从 S 中移除子字符串以使所有剩余字符相同的方法数量
原文:https://www . geeksforgeeks . org/从 s 中移除子字符串的方法数，以便所有剩余字符都相同/
给定一个仅由小写英文字母组成的字符串 str ，任务是计算从 str 中精确删除一个子字符串的方法数量，以便所有剩余的字符都相同。 注意:在字符串中至少有两个不同的字符，我们也可以去掉整个字符串。 举例:
输入: str = "abaa" 输出: 6 我们可以移除以下子字符串以满足给定条件: str[0:1]、str[0:2]、str[0:3]、str[1:1]、str[1:2]和 str[1:3] 输入:str = " geeksforgeks " 输出: 3 我们移除完整字符串。 除了第一个，我们全部移除。 我们移除除最后一个 以外的所有
进场:
- 将字符串中相同字符的前缀和后缀的长度存储在变量 count_left 和 count_right 中。
- 很明显，这个前缀和后缀不会重叠，因为字符串中至少有两个不同的字符。
- 现在有两种情况:
- 当str[0]= str[n–1]时，前缀的每个字符(包括前缀结束后的字符)将作为子字符串的起始字符，后缀的每个字符(包括后缀开始前的字符)将作为子字符串的结束字符。因此，总的有效子串将是计数=(计数 _ 左+ 1) *(计数 _ 右+ 1) 。
- 当 str[0]！= str[n–1]。然后计数=计数 _ 左+计数 _ 右+ 1 。
以下是上述方法的实现:
C++
// C++ program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
int no_of_ways(string s)
{
int n = s.length();
// To store the count of prefix and suffix
int count_left = 0, count_right = 0;
// Loop to count prefix
for (int i = 0; i < n; ++i) {
if (s[i] == s[0]) {
++count_left;
}
else
break;
}
// Loop to count suffix
for (int i = n - 1; i >= 0; --i) {
if (s[i] == s[n - 1]) {
++count_right;
}
else
break;
}
// First and last characters of the
// string are same
if (s[0] == s[n - 1])
return ((count_left + 1) * (count_right + 1));
// Otherwise
else
return (count_left + count_right + 1);
}
// Driver function
int main()
{
string s = "geeksforgeeks";
cout << no_of_ways(s);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
import java.util.*;
class solution
{
// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
static int no_of_ways(String s)
{
int n = s.length();
// To store the count of prefix and suffix
int count_left = 0, count_right = 0;
// Loop to count prefix
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == s.charAt(0)) {
++count_left;
}
else
break;
}
// Loop to count suffix
for (int i = n - 1; i >= 0; --i) {
if (s.charAt(i) == s.charAt(n - 1)) {
++count_right;
}
else
break;
}
// First and last characters of the
// string are same
if (s.charAt(0) == s.charAt(n - 1))
return ((count_left + 1) * (count_right + 1));
// Otherwise
else
return (count_left + count_right + 1);
}
// Driver function
public static void main(String args[])
{
String s = "geeksforgeeks";
System.out.println(no_of_ways(s));
}
}
Python 3
# Python 3 program to count number of
# ways of removing a substring from a
# string such that all remaining
# characters are equal
# Function to return the number of
# ways of removing a sub-string from
# s such that all the remaining
# characters are same
def no_of_ways(s):
n = len(s)
# To store the count of
# prefix and suffix
count_left = 0
count_right = 0
# Loop to count prefix
for i in range(0, n, 1):
if (s[i] == s[0]):
count_left += 1
else:
break
# Loop to count suffix
i = n - 1
while(i >= 0):
if (s[i] == s[n - 1]):
count_right += 1
else:
break
i -= 1
# First and last characters of
# the string are same
if (s[0] == s[n - 1]):
return ((count_left + 1) *
(count_right + 1))
# Otherwise
else:
return (count_left + count_right + 1)
# Driver Code
if __name__ == '__main__':
s = "geeksforgeeks"
print(no_of_ways(s))
# This code is contributed by
# Sahil_shelengia
C
// C# program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
using System;
class GFG
{
// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
static int no_of_ways(string s)
{
int n = s.Length;
// To store the count of prefix and suffix
int count_left = 0, count_right = 0;
// Loop to count prefix
for (int i = 0; i < n; ++i)
{
if (s[i] == s[0])
{
++count_left;
}
else
break;
}
// Loop to count suffix
for (int i = n - 1; i >= 0; --i)
{
if (s[i] == s[n - 1])
{
++count_right;
}
else
break;
}
// First and last characters of the
// string are same
if (s[0] == s[n - 1])
return ((count_left + 1) *
(count_right + 1));
// Otherwise
else
return (count_left + count_right + 1);
}
// Driver Code
public static void Main()
{
string s = "geeksforgeeks";
Console.WriteLine(no_of_ways(s));
}
}
// This code is contributed
// by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
// Function to return the number of ways
// of removing a sub-string from $s such
// that all the remaining characters are same
function no_of_ways($s)
{
$n = strlen($s);
// To store the count of prefix and suffix
$count_left = 0;
$count_right = 0;
// Loop to count prefix
for ($i = 0; $i < $n; ++$i)
{
if ($s[$i] == $s[0])
{
++$count_left;
}
else
break;
}
// Loop to count suffix
for ($i = $n - 1; $i >= 0; --$i)
{
if ($s[$i] == $s[$n - 1])
{
++$count_right;
}
else
break;
}
// First and last characters of the
// string are same
if ($s[0] == $s[$n - 1])
return (($count_left + 1) *
($count_right + 1));
// Otherwise
else
return ($count_left + $count_right + 1);
}
// Driver Code
$s = "geeksforgeeks";
echo no_of_ways($s);
// This code is contributed by ihritik
?>
java 描述语言
<script>
// JavaScript program to count number of ways of
// removing a substring from a string such
// that all remaining characters are equal
// Function to return the number of ways
// of removing a sub-string from s such
// that all the remaining characters are same
function no_of_ways(s)
{
let n = s.length;
// To store the count of prefix and suffix
let count_left = 0, count_right = 0;
// Loop to count prefix
for (let i = 0; i < n; ++i)
{
if (s[i] == s[0])
{
++count_left;
}
else
break;
}
// Loop to count suffix
for (let i = n - 1; i >= 0; --i)
{
if (s[i] == s[n - 1])
{
++count_right;
}
else
break;
}
// First and last characters of the
// string are same
if (s[0] == s[n - 1])
return ((count_left + 1) *
(count_right + 1));
// Otherwise
else
return (count_left + count_right + 1);
}
let s = "geeksforgeeks";
document.write(no_of_ways(s));
</script>
Output:
3
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