根据给定的编码技术从结果字符串重建原始字符串

长度为N的二进制字符串S由N个字符的字符串P和整数X构成。 S的第i个字符的选择如下：

• 如果字符P _{i- X} 存在且等于1，则S _i为1
• 如果字符P _{i+ X}存在且等于 1，则S _i为 1
• 如果上述两个条件都为假，则S _i为 0。

给定结果字符串S和整数X ，重构原始字符串P 。如果没有字符串P可以产生字符串S ，则输出 -1。

例子：

Input: S = “10011”, X = 2
Output: “01100”
Explanation: The input string S = “10011” can be constructed from the output string P = “01100”.

Input: S = “11101100111”, X = 3
Output: -1
Explanation: The input string S = “11101100111” cannot be constructed from any output string P.

方法：这个任务可以通过一个全为1的辅助字符串来解决。请按照以下步骤解决问题：

• 将字符串P初始化为所有字符为 '1'。
• 现在使用循环遍历字符串S并根据给定条件在P中的正确位置放置零：
  • 如果S [i]等于'0'且P [i- X ]存在，i,e,(i- X )>=0，则令P [i- X ]='0'。
  • 如果S [i] 等于 '0' 且P [i+ X ] 存在，即 (i+ X )< N ，则令P [i+ X ] = '0'。
• 初始化一个变量flag = 1 来判断字符串P是否存在。
• 要检查构造的字符串P的正确性，请使用 for 循环遍历字符串S ：
  • 如果S [i]=='1' 并且P [i- X ] 或P [i+ X ] 存在并且等于'1'，那么字符串P到目前为止是正确的并且可以继续遍历。
  • 如果S [i]=='1' 且P [i- X ] 或P [i+ X ] 不存在或不等于 1，则设置flag = 0，输出 -1 并退出循环。
• 如果遍历字符串S后flag等于 0，则输出原始字符串P。

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 original string P
void findString(string S, int X)
{
    // Stores the size of string S
    int N = S.size();
 
    // Each character of string
    // P is set to '1'
    string P(N, '1');
 
    // Loop to put zeroes at
    // the correct positions in P
    for (int i = 0; i < N; ++i) {
 
        // If the character is '0'
        if (S[i] == '0') {
 
            // If P[i-X] exists
            if (i - X >= 0) {
 
                // Set P[i-X] to '0'
                P[i - X] = '0';
            }
 
            // If P[i+X] exists
            if (i + X < N) {
 
                // Set P[i+X] to '0'
                P[i + X] = '0';
            }
        }
    }
 
    // Set flag to 1
    int flag = 1;
 
    // Loop to cross check
    // if P exists or not
    for (int i = 0; i < N; ++i) {
 
        // If the character is '1'
        if (S[i] == '1') {
 
            // If P[i-X] exists and
            // is equal to '1' or if
            // P[i+X] exists and
            // is equal to '1'
            if ((i - X >= 0
                 && P[i - X] == '1')
                || (i + X < N
                    && P[i + X] == '1')) {
                continue;
            }
            else {
 
                // Set flag to 0
                flag = 0;
 
                // Output -1
                cout << -1;
 
                // Break from the loop
                break;
            }
        }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
        // Output string P
        cout << P;
    }
}
 
// Driver Code
int main()
{
    // Given input
    string S = "10011";
    int X = 2;
 
    // Function Call
    findString(S, X);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to find the original string P
  static void findString(String S, int X)
  {
 
    // Stores the size of string S
    int N = S.length();
 
    // Each character of string
    // converted to char Array P
    // is set to '1'
    char[] P = new char[N];
    for (int i = 0; i < N; ++i) {
      P[i] = '1';
    }
 
    // Loop to put zeroes at
    // the correct positions in P
    for (int i = 0; i < N; ++i) {
 
      // If the character is '0'
      if (S.charAt(i) == '0') {
 
        // If P[i-X] exists
        if (i - X >= 0) {
 
          // Set P[i-X] to '0'
          P[i - X] = '0';
        }
 
        // If P[i+X] exists
        if (i + X < N) {
 
          // Set P[i+X] to '0'
          P[i + X] = '0';
        }
      }
    }
 
    // Set flag to 1
    int flag = 1;
 
    // Loop to cross check
    // if P exists or not
    for (int i = 0; i < N; ++i) {
 
      // If the character is '1'
      if (S.charAt(i) == '1') {
 
        // If P[i-X] exists and
        // is equal to '1' or if
        // P[i+X] exists and
        // is equal to '1'
        if ((i - X >= 0 && P[i - X] == '1')
            || (i + X < N && P[i + X] == '1')) {
          continue;
        }
        else {
 
          // Set flag to 0
          flag = 0;
 
          // Output -1
          System.out.print(-1);
 
          // Break from the loop
          break;
        }
      }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
      // Output string P
      String p = new String(P);
      System.out.print(p);
    }
  }
 
  // Driver Code
  public static void main(String args[])
  {
 
    // Given input
    String S = "10011";
    int X = 2;
 
