LeetCode #1542 — HARD

Find Longest Awesome Substring

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s. An awesome substring is a non-empty substring of s such that we can make any number of swaps in order to make it a palindrome.

Return the length of the maximum length awesome substring of s.

Example 1:

Input: s = "3242415"
Output: 5
Explanation: "24241" is the longest awesome substring, we can form the palindrome "24142" with some swaps.

Example 2:

Input: s = "12345678"
Output: 1

Example 3:

Input: s = "213123"
Output: 6
Explanation: "213123" is the longest awesome substring, we can form the palindrome "231132" with some swaps.

Constraints:

1 <= s.length <= 10⁵
s consists only of digits.

Step 01

Brute Force Baseline

Problem summary: You are given a string s. An awesome substring is a non-empty substring of s such that we can make any number of swaps in order to make it a palindrome. Return the length of the maximum length awesome substring of s.

Baseline thinking

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

Pattern signal: Hash Map · Bit Manipulation

Example 1

"3242415"

Example 2

"12345678"

Example 3

"213123"

Step 02

Core Insight

What unlocks the optimal approach

Given the character counts, under what conditions can a palindrome be formed ?
From left to right, use bitwise xor-operation to compute for any prefix the number of times modulo 2 of each digit. (mask ^= (1<<(s[i]-'0')).
Expected complexity is O(n*A) where A is the alphabet (10).

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

Largest constraint values

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 #1542: Find Longest Awesome Substring
class Solution {
    public int longestAwesome(String s) {
        int[] d = new int[1024];
        int st = 0, ans = 1;
        Arrays.fill(d, -1);
        d[0] = 0;
        for (int i = 1; i <= s.length(); ++i) {
            int v = s.charAt(i - 1) - '0';
            st ^= 1 << v;
            if (d[st] >= 0) {
                ans = Math.max(ans, i - d[st]);
            } else {
                d[st] = i;
            }
            for (v = 0; v < 10; ++v) {
                if (d[st ^ (1 << v)] >= 0) {
                    ans = Math.max(ans, i - d[st ^ (1 << v)]);
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1542: Find Longest Awesome Substring
func longestAwesome(s string) int {
	d := [1024]int{}
	d[0] = 1
	st, ans := 0, 1
	for i, c := range s {
		i += 2
		st ^= 1 << (c - '0')
		if d[st] > 0 {
			ans = max(ans, i-d[st])
		} else {
			d[st] = i
		}
		for v := 0; v < 10; v++ {
			if d[st^(1<<v)] > 0 {
				ans = max(ans, i-d[st^(1<<v)])
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #1542: Find Longest Awesome Substring
class Solution:
    def longestAwesome(self, s: str) -> int:
        st = 0
        d = {0: -1}
        ans = 1
        for i, c in enumerate(s):
            v = int(c)
            st ^= 1 << v
            if st in d:
                ans = max(ans, i - d[st])
            else:
                d[st] = i
            for v in range(10):
                if st ^ (1 << v) in d:
                    ans = max(ans, i - d[st ^ (1 << v)])
        return ans

// Accepted solution for LeetCode #1542: Find Longest Awesome Substring
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn longest_awesome(s: String) -> i32 {
        let mut mask = 0;
        let mut hm: HashMap<u32, usize> = HashMap::new();
        hm.insert(0, 0);
        let mut res = 0;
        for (i, b) in s.bytes().enumerate() {
            mask ^= 1 << (b - b'0');
            if let Some(j) = hm.get(&mask) {
                res = res.max(i + 1 - j);
            }
            for k in 0..10 {
                if let Some(j) = hm.get(&(mask ^ (1 << k))) {
                    res = res.max(i + 1 - j);
                }
            }
            hm.entry(mask).or_insert(i + 1);
        }
        res as i32
    }
}

#[test]
fn test() {
    let s = "3242415".to_string();
    let res = 5;
    assert_eq!(Solution::longest_awesome(s), res);
    let s = "12345678".to_string();
    let res = 1;
    assert_eq!(Solution::longest_awesome(s), res);
    let s = "213123".to_string();
    let res = 6;
    assert_eq!(Solution::longest_awesome(s), res);
    let s = "00".to_string();
    let res = 2;
    assert_eq!(Solution::longest_awesome(s), res);
}

// Accepted solution for LeetCode #1542: Find Longest Awesome Substring
function longestAwesome(s: string): number {
    const d: number[] = Array(1024).fill(-1);
    let [st, ans] = [0, 1];
    d[0] = 0;

    for (let i = 1; i <= s.length; ++i) {
        const v = s.charCodeAt(i - 1) - '0'.charCodeAt(0);
        st ^= 1 << v;

        if (d[st] >= 0) {
            ans = Math.max(ans, i - d[st]);
        } else {
            d[st] = i;
        }

        for (let v = 0; v < 10; ++v) {
            if (d[st ^ (1 << v)] >= 0) {
                ans = Math.max(ans, i - d[st ^ (1 << v)]);
            }
        }
    }

    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(n × C)

Space

O(2^C)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

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.