LeetCode #2231 — EASY

Largest Number After Digit Swaps by Parity

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).

Return the largest possible value of num after any number of swaps.

Example 1:

Input: num = 1234
Output: 3412
Explanation: Swap the digit 3 with the digit 1, this results in the number 3214.
Swap the digit 2 with the digit 4, this results in the number 3412.
Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.

Example 2:

Input: num = 65875
Output: 87655
Explanation: Swap the digit 8 with the digit 6, this results in the number 85675.
Swap the first digit 5 with the digit 7, this results in the number 87655.
Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.

Constraints:

1 <= num <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits). Return the largest possible value of num after any number of swaps.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

Example 2

Core Insight

What unlocks the optimal approach

The bigger digit should appear first (more to the left) because it contributes more to the value of the number.
Get all the even digits, as well as odd digits. Sort them separately.
Reconstruct the number by giving the earlier digits the highest available digit of the same parity.

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 #2231: Largest Number After Digit Swaps by Parity
class Solution {
    public int largestInteger(int num) {
        char[] s = String.valueOf(num).toCharArray();
        int[] cnt = new int[10];
        for (char c : s) {
            ++cnt[c - '0'];
        }
        int[] idx = {8, 9};
        int ans = 0;
        for (char c : s) {
            int x = c - '0';
            while (cnt[idx[x & 1]] == 0) {
                idx[x & 1] -= 2;
            }
            ans = ans * 10 + idx[x & 1];
            cnt[idx[x & 1]]--;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2231: Largest Number After Digit Swaps by Parity
func largestInteger(num int) int {
	s := []byte(fmt.Sprint(num))
	cnt := [10]int{}

	for _, c := range s {
		cnt[c-'0']++
	}

	idx := [2]int{8, 9}
	ans := 0

	for _, c := range s {
		x := int(c - '0')
		for cnt[idx[x&1]] == 0 {
			idx[x&1] -= 2
		}
		ans = ans*10 + idx[x&1]
		cnt[idx[x&1]]--
	}

	return ans
}

# Accepted solution for LeetCode #2231: Largest Number After Digit Swaps by Parity
class Solution:
    def largestInteger(self, num: int) -> int:
        nums = [int(c) for c in str(num)]
        cnt = Counter(nums)
        idx = [8, 9]
        ans = 0
        for x in nums:
            while cnt[idx[x & 1]] == 0:
                idx[x & 1] -= 2
            ans = ans * 10 + idx[x & 1]
            cnt[idx[x & 1]] -= 1
        return ans

// Accepted solution for LeetCode #2231: Largest Number After Digit Swaps by Parity
/**
 * [2231] Largest Number After Digit Swaps by Parity
 *
 * You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).
 * Return the largest possible value of num after any number of swaps.
 *  
 * Example 1:
 *
 * Input: num = 1234
 * Output: 3412
 * Explanation: Swap the digit 3 with the digit 1, this results in the number 3214.
 * Swap the digit 2 with the digit 4, this results in the number 3412.
 * Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
 * Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.
 *
 * Example 2:
 *
 * Input: num = 65875
 * Output: 87655
 * Explanation: Swap the digit 8 with the digit 6, this results in the number 85675.
 * Swap the first digit 5 with the digit 7, this results in the number 87655.
 * Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.
 *
 *  
 * Constraints:
 *
 * 	1 <= num <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/largest-number-after-digit-swaps-by-parity/
// discuss: https://leetcode.com/problems/largest-number-after-digit-swaps-by-parity/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn largest_integer(num: i32) -> i32 {
        let mut num = num as usize;
        let mut evens_odds = [
            std::collections::BinaryHeap::new(),
            std::collections::BinaryHeap::new(),
        ];
        let mut digits = Vec::new();

        while num > 0 {
            let dig = num % 10;
            evens_odds[dig % 2].push(dig as i32);
            digits.push(dig);
            num /= 10;
        }

        digits
            .iter()
            .rev()
            .fold(0, |res, &dig| 10 * res + evens_odds[dig % 2].pop().unwrap())
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2231_example_1() {
        let num = 1234;

        let result = 3412;

        assert_eq!(Solution::largest_integer(num), result);
    }

    #[test]
    fn test_2231_example_2() {
        let num = 65875;

        let result = 87655;

        assert_eq!(Solution::largest_integer(num), result);
    }
}

// Accepted solution for LeetCode #2231: Largest Number After Digit Swaps by Parity
function largestInteger(num: number): number {
    const s = num.toString().split('');
    const cnt = Array(10).fill(0);

    for (const c of s) {
        cnt[+c]++;
    }

    const idx = [8, 9];
    let ans = 0;

    for (const c of s) {
        const x = +c;
        while (cnt[idx[x % 2]] === 0) {
            idx[x % 2] -= 2;
        }
        ans = ans * 10 + idx[x % 2];
        cnt[idx[x % 2]]--;
    }

    return ans;
}

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(log \textitnum)

Space

O(log \textitnum)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.