查找解密字符串的第 k 个字符|套装 – 2

给定一个编码字符串，其中子字符串的重复表示为子字符串，后跟子字符串的计数。例如，如果加密字符串是“ab2cd2”并且k=4，那么输出将是“b”，因为解密后的字符串是“ababcdcd”并且第4个字符是“b”。
注意：加密子串的频率可以超过一位。例如，在“ab12c3”中，ab 重复了 12 次。子串的频率中不存在前导 0。

例子：

Input:  "a2b2c3", k = 5
Output:  c
Decrypted string is "aabbccc"

Input:  "ab4c2ed3", k = 9
Output :  c
Decrypted string is "ababababccededed"

上一篇文章中讨论的解决方案需要额外的 O(n) 空间。以下帖子讨论了需要恒定空间的解决方案。逐步算法是：

1. 查找当前子字符串的长度。使用两个指针。将一个指针固定在子字符串的开头并移动另一个指针，直到找不到数字为止。
2. 通过进一步移动第二个指针直到找不到字母来查找重复频率。
3. 如果通过将频率乘以它的原始长度来查找子字符串的长度。
4. 如果此长度小于 k，则所需字符位于后面的子字符串中。从 k 中减去此长度以计算需要覆盖的字符数。
5. 如果长度小于或等于 k，则所需字符位于当前子字符串中。由于 k 是 1-indexed ，因此将其减 1，然后将其 mod 与原始子字符串长度。所需字符是从第一个指针指向的子字符串开始的第 k 个字符。

下面是上述方法的实现：

C++

// C++ program to find K'th character in
// decrypted string
#include 
using namespace std;
 
// Function to find K'th character in
// Encoded String
char encodedChar(string str, int k)
{
    int i, j;
 
    int n = str.length();
 
    // To store length of substring
    int len;
 
    // To store length of substring when
    // it is repeated
    int num;
 
    // To store frequency of substring
    int freq;
 
    i = 0;
 
    while (i < n) {
        j = i;
        len = 0;
        freq = 0;
 
        // Find length of substring by
        // traversing the string until
        // no digit is found.
        while (j < n && isalpha(str[j])) {
            j++;
            len++;
        }
 
        // Find frequency of preceding substring.
        while (j < n && isdigit(str[j])) {
            freq = freq * 10 + (str[j] - '0');
            j++;
        }
 
        // Find length of substring when
        // it is repeated.
        num = freq * len;
 
        // If length of repeated substring is less than
        // k then required character is present in next
        // substring. Subtract length of repeated
        // substring from k to keep account of number of
        // characters required to be visited.
        if (k > num) {
            k -= num;
            i = j;
        }
 
        // If length of repeated substring is
        // more or equal to k then required
        // character lies in current substring.
        else {
            k--;
            k %= len;
            return str[i + k];
        }
    }
 
    // This is for the case when there
    // are no repetition in string.
    // e.g. str="abced".
    return str[k - 1];
}
 
// Driver Code
int main()
{
    string str = "abced";
    int k = 4;
 
    cout << encodedChar(str, k) << endl;
 
    return 0;
}

Java

// Java program to find K'th character in
// decrypted string
import java.util.*;
 
class GFG
{
 
// Function to find K'th character in
// Encoded String
static char encodedChar(char[] str, int k)
{
    int i, j;
 
    int n = str.length;
 
    // To store length of substring
    int len;
 
    // To store length of substring when
    // it is repeated
    int num;
 
    // To store frequency of substring
    int freq;
 
    i = 0;
 
    while (i < n)
    {
        j = i;
        len = 0;
        freq = 0;
 
        // Find length of substring by
        // traversing the string until
        // no digit is found.
        while (j < n && Character.isAlphabetic(str[j]))
        {
            j++;
            len++;
        }
 
        // Find frequency of preceding substring.
        while (j < n && Character.isDigit(str[j]))
        {
            freq = freq * 10 + (str[j] - '0');
            j++;
        }
 
        // Find length of substring when
        // it is repeated.
        num = freq * len;
 
        // If length of repeated substring is less than
        // k then required character is present in next
        // substring. Subtract length of repeated
        // substring from k to keep account of number of
        // characters required to be visited.
        if (k > num)
        {
            k -= num;
            i = j;
        }
 
        // If length of repeated substring is
        // more or equal to k then required
        // character lies in current substring.
        else
        {
            k--;
            k %= len;
            return str[i + k];
        }
    }
 
    // This is for the case when there
    // are no repetition in string.
    // e.g. str="abced".
    return str[k - 1];
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "abced";
    int k = 4;
 
