与英文字母距离相同的字符对的计数
原文:https://www . geesforgeks . org/count-characters-string-distance-English-alphabets/
给定一个字符串，任务是计算元素距离与英文字母相同的对的数量。 注:考虑字符间的绝对距离。 示例:
Input: str = "geeksforgeeks"
Output: 4
Explanation: In this (g, s), (e, g), (e, k), (e, g)
are the pairs that are at same distances as
in English alphabets.
Input: str = "observation"
Output: 4
Explanation: (b, i), (s, v), (o, n), (v, t) are
at same distances as in English alphabets.
一个简单的解决方案是考虑生成所有对，并将对字符与它们之间的距离进行比较。如果一对的距离相同，则结果递增。
C++
// A Simple C++ program to find pairs with distance
// equal to English alphabet distance
#include <bits/stdc++.h>
using namespace std;
// Function to count pairs
int countPairs(string str)
{
int result = 0;
int n = str.length();
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
// Increment count if characters are at
// same distance
if (abs(str[i] - str[j]) == abs(i - j))
result++;
return result;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
cout << countPairs(str);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A Simple Java program to find pairs with distance
// equal to English alphabet distance
class Test {
// Method to count pairs
static int countPairs(String str)
{
int result = 0;
int n = str.length();
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
// Increment count if characters
// are at same distance
if (Math.abs(str.charAt(i) - str.charAt(j)) ==
Math.abs(i - j))
result++;
return result;
}
// Driver method
public static void main(String args[])
{
String str = "geeksforgeeks";
System.out.println(countPairs(str));
}
}
Python 3
# Simple Python3 program to find pairs with
# distance equal to English alphabet distance
# Function to count pairs
def countPairs(str1):
result = 0;
n = len(str1)
for i in range(0, n):
for j in range(i + 1, n):
# Increment count if characters
# are at same distance
if (abs(ord(str1[i]) -
ord(str1[j])) == abs(i - j)):
result += 1;
return result;
# Driver code
if __name__ == "__main__":
str1 = "geeksforgeeks";
print(countPairs(str1));
# This code is contributed
# by Sairahul099
C
// A Simple C# program to find pairs with distance
// equal to English alphabet distance
using System;
class Test {
// Method to count pairs
static int countPairs(string str)
{
int result = 0;
int n = str.Length;
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
// Increment count if characters
// are at same distance
if (Math.Abs(str[i] - str[j]) == Math.Abs(i - j))
result++;
return result;
}
// Driver method
public static void Main()
{
string str = "geeksforgeeks";
Console.WriteLine(countPairs(str));
}
}
// This Code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for Count
// of character pairs at
// same distance as in
// English alphabets
// Function to count pairs
function countPairs($str)
{
$result = 0;
$n = strlen($str);
for ($i = 0; $i < $n; $i++)
for ($j = $i + 1;
$j < $n; $j++)
// Increment count if
// characters are at
// same distance
if (abs(ord($str[$i]) -
ord($str[$j])) ==
abs($i - $j))
$result++;
return $result;
}
// Driver code
$str = "geeksforgeeks";
echo countPairs($str);
// This code is contributed by Sam007
?>
java 描述语言
<script>
// A Simple Javascript program to find pairs with distance
// equal to English alphabet distance
// Method to count pairs
function countPairs(str)
{
let result = 0;
let n = str.length;
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
// Increment count if characters
// are at same distance
if (Math.abs(str[i].charCodeAt() - str[j].charCodeAt()) == Math.abs(i - j))
result++;
return result;
}
let str = "geeksforgeeks";
document.write(countPairs(str));
// This code is contributed by divyesh072019.
</script>
输出:
4
上述方法的时间复杂度为 O(n 2 ) 。上述方法可以通过使用只能有 26 个字母的事实来优化，即，不是检查一个元素直到字符串的长度，而是只检查从当前索引到第 26 个索引。
C++
// An optimized C++ program to find pairs with distance
// equal to English alphabet distance
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
// Function to count pairs with distance
// equal to English alphabet distance
int countPairs(string str)
{
int result = 0;
int n = str.length();
for (int i = 0; i < n; i++)
// This loop runs at most 26 times
for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++)
if ((abs(str[i + j] - str[i]) == j))
result++;
return result;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
cout << countPairs(str);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// An optimized Java program to find pairs with distance
// equal to English alphabet distance
class Test {
static final int MAX_CHAR = 26;
// Method to count pairs with distance
// equal to English alphabet distance
static int countPairs(String str)
{
int result = 0;
int n = str.length();
for (int i = 0; i < n; i++)
// This loop runs at most 26 times
for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++)
if ((Math.abs(str.charAt(i + j) - str.charAt(i)) == j))
result++;
return result;
}
// Driver method
public static void main(String args[])
{
String str = "geeksforgeeks";
System.out.println(countPairs(str));
}
}
Python 3
# An optimized C++ program to find pairs with
# distance equal to English alphabet distance
MAX_CHAR = 26
# Function to count pairs with distance
# equal to English alphabet distance
def countPairs(str1):
result = 0;
n = len(str1)
for i in range(0, n):
# This loop runs at most 26 times
for j in range(1, MAX_CHAR + 1):
if((i + j) < n):
if ((abs(ord(str1[i + j]) -
ord(str1[i])) == j)):
result += 1;
return result
# Driver code
if __name__ == "__main__":
str1 = "geeksforgeeks";
print(countPairs(str1))
# This code is contributed
# by Sairahul099
C
// An optimized C# program to find pairs with distance
// equal to English alphabet distance
using System;
class Test {
static int MAX_CHAR = 26;
// Method to count pairs with distance
// equal to English alphabet distance
static int countPairs(string str)
{
int result = 0;
int n = str.Length;
for (int i = 0; i < n; i++)
// This loop runs at most 26 times
for (int j = 1; (i + j) < n && j <= MAX_CHAR; j++)
if ((Math.Abs(str[i + j] - str[i]) == j))
result++;
return result;
}
// Driver method
public static void Main()
{
string str = "geeksforgeeks";
Console.WriteLine(countPairs(str));
}
}
// This Code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// An optimized PHP program
// to find pairs with distance
// equal to English alphabet distance
// Function to count pairs
// with distance equal to
// English alphabet distance
function countPairs($str)
{
$result = 0;
$n = strlen($str);
for ($i = 0; $i < $n; $i++)
// This loop runs at
// most 26 times
for ($j = 1; ($i + $j) < $n &&
$j <= 26; $j++)
if ((abs(ord($str[$i + $j]) -
ord($str[$i]) ) == $j))
$result++;
return $result;
}
// Driver code
$str = "geeksforgeeks";
echo countPairs($str);
// This code is contributed by Sam007
?>
java 描述语言
<script>
// An optimized Javascript program to find pairs with distance
// equal to English alphabet distance
let MAX_CHAR = 26;
// Method to count pairs with distance
// equal to English alphabet distance
function countPairs(str)
{
let result = 0;
let n = str.length;
for (let i = 0; i < n; i++)
// This loop runs at most 26 times
for (let j = 1; (i + j) < n && j <= MAX_CHAR; j++)
if ((Math.abs(str[i + j].charCodeAt() - str[i].charCodeAt()) == j))
result++;
return result;
}
let str = "geeksforgeeks";
document.write(countPairs(str));
</script>
输出:
4
在字母表大小不变的假设下，优化解的时间复杂度为 O(n) 。 本文由 萨希尔·查布拉(akku) 供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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