##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/1542.html

# 1542. Find Longest Awesome Substring (Hard)

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 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.


Example 4:

Input: s = "00"
Output: 2


Constraints:

• 1 <= s.length <= 10^5
• s consists only of digits.

Related Topics: String, Bit Manipulation

## Solution 1.

• class Solution {
public int longestAwesome(String s) {
if (s == null || s.length() == 0)
return 0;
Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
int length = s.length();
int[][] counts = new int[length][10];
int digit0 = s.charAt(0) - '0';
counts[0][digit0] = 1;
int num0 = countsToNum(counts[0]);
List<Integer> list0 = new ArrayList<Integer>();
map.put(num0, list0);
for (int i = 1; i < length; i++) {
int digit = s.charAt(i) - '0';
for (int j = 0; j <= 9; j++)
counts[i][j] = counts[i - 1][j];
counts[i][digit] = 1 - counts[i][digit];
int num = countsToNum(counts[i]);
List<Integer> list = map.getOrDefault(num, new ArrayList<Integer>());
map.put(num, list);
}
int maxLength = 1;
Set<Integer> visited = new HashSet<Integer>();
for (int i = length - 1; i > 0; i--) {
int num = countsToNum(counts[i]);
if (num == 0 || (num & (num - 1)) == 0)
maxLength = Math.max(maxLength, i + 1);
int[] curCounts = new int[10];
System.arraycopy(counts[i], 0, curCounts, 0, 10);
List<Integer> list = map.get(num);
if (list.size() > 1)
maxLength = Math.max(maxLength, i - list.get(0));
for (int j = 0; j <= 9; j++) {
curCounts[j] = 1 - curCounts[j];
int newNum = countsToNum(curCounts);
if (map.containsKey(newNum)) {
List<Integer> newList = map.get(newNum);
maxLength = Math.max(maxLength, i - newList.get(0));
}
curCounts[j] = 1 - curCounts[j];
}
}
}
return maxLength;
}

public int countsToNum(int[] counts) {
int num = 0;
int base = 1;
int length = counts.length;
for (int i = 0; i < length; i++) {
num += counts[i] * base;
base *= 2;
}
return num;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/find-longest-awesome-substring/
// Time: O(N)
// Space: O(1) since there are at most 2^10=1024 states.
class Solution {
public:
int longestAwesome(string s) {
unordered_map<int, int> m{ {0,-1} }; // mask -> index of first occurrence
int ans = 0;
for (int i = 0, mask = 0; i < s.size(); ++i) {
mask ^= 1 << (s[i] - '0');
for (int j = 0; j < 10; ++j) {
int prev = mask ^ (1 << j);
if (m.count(prev)) ans = max(ans, i - m[prev]);
}
}
return ans;
}
};

• 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


• 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
}

func max(a, b int) int {
if a > b {
return a
}
return b
}