计算生成具有 0、1 和 2 的 N 长度数组的方法，以便所有相邻成对乘积的总和为 K

给定两个整数N和K ，任务是找到可以通过使用值0 、 1和2任意次数生成的N长度数组的数量，使得数组的所有相邻成对乘积的总和是K 。

例子：

Input: N = 4, K = 3
Output: 5
Explanation: All possible arrangements are: 

1. arr[] = {2, 1, 1, 0}, Adjacent pairwise product sum = 2 * 1 + 1 * 1 + 1 * 0 = 3.
2. arr[] = {0, 2, 1, 1}, Adjacent pairwise product sum = 0 * 2 + 2 * 1 + 1 * 1 = 3.
3. arr[] = {1, 1, 2, 0}, Adjacent pairwise product sum = 1 * 1 + 1 * 2 + 2 * 0 = 3.
4. arr[] = {0, 1, 1, 2}, Adjacent pairwise product sum is 0 * 1 + 1 * 1 + 1 * 2 = 3.
5. arr[] = {1, 1, 1, 1}, Adjacent pairwise product sum = 1*1 + 1*1 + 1*1 = 3.

Input: N = 10, K = 9
Output: 3445

朴素方法：最简单的方法是生成数组的所有可能排列，其值可以是0 、 1或2 ，并对相邻成对乘积和为K 的那些数组进行计数。打印此类安排的计数。

时间复杂度： O(N*3 ^N )
辅助空间： O(N)

高效的方法：为了优化上述方法，最佳思路是使用动态规划。重叠的子问题可以存储在dp[][][] 表中，其中dp[i][remaining][previous]存储从位置‘i’到位置(N – 1)的答案，其中‘remaining’作为要添加的剩余值和“前一个”作为放置在位置(i – 1) 中的数字。对于任何位置“i”，可能存在三种情况：

• 将“0”分配给位置“i”。
• 将“1”分配给位置“i”。
• 将“2”分配给位置“i”。

请按照以下步骤解决问题：

• 初始化dp[][][]以存储当前位置、要添加的剩余值和前一个位置的元素。
• 过渡状态如下：

dp[i][remaining_sum][previous_element] = dp(assign 0 to pos ‘i’) + dp(assign 1 to ‘i’ ) + dp(assign 2 to ‘i’) 

• 递归求解上述递推关系，将每个状态的结果存入dp表。对于重叠，子问题使用dp 表中存储的结果。
• 在上述递归调用结束后，打印具有相邻成对积的数组总数为K的函数返回。

以下是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find total number of
// possible arrangements of array
int waysForPairwiseSumToBeK(
    int i, int rem, int previous,
    int N, int dp[][15][3])
{
    // Base Case
    if (i == N) {
        if (rem == 0)
            return 1;
        else
            return 0;
    }
 
    // If rem exceeds 'k' return 0
    if (rem < 0)
        return 0;
 
    // Return the already calculated
    // states
    if (dp[i][rem][previous] != -1)
        return dp[i][rem][previous];
 
    int ways = 0;
 
    // Place a '0' at current position
    ways += waysForPairwiseSumToBeK(
        i + 1, rem, 0, N, dp);
 
    // Place a '1' at current position
    // Add it to previous value
    ways += waysForPairwiseSumToBeK(
        i + 1, rem - (previous), 1, N, dp);
 
    // Place a '2' at current position.
    // Add it to previous value.
    ways += waysForPairwiseSumToBeK(
        i + 1, rem - (2 * previous), 2, N, dp);
 
    // Store the current state result
    // return the same result
    return dp[i][rem][previous] = ways;
}
 
// Function to find number of possible
// arrangements of array with 0, 1, and
// 2 having pairwise product sum K
void countOfArrays(int i, int rem,
                   int previous, int N)
{
    // Store the overlapping states
    int dp[15][15][3];
 
    // Initialize dp table with -1
    memset(dp, -1, sizeof dp);
 
    // Stores total number of ways
    int totWays
        = waysForPairwiseSumToBeK(
            i, rem, previous, N, dp);
 
