Description#
You are given a string s consisting only of characters 'a' and 'b'.
You can delete any number of characters in s to make s balanced. s is balanced if there is no pair of indices (i,j) such that i < j and s[i] = 'b' and s[j]= 'a'.
Return the minimum number of deletions needed to make s balanced.
Example 1:
Input: s = "aababbab"
Output: 2
Explanation: You can either:
Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
Example 2:
Input: s = "bbaaaaabb"
Output: 2
Explanation: The only solution is to delete the first two characters.
Constraints:
1 <= s.length <= 105s[i] is 'a' or 'b'.
Solutions#
Solution 1#
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| class Solution:
def minimumDeletions(self, s: str) -> int:
n = len(s)
f = [0] * (n + 1)
b = 0
for i, c in enumerate(s, 1):
if c == 'b':
f[i] = f[i - 1]
b += 1
else:
f[i] = min(f[i - 1] + 1, b)
return f[n]
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| class Solution {
public int minimumDeletions(String s) {
int n = s.length();
int[] f = new int[n + 1];
int b = 0;
for (int i = 1; i <= n; ++i) {
if (s.charAt(i - 1) == 'b') {
f[i] = f[i - 1];
++b;
} else {
f[i] = Math.min(f[i - 1] + 1, b);
}
}
return f[n];
}
}
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| class Solution {
public:
int minimumDeletions(string s) {
int n = s.size();
int f[n + 1];
memset(f, 0, sizeof(f));
int b = 0;
for (int i = 1; i <= n; ++i) {
if (s[i - 1] == 'b') {
f[i] = f[i - 1];
++b;
} else {
f[i] = min(f[i - 1] + 1, b);
}
}
return f[n];
}
};
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| func minimumDeletions(s string) int {
n := len(s)
f := make([]int, n+1)
b := 0
for i, c := range s {
i++
if c == 'b' {
f[i] = f[i-1]
b++
} else {
f[i] = min(f[i-1]+1, b)
}
}
return f[n]
}
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| function minimumDeletions(s: string): number {
const n = s.length;
const f = new Array(n + 1).fill(0);
let b = 0;
for (let i = 1; i <= n; ++i) {
if (s.charAt(i - 1) === 'b') {
f[i] = f[i - 1];
++b;
} else {
f[i] = Math.min(f[i - 1] + 1, b);
}
}
return f[n];
}
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Solution 2#
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| class Solution:
def minimumDeletions(self, s: str) -> int:
ans = b = 0
for c in s:
if c == 'b':
b += 1
else:
ans = min(ans + 1, b)
return ans
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| class Solution {
public int minimumDeletions(String s) {
int n = s.length();
int ans = 0, b = 0;
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == 'b') {
++b;
} else {
ans = Math.min(ans + 1, b);
}
}
return ans;
}
}
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| class Solution {
public:
int minimumDeletions(string s) {
int ans = 0, b = 0;
for (char& c : s) {
if (c == 'b') {
++b;
} else {
ans = min(ans + 1, b);
}
}
return ans;
}
};
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| func minimumDeletions(s string) int {
ans, b := 0, 0
for _, c := range s {
if c == 'b' {
b++
} else {
ans = min(ans+1, b)
}
}
return ans
}
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| function minimumDeletions(s: string): number {
const n = s.length;
let ans = 0,
b = 0;
for (let i = 0; i < n; ++i) {
if (s.charAt(i) === 'b') {
++b;
} else {
ans = Math.min(ans + 1, b);
}
}
return ans;
}
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Solution 3#
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| class Solution:
def minimumDeletions(self, s: str) -> int:
lb, ra = 0, s.count('a')
ans = len(s)
for c in s:
ra -= c == 'a'
ans = min(ans, lb + ra)
lb += c == 'b'
return ans
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| class Solution {
public int minimumDeletions(String s) {
int lb = 0, ra = 0;
int n = s.length();
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == 'a') {
++ra;
}
}
int ans = n;
for (int i = 0; i < n; ++i) {
ra -= (s.charAt(i) == 'a' ? 1 : 0);
ans = Math.min(ans, lb + ra);
lb += (s.charAt(i) == 'b' ? 1 : 0);
}
return ans;
}
}
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| class Solution {
public:
int minimumDeletions(string s) {
int lb = 0, ra = count(s.begin(), s.end(), 'a');
int ans = ra;
for (char& c : s) {
ra -= c == 'a';
ans = min(ans, lb + ra);
lb += c == 'b';
}
return ans;
}
};
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| func minimumDeletions(s string) int {
lb, ra := 0, strings.Count(s, "a")
ans := ra
for _, c := range s {
if c == 'a' {
ra--
}
if t := lb + ra; ans > t {
ans = t
}
if c == 'b' {
lb++
}
}
return ans
}
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| function minimumDeletions(s: string): number {
let lb = 0,
ra = 0;
const n = s.length;
for (let i = 0; i < n; ++i) {
if (s.charAt(i) === 'a') {
++ra;
}
}
let ans = n;
for (let i = 0; i < n; ++i) {
ra -= s.charAt(i) === 'a' ? 1 : 0;
ans = Math.min(ans, lb + ra);
lb += s.charAt(i) === 'b' ? 1 : 0;
}
return ans;
}
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