Move from brute-force thinking to an efficient approach using two pointers strategy.
Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the following algorithm on the string any number of times:
s where all the characters in the prefix are equal.s where all the characters in this suffix are equal.Return the minimum length of s after performing the above operation any number of times (possibly zero times).
Example 1:
Input: s = "ca" Output: 2 Explanation: You can't remove any characters, so the string stays as is.
Example 2:
Input: s = "cabaabac" Output: 0 Explanation: An optimal sequence of operations is: - Take prefix = "c" and suffix = "c" and remove them, s = "abaaba". - Take prefix = "a" and suffix = "a" and remove them, s = "baab". - Take prefix = "b" and suffix = "b" and remove them, s = "aa". - Take prefix = "a" and suffix = "a" and remove them, s = "".
Example 3:
Input: s = "aabccabba" Output: 3 Explanation: An optimal sequence of operations is: - Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb". - Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
Constraints:
1 <= s.length <= 105s only consists of characters 'a', 'b', and 'c'.Problem summary: Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the following algorithm on the string any number of times: Pick a non-empty prefix from the string s where all the characters in the prefix are equal. Pick a non-empty suffix from the string s where all the characters in this suffix are equal. The prefix and the suffix should not intersect at any index. The characters from the prefix and suffix must be the same. Delete both the prefix and the suffix. Return the minimum length of s after performing the above operation any number of times (possibly zero times).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
"ca"
"cabaabac"
"aabccabba"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1750: Minimum Length of String After Deleting Similar Ends
class Solution {
public int minimumLength(String s) {
int i = 0, j = s.length() - 1;
while (i < j && s.charAt(i) == s.charAt(j)) {
while (i + 1 < j && s.charAt(i) == s.charAt(i + 1)) {
++i;
}
while (i < j - 1 && s.charAt(j) == s.charAt(j - 1)) {
--j;
}
++i;
--j;
}
return Math.max(0, j - i + 1);
}
}
// Accepted solution for LeetCode #1750: Minimum Length of String After Deleting Similar Ends
func minimumLength(s string) int {
i, j := 0, len(s)-1
for i < j && s[i] == s[j] {
for i+1 < j && s[i] == s[i+1] {
i++
}
for i < j-1 && s[j] == s[j-1] {
j--
}
i, j = i+1, j-1
}
return max(0, j-i+1)
}
# Accepted solution for LeetCode #1750: Minimum Length of String After Deleting Similar Ends
class Solution:
def minimumLength(self, s: str) -> int:
i, j = 0, len(s) - 1
while i < j and s[i] == s[j]:
while i + 1 < j and s[i] == s[i + 1]:
i += 1
while i < j - 1 and s[j - 1] == s[j]:
j -= 1
i, j = i + 1, j - 1
return max(0, j - i + 1)
// Accepted solution for LeetCode #1750: Minimum Length of String After Deleting Similar Ends
impl Solution {
pub fn minimum_length(s: String) -> i32 {
let s = s.as_bytes();
let n = s.len();
let mut start = 0;
let mut end = n - 1;
while start < end && s[start] == s[end] {
while start + 1 < end && s[start] == s[start + 1] {
start += 1;
}
while start < end - 1 && s[end] == s[end - 1] {
end -= 1;
}
start += 1;
end -= 1;
}
(0).max(end - start + 1) as i32
}
}
// Accepted solution for LeetCode #1750: Minimum Length of String After Deleting Similar Ends
function minimumLength(s: string): number {
let i = 0;
let j = s.length - 1;
while (i < j && s[i] === s[j]) {
while (i + 1 < j && s[i + 1] === s[i]) {
++i;
}
while (i < j - 1 && s[j - 1] === s[j]) {
--j;
}
++i;
--j;
}
return Math.max(0, j - i + 1);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.