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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

• Last Updated : 16 Jul, 2021

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 ``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[])``{``    ``// 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;` `    ``// 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

 ``
Output:
`5`

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

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