    // Print number of ways
    cout << totWays << ' ';
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 3;
 
    // Function Call
    countOfArrays(0, K, 0, N);
 
    return 0;
}

Java

// Java program for the
// above approach
import java.util.*;
class solution{
   
// Function to find total number of
// possible arrangements of array
static int waysForPairwiseSumToBeK(int i, int rem,
                                   int previous,
                                   int N, int [][][]dp)
{
  // Base Case
  if (i == N)
  {
    if (rem == 0)
      return 1;
    else
      return 0;
  }
 
  // If rem exceeds
  // 'k' return 0
  if (rem < 0)
    return 0;
 
  // Return the already
  // calculated states
  if (dp[i][rem][previous] != -1)
    return dp[i][rem][previous];
 
  int ways = 0;
 
  // Place a '0' at current position
  ways += waysForPairwiseSumToBeK(i + 1, rem,
                                  0, N, dp);
 
  // Place a '1' at current position
  // Add it to previous value
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (previous),
                                  1, N, dp);
 
  // Place a '2' at current position.
  // Add it to previous value.
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (2 * previous),
                                  2, N, dp);
 
  // Store the current state result
  // return the same result
  dp[i][rem][previous] = ways;
  return ways;
}
 
// Function to find number of possible
// arrangements of array with 0, 1, and
// 2 having pairwise product sum K
static void countOfArrays(int i, int rem,
                          int previous, int N)
{
  // Store the overlapping states
  int [][][]dp = new int[15][15][3];
 
  // Initialize dp table with -1
  for(int p = 0; p < 15; p++)
  {
    for(int q = 0; q < 15; q++)
    {     
      for(int r = 0; r < 3; r++)
        dp[p][q][r] = -1;
    }
  }
 
  // Stores total number of ways
  int totWays = waysForPairwiseSumToBeK(i, rem,
                                        previous,
                                        N, dp);
  // Print number of ways
  System.out.print(totWays);
}
 
// Driver Code
public static void main(String args[])
{
  // Given N and K
  int N = 4, K = 3;
 
  // Function Call
  countOfArrays(0, K, 0, N);
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Python3

# Pyhton3 program for the above approach
 
# Function to find total number of
# possible arrangements of array
def waysForPairwiseSumToBeK(i, rem, previous, N, dp):
     
    # Base Case
    if (i == N):
        if (rem == 0):
            return 1
        else:
            return 0
 
    # If rem exceeds 'k' return 0
    if (rem < 0):
        return 0
         
    # Return the already calculated
    # states
    if (dp[i][rem][previous] != -1):
        return dp[i][rem][previous]
         
    ways = 0
 
    # Place a '0' at current position
    ways += waysForPairwiseSumToBeK(i + 1, rem,
                                    0, N, dp)
 
    # Place a '1' at current position
    # Add it to previous value
    ways += waysForPairwiseSumToBeK(i + 1,
                                  rem - (previous),
                                  1, N, dp)
 
    # Place a '2' at current position.
    # Add it to previous value.
    ways += waysForPairwiseSumToBeK(i + 1,
                             rem - (2 * previous),
                             2, N, dp)
 
    # Store the current state result
    # return the same result
    dp[i][rem][previous] = ways
     
    return ways
 
# Function to find number of possible
# arrangements of array with 0, 1, and
# 2 having pairwise product sum K
def countOfArrays(i, rem, previous, N):
     
    # Store the overlapping states
    dp = [[[-1 for i in range(3)]
               for j in range(15)]
               for k in range(15)]
 
    # Stores total number of ways
    totWays = waysForPairwiseSumToBeK(i, rem,
                                      previous,
                                      N, dp)
 
    # Print number of ways
    print(totWays, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given N and K
    N = 4
    K = 3
 
    # Function Call
    countOfArrays(0, K, 0, N)
 
# This code is contributed by bgangwar59

C#

// C# program for the
// above approach
using System;
 
class GFG{
   
// Function to find total number of
// possible arrangements of array
static int waysForPairwiseSumToBeK(int i, int rem,
                                   int previous,
                                   int N, int [,,]dp)
{
   
