Question

1653. Minimum Deletions to Make String Balanced

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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 <= 105
• s[i] is 'a' or 'b'​​.

Solutions

Solution 1: Dynamic Programming

We define $f[i]$ as the minimum number of characters to be deleted in the first $i$ characters to make the string balanced. Initially, $f[0]=0$. The answer is $f[n]$.

We traverse the string $s$, maintaining a variable $b$, which represents the number of character 'b' in the characters before the current position.

• If the current character is 'b', it does not affect the balance of the first $i$ characters, so $f[i]=f[i-1]$, then we update $b \leftarrow b+1$.
• If the current character is 'a', we can choose to delete the current character, so $f[i]=f[i-1]+1$; or we can choose to delete the previous character 'b', so $f[i]=b$. Therefore, we take the minimum of the two, that is, $f[i]=\min(f[i-1]+1,b)$.

In summary, we can get the state transition equation:

$$ f[i]=\begin{cases} f[i-1], & s[i-1]='b'\ \min(f[i-1]+1,b), & s[i-1]='a' \end{cases} $$

The final answer is $f[n]$.

We notice that the state transition equation is only related to the previous state and the variable $b$, so we can just use an answer variable $ans$ to maintain the current $f[i]$, and there is no need to allocate an array $f$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

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]

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];
  }
}

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];
  }
};

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]
}

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];
}

Solution 2: Enumeration + Prefix Sum

We can enumerate each position $i$ in the string $s$, dividing the string $s$ into two parts, namely $s[0,..,i-1]$ and $s[i+1,..n-1]$. To make the string balanced, the number of characters we need to delete at the current position $i$ is the number of character 'b' in $s[0,..,i-1]$ plus the number of character 'a' in $s[i+1,..n-1]$.

Therefore, we maintain two variables $lb$ and $ra$ to represent the number of character 'b' in $s[0,..,i-1]$ and the number of character 'a' in $s[i+1,..n-1]$ respectively. The number of characters we need to delete is $lb+ra$. During the enumeration process, we update the variables $lb$ and $ra$.

The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the string $s$.

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

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;
  }
}

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;
  }
};

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
}

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;
}

Solution 3

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

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;
  }
}

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;
  }
};

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
}

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;
}