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Count ways to generate N-length array with 0s, 1s, and 2s such that sum of all adjacent pairwise products is K

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Given two integers N and K, the task is to find the number of N-length arrays that can be generated by using the values 0, 1, and 2 any number of times, such that the sum of all adjacent pairwise products of the array is K.

Examples:

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

Naive Approach: The simplest approach is to generate all possible arrangements of the array whose value can be 0, 1, or 2 and count those arrays whose adjacent pairwise product sum is K. Print the count of such arrangements. 

Time Complexity: O(N*3N )
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the optimal idea is to use Dynamic Programming. The overlapping subproblems can be stored in a dp[][][] table where dp[i][remaining][previous] stores the answer for up to position (N – 1) from position ‘i’ with ‘remaining’ as the remaining value to be added and ‘previous’ as the number placed in the position (i – 1). There can be three cases possible for any position ‘i’:

  • Assign ‘0’ to position ‘i’.
  • Assign ‘1’ to position ‘i’.
  • Assign ‘2’ to position ‘i’.

Follow the steps below to solve the problem:

  • Initialize the dp[][][] to store the current position, remaining value to be added, and element at the previous position.
  • The transition state is as follows :

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

  • Solve the above recurrence relation recursively and store the result for each state in the dp table. For overlapping, subproblems use the stored result in the dp table.
  • After the above recursive calls end, print the total number of arrays having adjacent pairwise products of the array is K return by the function.

Below is an implementation of the above approach : 

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
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

# Python3 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

<script>
// 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
</script>

                    

Output
5


Time Complexity: O(N*K)
Auxiliary Space: O(N*K)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

  • Create a table to store the solution of the subproblems.
  • Initialize the table with base cases
  • Fill up the table iteratively
  • Return the final solution

Implementation :

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
 
// Function to find total number of
// possible arrangements of array
int waysForPairwiseSumToBeK(int N, int K)
{
    int dp[N+1][K+1][3];
    memset(dp, 0, sizeof dp);
     
    // find solution of current problema through
    // subproblems iteratively
    for(int i = 0; i <= N; i++) {
        for(int rem = 0; rem <= K; rem++) {
            for(int previous = 0; previous <= 2; previous++) {
                // Base case
                if(i == 0) {
                    if(rem == 0) dp[i][rem][previous] = 1;
                    else dp[i][rem][previous] = 0;
                }
                else {
                    int ways = 0;
                    // Place a '0' at current position
                    ways += dp[i-1][rem][0];
                     
                    // Place a '1' at current position
                     // Add it to previous value
                    if(rem - previous >= 0)
                        ways += dp[i-1][rem-previous][1];
                         
                    // Place a '2' at current position.
                    // Add it to previous value.
                    if(rem - (2 * previous) >= 0)
                        ways += dp[i-1][rem-(2*previous)][2];
                   
                      // store computaion of subproblem
                    // in DP and iterate further
                    dp[i][rem][previous] = ways;
                }
            }
        }
    }
    // return answer
    return dp[N][K][0];
}
 
// Driver code
int main()
{
    int N = 4, K = 3;
    // function call
    int totWays = waysForPairwiseSumToBeK(N, K);
    cout << totWays << ' ';
 
    return 0;
}

                    

Java

public class PairwiseSum {
    // Function to find total number of
    // possible arrangements of array
    static int waysForPairwiseSumToBeK(int N, int K) {
        int[][][] dp = new int[N + 1][K + 1][3];
         
        // Initialize the dp array with zeros
        for (int i = 0; i <= N; i++) {
            for (int rem = 0; rem <= K; rem++) {
                for (int previous = 0; previous <= 2; previous++) {
                    dp[i][rem][previous] = 0;
                }
            }
        }
         
        // Find solution of current problem through subproblems iteratively
        for (int i = 0; i <= N; i++) {
            for (int rem = 0; rem <= K; rem++) {
                for (int previous = 0; previous <= 2; previous++) {
                    // Base case
                    if (i == 0) {
                        if (rem == 0) dp[i][rem][previous] = 1;
                        else dp[i][rem][previous] = 0;
                    } else {
                        int ways = 0;
                         
                        // Place a '0' at current position
                        ways += dp[i - 1][rem][0];
                         
                        // Place a '1' at current position
                        // Add it to previous value
                        if (rem - previous >= 0)
                            ways += dp[i - 1][rem - previous][1];
                         
                        // Place a '2' at current position
                        // Add it to previous value
                        if (rem - (2 * previous) >= 0)
                            ways += dp[i - 1][rem - (2 * previous)][2];
                         
                        // Store computation of subproblem
                        // in DP and iterate further
                        dp[i][rem][previous] = ways;
                    }
                }
            }
        }
        // Return answer
        return dp[N][K][0];
    }
 
