[i] > 0 {
            cnt[i] >>= 1
            s := strings.Repeat(strconv.Itoa(i), cnt[i])
            ans = s + ans + s
        }
    }
    ans = strings.Trim(ans, "0")
    if ans == "" {
        return "0"
    }
    return ans
}

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function largestPalindromic(num: string): string {
    const count = new Array(10).fill(0);
    for (const c of num) {
        count[c]++;
    }
    while (count.reduce((r, v) => (v % 2 === 1 ? r + 1 : r), 0) > 1) {
        for (let i = 0; i < 10; i++) {
            if (count[i] % 2 === 1) {
                count[i]--;
                break;
            }
        }
    }

    let res = [];
    let oddIndex = -1;
    for (let i = 9; i >= 0; i--) {
        if (count[i] % 2 == 1) {
            oddIndex = i;
            count[i] -= 1;
        }
        res.push(...new Array(count[i] >> 1).fill(i));
    }
    if (oddIndex !== -1) {
        res.push(oddIndex);
    }
    const n = res.length;
    for (let i = 0; i < n; i++) {
        if (res[i] !== 0) {
            res = res.slice(i);
            if (oddIndex !== -1) {
                res.push(...[...res.slice(0, res.length - 1)].reverse());
            } else {
                res.push(...[...res].reverse());
            }
            return res.join('');
        }
    }

    return '0';
}

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