    System.out.println(encodedChar(str.toCharArray(), k));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find K'th
# character in decrypted string
 
# Function to find K'th character
# in Encoded String
def encodedChar(string, k):
 
    n = len(string)
 
    i = 0
    while i < n:
        j = i
        length = 0
        freq = 0
 
        # Find length of substring by
        # traversing the string until
        # no digit is found.
        while j < n and string[j].isalpha():
            j += 1
            length += 1
 
        # Find frequency of preceding substring.
        while j < n and string[j].isdigit():
            freq = freq * 10 + int(string[j])
            j += 1
 
        # Find the length of the substring
        # when it is repeated.
        num = freq * length
 
        # If the length of the repeated substring
        # is less than k then required character
        # is present in next substring. Subtract
        # the length of repeated substring from
        # k to keep account of the number
        # of characters required to be visited.
        if k > num:
            k -= num
            i = j
 
        # If length of repeated substring is
        # more or equal to k then required
        # character lies in current substring.
        else:
            k -= 1
            k %= length
            return string[i + k]
 
    # This is for the case when there are no
    # repetition in string. e.g. str="abced".
    return string[k - 1]
 
# Driver Code
if __name__ == "__main__":
 
    string = "abced"
    k = 4
 
    print(encodedChar(string, k))
 
# This code is contributed
# by Rituraj Jain

C#

// C# program to find K'th character in
// decrypted string
using System;
 
class GFG
{
 
// Function to find K'th character in
// Encoded String
static char encodedChar(char[] str, int k)
{
    int i, j;
 
    int n = str.Length;
 
    // To store length of substring
    int len;
 
    // To store length of substring when
    // it is repeated
    int num;
 
    // To store frequency of substring
    int freq;
 
    i = 0;
 
    while (i < n)
    {
        j = i;
        len = 0;
        freq = 0;
 
        // Find length of substring by
        // traversing the string until
        // no digit is found.
        while (j < n && char.IsLetter(str[j]))
        {
            j++;
            len++;
        }
 
        // Find frequency of preceding substring.
        while (j < n && char.IsDigit(str[j]))
        {
            freq = freq * 10 + (str[j] - '0');
            j++;
        }
 
        // Find length of substring when
        // it is repeated.
        num = freq * len;
 
        // If length of repeated substring is less than
        // k then required character is present in next
        // substring. Subtract length of repeated
        // substring from k to keep account of number of
        // characters required to be visited.
        if (k > num)
        {
            k -= num;
            i = j;
        }
 
        // If length of repeated substring is
        // more or equal to k then required
        // character lies in current substring.
        else
        {
            k--;
            k %= len;
            return str[i + k];
        }
    }
 
    // This is for the case when there
    // are no repetition in string.
    // e.g. str="abced".
    return str[k - 1];
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "abced";
    int k = 4;
 
    Console.WriteLine(encodedChar(str.ToCharArray(), k));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript program to find K'th character in
// decrypted string
 
// Function to find K'th character in
// Encoded String
function encodedChar(str, k)
{
    var i, j;
 
    var n = str.length;
 
    // To store length of substring
    var len;
 
    // To store length of substring when
    // it is repeated
    var num;
 
    // To store frequency of substring
    var freq;
 
    i = 0;
 
    while (i < n)
    {
        j = i;
        len = 0;
        freq = 0;
 
        // Find length of substring by
        // traversing the string until
        // no digit is found.
        while (j < n && str[j].match(/^[0-9a-z]+$/))
        {
            j++;
            len++;
        }
 
        // Find frequency of preceding substring.
        while (j < n && str[j].match(/^[0-9]+$/))
        {
            freq = freq * 10 + (str[j] - '0');
            j++;
        }
 
        // Find length of substring when
        // it is repeated.
        num = freq * len;
 
        // If length of repeated substring is less than
        // k then required character is present in next
        // substring. Subtract length of repeated
        // substring from k to keep account of number of
        // characters required to be visited.
        if (k > num)
        {
            k -= num;
            i = j;
        }
 
        // If length of repeated substring is
        // more or equal to k then required
        // character lies in current substring.
        else
        {
            k--;
            k %= len;
            return str[i + k];
        }
    }
 
    // This is for the case when there
    // are no repetition in string.
    // e.g. str="abced".
    return str[k - 1];
}
 
// Driver Code
var str = "abced";
var k = 4;
 
document.write(encodedChar(str, k));
 
// This code is contributed by rutvik_56
 

输出：

e

时间复杂度： O(n)
辅助空间： O(1)