给定一个字符串S ,任务是找到所需的最少字符删除,使得字符串S仅包含两个交替字符。
例子:
Input: S = “adebbeeaebd”
Output: 7
Explanation: Removing all occurrences of ‘b’ and ‘e’ modifies the string to “adad”, which consist of alternating occurrences of ‘a’ and ‘d’.
Input: S = “abccd”
Output: 3
Explanation: Removing all occurrences of ‘c’ and ‘d’ modifies the string to “ab”, which consist of alternating ‘a’ and ‘b’.
方法:该问题可以通过生成所有可能的26对2英文字母并在字符串S 中找到具有最大交替出现长度的对,例如len来解决。然后,通过从N 中减去len来打印需要删除的字符数。
请按照以下步骤解决给定的问题:
- 初始化一个变量,比如len,以存储一对字符交替出现的最大长度。
- 迭代每对可能的英文字母,并对每一对执行以下操作:
- 遍历字符串S的字符,并找到长,说newlen从字符串S两个字符交替出现。
- 检查len是否小于newlen 。如果发现为真,则更新len = newlen 。
- 最后,打印N – len作为所需答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
int findLength(string s, char i, char j)
{
// Stores the next character
// for alternating sequence
char required = i;
// Stores the length of alternating
// occurrences of a pair of characters
int length = 0;
// Traverse the given string
for (char curr : s) {
// If current character is same
// as the required character
if (curr == required) {
// Increase length by 1
length += 1;
// Reset required character
if (required == i)
required = j;
else
required = i;
}
}
// Return the length
return length;
}
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
int minimumDeletions(string S)
{
// Stores maximum length
// of alternating sequence
// of two characters
int len = 0;
// Stores length of the string
int n = S.length();
// Generate every pair
// of English alphabets
for (char i = 'a'; i <= 'z'; i++) {
for (char j = i + 1; j <= 'z'; j++) {
// Function call to find length
// of alternating sequence for
// currrent pair of characters
int newLen = findLength(S, i, j);
// Update len to store the maximum
// of len and newLen in len
len = max(len, newLen);
}
}
// Return n - len as the final result
return n - len;
}
// Driver Code
int main()
{
// Given Input
string S = "adebbeeaebd";
// Function call to find minimum
// characters required to be removed
// from S to make it an alternating
// sequence of a pair of characters
cout << minimumDeletions(S);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
static int findLength(String s, char i, char j)
{
// Stores the next character
// for alternating sequence
char required = i;
// Stores the length of alternating
// occurrences of a pair of characters
int length = 0;
// Traverse the given string
for(int k = 0; k < s.length(); k++)
{
char curr = s.charAt(k);
// If current character is same
// as the required character
if (curr == required)
{
// Increase length by 1
length += 1;
// Reset required character
if (required == i)
required = j;
else
required = i;
}
}
// Return the length
return length;
}
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
static int minimumDeletions(String S)
{
// Stores maximum length
// of alternating sequence
// of two characters
int len = 0;
// Stores length of the string
int n = S.length();
// Generate every pair
// of English alphabets
for(int i = 0; i < 26; i++)
{
for(int j = i + 1; j < 26; j++)
{
// Function call to find length
// of alternating sequence for
// currrent pair of characters
int newLen = findLength(S, (char)(i + 97),
(char)(j + 97));
// Update len to store the maximum
// of len and newLen in len
len = Math.max(len, newLen);
}
}
// Return n - len as the final result
return n - len;
}
// Driver Code
public static void main (String[] args)
{
// Given Input
String S = "adebbeeaebd";
// Function call to find minimum
// characters required to be removed
// from S to make it an alternating
// sequence of a pair of characters
System.out.print(minimumDeletions(S));
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach
# Function to find the maximum
# length of alternating occurrences
# of a pair of characters in a string s
def findLength(s, i, j):
# Stores the next character
# for alternating sequence
required = i
# Stores the length of alternating
# occurrences of a pair of characters
length = 0
# Traverse the given string
for curr in s:
# If current character is same
# as the required character
if (curr == required):
# Increase length by 1
length += 1
# Reset required character
if (required == i):
required = j
else:
required = i
# Return the length
return length
# Function to find minimum characters
# required to be deleted from S to
# obtain an alternating sequence
def minimumDeletions(S):
# Stores maximum length
# of alternating sequence
# of two characters
len1 = 0
# Stores length of the string
n = len(S)
# Generate every pair
# of English alphabets
for i in range(0, 26, 1):
for j in range(i + 1, 26, 1):
# Function call to find length
# of alternating sequence for
# currrent pair of characters
newLen = findLength(S, chr(i + 97),
chr(j + 97))
# Update len to store the maximum
# of len and newLen in len
len1 = max(len1, newLen)
# Return n - len as the final result
return n - len1
# Driver Code
if __name__ == '__main__':
# Given Input
S = "adebbeeaebd"
# Function call to find minimum
# characters required to be removed
# from S to make it an alternating
# sequence of a pair of characters
print(minimumDeletions(S))
# This code is contributed by ipg2016107
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum
// length of alternating occurrences
// of a pair of characters in a string s
static int findLength(string s, char i, char j)
{
// Stores the next character
// for alternating sequence
char required = i;
// Stores the length of alternating
// occurrences of a pair of characters
int length = 0;
// Traverse the given string
for(int k = 0; k < s.Length; k++)
{
char curr = s[k];
// If current character is same
// as the required character
if (curr == required)
{
// Increase length by 1
length += 1;
// Reset required character
if (required == i)
required = j;
else
required = i;
}
}
// Return the length
return length;
}
// Function to find minimum characters
// required to be deleted from S to
// obtain an alternating sequence
static int minimumDeletions(string S)
{
// Stores maximum length
// of alternating sequence
// of two characters
int len = 0;
// Stores length of the string
int n = S.Length;
// Generate every pair
// of English alphabets
for(int i = 0; i < 26; i++)
{
for(int j = i + 1; j < 26; j++)
{
// Function call to find length
// of alternating sequence for
// currrent pair of characters
int newLen = findLength(S, (char)(i + 97),
(char)(j + 97));
// Update len to store the maximum
// of len and newLen in len
len = Math.Max(len, newLen);
}
}
// Return n - len as the final result
return n - len;
}
// Driver Code
public static void Main(string[] args)
{
// Given Input
string S = "adebbeeaebd";
// Function call to find minimum
// characters required to be removed
// from S to make it an alternating
// sequence of a pair of characters
Console.WriteLine(minimumDeletions(S));
}
}
// This code is contributed by avijitmondal1998
Javascript
输出:
7
时间复杂度: O(N)
辅助空间: O(1)
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