    // Function Call
    findString(S, X);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to find the original string P
def findString(S, X):
 
    # Stores the size of string S
    N = len(S)
 
    # Each character of string
    # P is set to '1'
    P = ['1'] * N
 
    # Loop to put zeroes at
    # the correct positions in P
    for i in range(N):
 
        # If the character is '0'
        if (S[i] == '0'):
 
            # If P[i-X] exists
            if (i - X >= 0):
 
                # Set P[i-X] to '0'
                P[i - X] = '0';
 
            # If P[i+X] exists
            if (i + X < N):
 
                # Set P[i+X] to '0'
                P[i + X] = '0';
 
    # Set flag to 1
    flag = 1;
 
    # Loop to cross check
    # if P exists or not
    for i in range(N):
 
        # If the character is '1'
        if (S[i] == '1'):
 
            # If P[i-X] exists and
            # is equal to '1' or if
            # P[i+X] exists and
            # is equal to '1'
            if ((i - X >= 0 and P[i - X] == '1') or (i + X < N and P[i + X] == '1')):
                continue;
            else:
 
                # Set flag to 0
                flag = 0;
 
                # Output -1
                print(-1);
 
                # Break from the loop
                break;
 
    # If flag is equal to 1
    if (flag == 1):
 
        # Output string P
        print("".join(P));
 
# Driver Code
 
# Given input
S = "10011";
X = 2;
 
# Function Call
findString(S, X);
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG
{
 
  // Function to find the original string P
  static void findString(string S, int X)
  {
 
    // Stores the size of string S
    int N = S.Length;
 
    // Each character of string
    // converted to char Array P
    // is set to '1'
    char[] P = new char[N];
    for (int i = 0; i < N; ++i) {
      P[i] = '1';
    }
 
    // Loop to put zeroes at
    // the correct positions in P
    for (int i = 0; i < N; ++i) {
 
      // If the character is '0'
      if (S[i] == '0') {
 
        // If P[i-X] exists
        if (i - X >= 0) {
 
          // Set P[i-X] to '0'
          P[i - X] = '0';
        }
 
        // If P[i+X] exists
        if (i + X < N) {
 
          // Set P[i+X] to '0'
          P[i + X] = '0';
        }
      }
    }
 
    // Set flag to 1
    int flag = 1;
 
    // Loop to cross check
    // if P exists or not
    for (int i = 0; i < N; ++i) {
 
      // If the character is '1'
      if (S[i] == '1') {
 
        // If P[i-X] exists and
        // is equal to '1' or if
        // P[i+X] exists and
        // is equal to '1'
        if ((i - X >= 0 && P[i - X] == '1')
            || (i + X < N && P[i + X] == '1')) {
          continue;
        }
        else {
 
          // Set flag to 0
          flag = 0;
 
          // Output -1
          Console.Write(-1);
 
          // Break from the loop
          break;
        }
      }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
      // Output string P
      string p = new string(P);
      Console.Write(p);
    }
  }
 
  // Driver Code
  public static void Main()
  {
    // Given input
    string S = "10011";
    int X = 2;
 
    // Function Call
    findString(S, X);
  }
}
 
// This code is contributed by Taranpreet

Javascript

    // JavaScript program for the above approach
 
    // Function to find the original string P
    const findString = (S, X) => {
     
        // Stores the size of string S
        let N = S.length;
 
        // Each character of string
        // P is set to '1'
        let P = new Array(N).fill('1');
 
        // Loop to put zeroes at
        // the correct positions in P
        for (let i = 0; i < N; ++i) {
 
            // If the character is '0'
            if (S[i] == '0') {
 
                // If P[i-X] exists
                if (i - X >= 0) {
 
                    // Set P[i-X] to '0'
                    P[i - X] = '0';
                }
 
                // If P[i+X] exists
                if (i + X < N) {
 
                    // Set P[i+X] to '0'
                    P[i + X] = '0';
                }
            }
        }
 
        // Set flag to 1
        let flag = 1;
 
        // Loop to cross check
        // if P exists or not
        for (let i = 0; i < N; ++i) {
 
            // If the character is '1'
            if (S[i] == '1') {
 
                // If P[i-X] exists and
                // is equal to '1' or if
                // P[i+X] exists and
                // is equal to '1'
                if ((i - X >= 0
                    && P[i - X] == '1')
                    || (i + X < N
                        && P[i + X] == '1')) {
                    continue;
                }
                else {
 
                    // Set flag to 0
                    flag = 0;
 
                    // Output -1
                    document.write(-1);
 
                    // Break from the loop
                    break;
                }
            }
        }
 
        // If flag is equal to 1
        if (flag == 1) {
 
            // Output string P
            document.write(P.join(""));
        }
    }
 
    // Driver Code
 
    // Given input
    let S = "10011";
    let X = 2;
 
    // Function Call
    findString(S, X);
 
// This code is contributed by rakeshsahni
 

输出：

01100

时间复杂度： O(N)
辅助空间： O(N)