Description#
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.<ul>
<li>For example, the value of <code>'a'</code> is <code>1</code>, the value of <code>'b'</code> is <code>2</code>, and so on. The value of <code>'z'</code> is <code>26</code>.</li>
</ul>
</li>
<li>Otherwise, assuming <code>i</code> is the index where the character occurs in the string <code>chars</code>, then its value is <code>vals[i]</code>.</li>
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] <= 1000
Solutions#
Solution 1: Prefix sum + Maintain the minimum prefix sum#
According to the description of the problem, we traverse each character $c$ in the string $s$, obtain its corresponding value $v$, and then update the current prefix sum $tot=tot+v$. Then, the cost of the maximum cost substring ending with $c$ is $tot$ minus the minimum prefix sum $mi$, that is, $tot-mi$. We update the answer $ans=max(ans,tot-mi)$ and maintain the minimum prefix sum $mi=min(mi,tot)$.
After the traversal is over, return the answer $ans$.
The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the length of the string $s$; and $C$ is the size of the character set, which is $26$ in this problem.
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| 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
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| 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;
}
}
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| class Solution {
public:
int maximumCostSubstring(string s, string chars, vector<int>& vals) {
vector<int> d(26);
iota(d.begin(), d.end(), 1);
int m = chars.size();
for (int i = 0; i < m; ++i) {
d[chars[i] - 'a'] = vals[i];
}
int ans = 0, tot = 0, mi = 0;
for (char& c : s) {
int v = d[c - 'a'];
tot += v;
ans = max(ans, tot - mi);
mi = min(mi, tot);
}
return ans;
}
};
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| 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
}
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| 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;
}
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Solution 2: Convert to the maximum subarray sum problem#
We can consider the value $v$ of each character $c$ as an integer, so the actual problem is to solve the maximum subarray sum problem.
We use the variable $f$ to maintain the cost of the maximum cost substring ending with the current character $c$. Each time we traverse to a character $c$, we update $f=max(f, 0) + v$. Then we update the answer $ans=max(ans,f)$.
The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the length of the string $s$; and $C$ is the size of the character set, which is $26$ in this problem.
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| class Solution:
def maximumCostSubstring(self, s: str, chars: str, vals: List[int]) -> int:
d = {c: v for c, v in zip(chars, vals)}
ans = f = 0
for c in s:
v = d.get(c, ord(c) - ord('a') + 1)
f = max(f, 0) + v
ans = max(ans, f)
return ans
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| 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, f = 0;
int n = s.length();
for (int i = 0; i < n; ++i) {
int v = d[s.charAt(i) - 'a'];
f = Math.max(f, 0) + v;
ans = Math.max(ans, f);
}
return ans;
}
}
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| class Solution {
public:
int maximumCostSubstring(string s, string chars, vector<int>& vals) {
vector<int> d(26);
iota(d.begin(), d.end(), 1);
int m = chars.size();
for (int i = 0; i < m; ++i) {
d[chars[i] - 'a'] = vals[i];
}
int ans = 0, f = 0;
for (char& c : s) {
int v = d[c - 'a'];
f = max(f, 0) + v;
ans = max(ans, f);
}
return ans;
}
};
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| 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]
}
f := 0
for _, c := range s {
v := d[c-'a']
f = max(f, 0) + v
ans = max(ans, f)
}
return
}
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| 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 f = 0;
for (const c of s) {
f = Math.max(f, 0) + d[c.charCodeAt(0) - 97];
ans = Math.max(ans, f);
}
return ans;
}
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