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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a string s, a string chars of distinct characters and an integer array vals of the same length as chars.
The cost of the substring is the sum of the values of each character in the substring. The cost of an empty string is considered 0.
The value of the character is defined in the following way:
chars, then its value is its corresponding position (1-indexed) in the alphabet.
'a' is 1, the value of 'b' is 2, and so on. The value of 'z' is 26.i is the index where the character occurs in the string chars, then its value is vals[i].Return the maximum cost among all substrings of the string s.
Example 1:
Input: s = "adaa", chars = "d", vals = [-1000] Output: 2 Explanation: The value of the characters "a" and "d" is 1 and -1000 respectively. The substring with the maximum cost is "aa" and its cost is 1 + 1 = 2. It can be proven that 2 is the maximum cost.
Example 2:
Input: s = "abc", chars = "abc", vals = [-1,-1,-1] Output: 0 Explanation: The value of the characters "a", "b" and "c" is -1, -1, and -1 respectively. The substring with the maximum cost is the empty substring "" and its cost is 0. It can be proven that 0 is the maximum cost.
Constraints:
1 <= s.length <= 105s consist of lowercase English letters.1 <= chars.length <= 26chars consist of distinct lowercase English letters.vals.length == chars.length-1000 <= vals[i] <= 1000Problem summary: You are given a string s, a string chars of distinct characters and an integer array vals of the same length as chars. The cost of the substring is the sum of the values of each character in the substring. The cost of an empty string is considered 0. The value of the character is defined in the following way: If the character is not in the string chars, then its value is its corresponding position (1-indexed) in the alphabet. For example, the value of 'a' is 1, the value of 'b' is 2, and so on. The value of 'z' is 26. Otherwise, assuming i is the index where the character occurs in the string chars, then its value is vals[i]. Return the maximum cost among all substrings of the string s.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Dynamic Programming
"adaa" "d" [-1000]
"abc" "abc" [-1,-1,-1]
maximum-subarray)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
class Solution {
public int maximumCostSubstring(String s, String chars, int[] vals) {
int[] d = new int[26];
for (int i = 0; i < d.length; ++i) {
d[i] = i + 1;
}
int m = chars.length();
for (int i = 0; i < m; ++i) {
d[chars.charAt(i) - 'a'] = vals[i];
}
int ans = 0, tot = 0, mi = 0;
int n = s.length();
for (int i = 0; i < n; ++i) {
int v = d[s.charAt(i) - 'a'];
tot += v;
ans = Math.max(ans, tot - mi);
mi = Math.min(mi, tot);
}
return ans;
}
}
// Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
func maximumCostSubstring(s string, chars string, vals []int) (ans int) {
d := [26]int{}
for i := range d {
d[i] = i + 1
}
for i, c := range chars {
d[c-'a'] = vals[i]
}
tot, mi := 0, 0
for _, c := range s {
v := d[c-'a']
tot += v
ans = max(ans, tot-mi)
mi = min(mi, tot)
}
return
}
# Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
class Solution:
def maximumCostSubstring(self, s: str, chars: str, vals: List[int]) -> int:
d = {c: v for c, v in zip(chars, vals)}
ans = tot = mi = 0
for c in s:
v = d.get(c, ord(c) - ord('a') + 1)
tot += v
ans = max(ans, tot - mi)
mi = min(mi, tot)
return ans
// Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
// class Solution {
// public int maximumCostSubstring(String s, String chars, int[] vals) {
// int[] d = new int[26];
// for (int i = 0; i < d.length; ++i) {
// d[i] = i + 1;
// }
// int m = chars.length();
// for (int i = 0; i < m; ++i) {
// d[chars.charAt(i) - 'a'] = vals[i];
// }
// int ans = 0, tot = 0, mi = 0;
// int n = s.length();
// for (int i = 0; i < n; ++i) {
// int v = d[s.charAt(i) - 'a'];
// tot += v;
// ans = Math.max(ans, tot - mi);
// mi = Math.min(mi, tot);
// }
// return ans;
// }
// }
// Accepted solution for LeetCode #2606: Find the Substring With Maximum Cost
function maximumCostSubstring(s: string, chars: string, vals: number[]): number {
const d: number[] = Array.from({ length: 26 }, (_, i) => i + 1);
for (let i = 0; i < chars.length; ++i) {
d[chars.charCodeAt(i) - 97] = vals[i];
}
let ans = 0;
let tot = 0;
let mi = 0;
for (const c of s) {
tot += d[c.charCodeAt(0) - 97];
ans = Math.max(ans, tot - mi);
mi = Math.min(mi, tot);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
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.
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.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.