Algorithm
Problem Name: 1156. Swap For Longest Repeated Character Substring
You are given a string text. You can swap two of the characters in the text.
Return the length of the longest substring with repeated characters.
Example 1:
Input: text = "ababa" Output: 3 Explanation: We can swap the first 'b' with the last 'a', or the last 'b' with the first 'a'. Then, the longest repeated character substring is "aaa" with length 3.
Example 2:
Input: text = "aaabaaa" Output: 6 Explanation: Swap 'b' with the last 'a' (or the first 'a'), and we get longest repeated character substring "aaaaaa" with length 6.
Example 3:
Input: text = "aaaaa" Output: 5 Explanation: No need to swap, longest repeated character substring is "aaaaa" with length is 5.
Constraints:
1 <= text.length <= 2 * 104textconsist of lowercase English characters only.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public ListNode swapPairs(ListNode head) {
ListNode curr = head;
ListNode prev = null;
while (curr != null && curr.next != null) {
ListNode nextNode = curr.next;
if (curr == head) {
head = nextNode;
}
curr.next = nextNode.next;
nextNode.next = curr;
if (prev != null) {
prev.next = nextNode;
}
prev = curr;
curr = curr.next;
}
return head;
}
}
Input
text = "ababa"
Output
3
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxRepOpt1 = function(text) {
const count = [...text].reduce((a, c) => {
a[c] = a[c] || 0;
a[c]++;
return a;
}, {});
let ans = 0;
let i = 0;
while (i < text.length) {
let j = i;
const c = text.charAt(i);
while (j < text.length && text.charAt(j) === c) j++;
if (j - i < count[c]) {
let k = j + 1;
while (k < text.length && text.charAt(k) === c && k - i < count[c]) k++;
ans = Math.max(k - i, ans);
} else ans = Math.max(j - i, ans);
i = j;
}
return ans;
};
Input
text = "ababa"
Output
3
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxRepOpt1(self, S: str) -> int:
A = [[c, len(list(g))] for c, g in itertools.groupby(S)]
count = collections.Counter(S)
res = max(min(k + 1, count[c]) for c, k in A)
for i in range(1, len(A) - 1):
if A[i - 1][0] == A[i + 1][0] and A[i][1] == 1:
res = max(res, min(A[i - 1][1] + A[i + 1][1] + 1, count[A[i + 1][0]]))
return res
Input
text = "aaabaaa"
Output
6