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.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).
The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.
Example 1:
Input: s = "011101" Output: 5 Explanation: All possible ways of splitting s into two non-empty substrings are: left = "0" and right = "11101", score = 1 + 4 = 5 left = "01" and right = "1101", score = 1 + 3 = 4 left = "011" and right = "101", score = 1 + 2 = 3 left = "0111" and right = "01", score = 1 + 1 = 2 left = "01110" and right = "1", score = 2 + 1 = 3
Example 2:
Input: s = "00111" Output: 5 Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5
Example 3:
Input: s = "1111" Output: 3
Constraints:
2 <= s.length <= 500s consists of characters '0' and '1' only.Problem summary: Given a string s of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring). The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
"011101"
"00111"
"1111"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1422: Maximum Score After Splitting a String
class Solution {
public int maxScore(String s) {
int l = 0, r = 0;
int n = s.length();
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == '1') {
++r;
}
}
int ans = 0;
for (int i = 0; i < n - 1; ++i) {
l += (s.charAt(i) - '0') ^ 1;
r -= s.charAt(i) - '0';
ans = Math.max(ans, l + r);
}
return ans;
}
}
// Accepted solution for LeetCode #1422: Maximum Score After Splitting a String
func maxScore(s string) (ans int) {
l, r := 0, strings.Count(s, "1")
for _, c := range s[:len(s)-1] {
if c == '0' {
l++
} else {
r--
}
ans = max(ans, l+r)
}
return
}
# Accepted solution for LeetCode #1422: Maximum Score After Splitting a String
class Solution:
def maxScore(self, s: str) -> int:
l, r = 0, s.count("1")
ans = 0
for x in s[:-1]:
l += int(x) ^ 1
r -= int(x)
ans = max(ans, l + r)
return ans
// Accepted solution for LeetCode #1422: Maximum Score After Splitting a String
impl Solution {
pub fn max_score(s: String) -> i32 {
let mut l = 0;
let mut r = s.bytes().filter(|&b| b == b'1').count() as i32;
let mut ans = 0;
let cs = s.as_bytes();
for i in 0..s.len() - 1 {
l += ((cs[i] - b'0') ^ 1) as i32;
r -= (cs[i] - b'0') as i32;
ans = ans.max(l + r);
}
ans
}
}
// Accepted solution for LeetCode #1422: Maximum Score After Splitting a String
function maxScore(s: string): number {
let [l, r] = [0, 0];
for (const c of s) {
r += c === '1' ? 1 : 0;
}
let ans = 0;
for (let i = 0; i < s.length - 1; ++i) {
if (s[i] === '0') {
++l;
} else {
--r;
}
ans = Math.max(ans, l + r);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.