给定一个由小写字母组成的字符串str和一个非负整数K。任务是找到给定字符串ASCII值恰好为K的字符对数。
例子：
Input: str = “abcdab”, K = 0
Output: 2
(a, a) and (b, b) are the only valid pairs.
Input: str = “geeksforgeeks”, K = 1
Output: 8
(e, f), (e, f), (f, e), (f, e), (g, f),
(f, g), (s, r) and (r, s) are the valid pairs.
方法：将每个字符的频率存储在一个数组中。遍历此频率阵列以获得所需的答案。存在两种情况：
- 如果K = 0，则检查相似字符出现多次(即freq [i]> 1) 。如果是，则将(freq [i] *(freq [i] – 1))/ 2加到计数中。
- 如果K！= 0，则检查是否存在两个具有ASCII值差异的字符，如K所说的freq [i]和freq [j] 。然后将freq [i] * freq [j]添加到计数中。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 26
// Function to return the count of
// required pairs of characters
int countPairs(string str, int k)
{
// Length of the string
int n = str.size();
// To store the frequency
// of each character
int freq[MAX];
memset(freq, 0, sizeof freq);
// Update the frequency
// of each character
for (int i = 0; i < n; i++)
freq[str[i] - 'a']++;
// To store the required
// count of pairs
int cnt = 0;
// If ascii value difference is zero
if (k == 0) {
// If there exists similar characters
// more than once
for (int i = 0; i < MAX; i++)
if (freq[i] > 1)
cnt += ((freq[i] * (freq[i] - 1)) / 2);
}
else {
// If there exits characters with
// ASCII value difference as k
for (int i = 0; i < MAX; i++)
if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0)
cnt += (freq[i] * freq[i + k]);
;
}
// Return the required count
return cnt;
}
// Driver code
int main()
{
string str = "abcdab";
int k = 0;
cout << countPairs(str, k);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
static int MAX = 26;
// Function to return the count of
// required pairs of characters
static int countPairs(char[] str, int k)
{
// Length of the string
int n = str.length;
// To store the frequency
// of each character
int[] freq = new int[MAX];
// Update the frequency
// of each character
for (int i = 0; i < n; i++)
{
freq[str[i] - 'a']++;
}
// To store the required
// count of pairs
int cnt = 0;
// If ascii value difference is zero
if (k == 0)
{
// If there exists similar characters
// more than once
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 1)
{
cnt += ((freq[i] * (freq[i] - 1)) / 2);
}
}
}
else
{
// If there exits characters with
// ASCII value difference as k
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0)
{
cnt += (freq[i] * freq[i + k]);
}
}
;
}
// Return the required count
return cnt;
}
// Driver code
public static void main(String[] args)
{
String str = "abcdab";
int k = 0;
System.out.println(countPairs(str.toCharArray(), k));
}
}
/* This code contributed by PrinciRaj1992 */Python3
# Python3 implementation of the approach
MAX = 26
# Function to return the count of
# required pairs of characters
def countPairs(string, k) :
# Length of the string
n = len(string);
# To store the frequency
# of each character
freq = [0] * MAX;
# Update the frequency
# of each character
for i in range(n) :
freq[ord(string[i]) -
ord('a')] += 1;
# To store the required
# count of pairs
cnt = 0;
# If ascii value difference is zero
if (k == 0) :
# If there exists similar characters
# more than once
for i in range(MAX) :
if (freq[i] > 1) :
cnt += ((freq[i] *
(freq[i] - 1)) // 2);
else :
# If there exits characters with
# ASCII value difference as k
for i in range(MAX) :
if (freq[i] > 0 and
i + k < MAX and
freq[i + k] > 0) :
cnt += (freq[i] * freq[i + k]);
# Return the required count
return cnt;
# Driver code
if __name__ == "__main__" :
string = "abcdab";
k = 0;
print(countPairs(string, k));
# This code is contributed by AnkitRai01C#
// C# implementation of the approach
using System;
class GFG
{
static int MAX = 26;
// Function to return the count of
// required pairs of characters
static int countPairs(char[] str, int k)
{
// Length of the string
int n = str.Length;
// To store the frequency
// of each character
int[] freq = new int[MAX];
// Update the frequency
// of each character
for (int i = 0; i < n; i++)
{
freq[str[i] - 'a']++;
}
// To store the required
// count of pairs
int cnt = 0;
// If ascii value difference is zero
if (k == 0)
{
// If there exists similar characters
// more than once
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 1)
{
cnt += ((freq[i] * (freq[i] - 1)) / 2);
}
}
}
else
{
// If there exits characters with
// ASCII value difference as k
for (int i = 0; i < MAX; i++)
{
if (freq[i] > 0 && i + k < MAX && freq[i + k] > 0)
{
cnt += (freq[i] * freq[i + k]);
}
}
;
}
// Return the required count
return cnt;
}
// Driver code
public static void Main(String[] args)
{
String str = "abcdab";
int k = 0;
Console.WriteLine(countPairs(str.ToCharArray(), k));
}
}
// This code has been contributed by 29AjayKumar输出：
2