  // Base Case
  if (i == N)
  {
    if (rem == 0)
      return 1;
    else
      return 0;
  }
   
  // If rem exceeds
  // 'k' return 0
  if (rem < 0)
    return 0;
 
  // Return the already
  // calculated states
  if (dp[i, rem, previous] != -1)
    return dp[i, rem, previous];
 
  int ways = 0;
 
  // Place a '0' at current position
  ways += waysForPairwiseSumToBeK(i + 1, rem,
                                  0, N, dp);
 
  // Place a '1' at current position
  // Add it to previous value
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (previous),
                                  1, N, dp);
 
  // Place a '2' at current position.
  // Add it to previous value.
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (2 * previous),
                                   2, N, dp);
 
  // Store the current state result
  // return the same result
  dp[i, rem, previous] = ways;
   
  return ways;
}
 
// Function to find number of possible
// arrangements of array with 0, 1, and
// 2 having pairwise product sum K
static void countOfArrays(int i, int rem,
                          int previous, int N)
{
   
  // Store the overlapping states
  int [,,]dp = new int[ 15, 15, 3 ];
 
  // Initialize dp table with -1
  for(int p = 0; p < 15; p++)
  {
    for(int q = 0; q < 15; q++)
    {     
      for(int r = 0; r < 3; r++)
        dp[p, q, r] = -1;
    }
  }
 
  // Stores total number of ways
  int totWays = waysForPairwiseSumToBeK(i, rem,
                                        previous,
                                        N, dp);
   
  // Print number of ways
  Console.Write(totWays);
}
 
// Driver Code
public static void Main(String []args)
{
   
  // Given N and K
  int N = 4, K = 3;
 
  // Function Call
  countOfArrays(0, K, 0, N);
}
}
 
// This code is contributed by gauravrajput1

Javascript

// Javascript program for the
// above approach
 
// Function to find total number of
// possible arrangements of array
function waysForPairwiseSumToBeK(i,rem,previous,N,dp)
{
    // Base Case
  if (i == N)
  {
    if (rem == 0)
      return 1;
    else
      return 0;
  }
  
  // If rem exceeds
  // 'k' return 0
  if (rem < 0)
    return 0;
  
  // Return the already
  // calculated states
  if (dp[i][rem][previous] != -1)
    return dp[i][rem][previous];
  
  let ways = 0;
  
  // Place a '0' at current position
  ways += waysForPairwiseSumToBeK(i + 1, rem,
                                  0, N, dp);
  
  // Place a '1' at current position
  // Add it to previous value
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (previous),
                                  1, N, dp);
  
  // Place a '2' at current position.
  // Add it to previous value.
  ways += waysForPairwiseSumToBeK(i + 1, rem -
                                  (2 * previous),
                                  2, N, dp);
  
  // Store the current state result
  // return the same result
  dp[i][rem][previous] = ways;
  return ways;
}
 
// Function to find number of possible
// arrangements of array with 0, 1, and
// 2 having pairwise product sum K
function countOfArrays(i,rem,previous,N)
{
    // Store the overlapping states
  let dp = new Array(15);
  for(let i = 0; i < 15; i++)
  {
      dp[i] = new Array(15);
    for(let j = 0; j < 15; j++)
    {
        dp[i][j] = new Array(3);
        for(let k = 0; k < 3; k++)
            dp[i][j][k] = -1;
    }
  }
  
  // Stores total number of ways
  let totWays = waysForPairwiseSumToBeK(i, rem,
                                        previous,
                                        N, dp);
  // Print number of ways
  document.write(totWays);
}
 
// Driver Code
// Given N and K
let N = 4, K = 3;
 
// Function Call
countOfArrays(0, K, 0, N);
 
// This code is contributed by patel2127

输出：

5

时间复杂度： O(N*K)
辅助空间： O(N*K)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。