    // Driver code
    public static void main(String[] args) {
        int N = 4, K = 3;
        // Function call
        int totWays = waysForPairwiseSumToBeK(N, K);
        System.out.println(totWays);
    }
}

                    

Python3

# Python program for the above approach
 
# Function to find total number of
# possible arrangements of array
def waysForPairwiseSumToBeK(N, K):
    dp = [[[0 for _ in range(3)] for _ in range(K+1)] for _ in range(N+1)]
 
    # find solution of current problema through
    # subproblems iteratively
    for i in range(N+1):
        for rem in range(K+1):
            for previous in range(3):
                # Base case
                if i == 0:
                    if rem == 0:
                        dp[i][rem][previous] = 1
                    else:
                        dp[i][rem][previous] = 0
                else:
                    ways = 0
                    # Place a '0' at current position
                    ways += dp[i-1][rem][0]
                     
                    # Place a '1' at current position
                    # Add it to previous value
                    if rem - previous >= 0:
                        ways += dp[i-1][rem-previous][1]
                         
                    # Place a '2' at current position.
                    # Add it to previous value.
                    if rem - (2 * previous) >= 0:
                        ways += dp[i-1][rem-(2*previous)][2]
                   
                    # store computation of subproblem
                    # in DP and iterate further
                    dp[i][rem][previous] = ways
 
    # return answer
    return dp[N][K][0]
 
# Driver code
N = 4
K = 3
# function call
totWays = waysForPairwiseSumToBeK(N, K)
print(totWays)

                    

C#

using System;
 
public class PairwiseSum
{
    // Function to find total number of
    // possible arrangements of array
    static int WaysForPairwiseSumToBeK(int N, int K)
    {
        int[,,] dp = new int[N + 1, K + 1, 3];
 
        // Initialize the dp array with zeros
        for (int i = 0; i <= N; i++)
        {
            for (int rem = 0; rem <= K; rem++)
            {
                for (int previous = 0; previous <= 2; previous++)
                {
                    dp[i, rem, previous] = 0;
                }
            }
        }
 
        // Find solution of current problem through subproblems iteratively
        for (int i = 0; i <= N; i++)
        {
            for (int rem = 0; rem <= K; rem++)
            {
                for (int previous = 0; previous <= 2; previous++)
                {
                    // Base case
                    if (i == 0)
                    {
                        if (rem == 0) dp[i, rem, previous] = 1;
                        else dp[i, rem, previous] = 0;
                    }
                    else
                    {
                        int ways = 0;
 
                        // Place a '0' at current position
                        ways += dp[i - 1, rem, 0];
 
                        // Place a '1' at current position
                        // Add it to previous value
                        if (rem - previous >= 0)
                            ways += dp[i - 1, rem - previous, 1];
 
                        // Place a '2' at current position
                        // Add it to previous value
                        if (rem - (2 * previous) >= 0)
                            ways += dp[i - 1, rem - (2 * previous), 2];
 
                        // Store computation of subproblem
                        // in DP and iterate further
                        dp[i, rem, previous] = ways;
                    }
                }
            }
        }
        // Return answer
        return dp[N, K, 0];
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int N = 4, K = 3;
        // Function call
        int totWays = WaysForPairwiseSumToBeK(N, K);
        Console.WriteLine(totWays);
    }
}

                    

Javascript

// Function to find total number of possible arrangements of array
function waysForPairwiseSumToBeK(N, K) {
    let dp = new Array(N + 1).fill(0).map(() => new Array(K + 1).fill(0).map(() => new Array(3).fill(0)));
 
    // Find solution of current problem through subproblems iteratively
    for (let i = 0; i <= N; i++) {
        for (let rem = 0; rem <= K; rem++) {
            for (let previous = 0; previous <= 2; previous++) {
                // Base case
                if (i == 0) {
                    if (rem == 0) dp[i][rem][previous] = 1;
                    else dp[i][rem][previous] = 0;
                } else {
                    let ways = 0;
                    // Place a '0' at current position
                    ways += dp[i - 1][rem][0];
 
                    // Place a '1' at current position
                    // Add it to previous value
                    if (rem - previous >= 0)
                        ways += dp[i - 1][rem - previous][1];
 
                    // Place a '2' at current position.
                    // Add it to previous value.
                    if (rem - (2 * previous) >= 0)
                        ways += dp[i - 1][rem - (2 * previous)][2];
 
                    // Store computation of subproblem
                    // in DP and iterate further
                    dp[i][rem][previous] = ways;
                }
            }
        }
    }
    // Return answer
    return dp[N][K][0];
}
 
let N = 4,
    K = 3;
// Function call
let totWays = waysForPairwiseSumToBeK(N, K);
console.log(totWays);

                    

Output
5 


Time Complexity: O(N*K)
Auxiliary Space: O(N*K)



Last Updated : 04 Oct, 2023
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