LeetCode #2384 — MEDIUM

Largest Palindromic Number

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Greedy

Example 1

"444947137"

Example 2

"00009"

Core Insight

What unlocks the optimal approach

In order to form a valid palindrome, other than the middle digit in an odd-length palindrome, every digit needs to exist on both sides.
A longer palindrome implies a larger valued palindrome. For palindromes of the same length, the larger digits should occur first.
We can count the occurrences of each digit and build the palindrome starting from the ends. Starting from the larger digits, if there are still at least 2 occurrences of a digit, we can place these digits on each side.
Make sure to consider the special case for the center digit (if any) and zeroes. There should not be leading zeroes.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2384: Largest Palindromic Number
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;
    }
}

// Accepted solution for LeetCode #2384: Largest Palindromic Number
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
}

# Accepted solution for LeetCode #2384: Largest Palindromic Number
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'

// Accepted solution for LeetCode #2384: Largest Palindromic Number
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2384: Largest Palindromic Number
// 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;
//     }
// }

// Accepted solution for LeetCode #2384: Largest Palindromic Number
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';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.