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Count of Arrays of size N having absolute difference between adjacent elements at most 1

  • Last Updated : 16 Sep, 2021

Given a positive integer M and an array arr[] of size N and a few integers are missing in the array represented as -1, the task is to find the count of distinct arrays after replacing all -1 with the elements over the range [1, M] such that the absolute difference between any pair of adjacent elements is at most 1

Examples:

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Input: arr[] = {2, -1, 2}, M = 5
Output: 3
Explanation:
The arrays that follow the given conditions are {2, 1, 2}, {2, 2, 2} and {2, 3, 2}.



Input: arr[] = {4, -1, 2, 1, -1, -1}, M = 10
Output:  5

Approach: The given problem can be solved using Dynamic Programming based on the following observations:

  • Consider a 2D array, say dp[][] where dp[i][j] represents the count of valid arrays of length i+1 having their last element as j.
  • Since the absolute difference between any adjacent elements must be at most 1, so for any integer j, the valid adjacent integer can be j-1, j, and j+1. Therefore, any state dp[i][j] can be calculated using the following relation:

  dp[ i ][ j ] = dp[ i-1 ][ j ] + dp[ i-1 ][ j-1 ] + dp[ i-1 ][ j+1 ]

  • In cases where arr[i] = -1, calculate dp[i][j] = (dp[i-1][j] + dp[i-1][j-1] + dp[i-1][j+1]) for all values of j in the range [1, M].
  • In cases where arr[i] != -1, calculate dp[i][j] = (dp[i-1][j] + dp[i-1][j-1] + dp[i-1][j+1]) for j = arr[i].
  • Note that in cases where arr[0] = -1, all values in range [1, M] are reachable as the 1st array element, therefore initialize dp[0][j] = 1 for all j in the range [1, M] otherwise initialize dp[0][arr[0]] = 1.
  • The required answer will be the sum of all values of dp[N – 1][j] for all j in the range [1, M].

Below is the implementation of the above approach:

C++




// C++ program of the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the count of possible
// arrays such that the absolute difference
// between any adjacent elements is atmost 1
int countArray(int arr[], int N, int M)
{
    // Stores the dp states where dp[i][j]
    // represents count of arrays of length
    // i+1 having their last element as j
    int dp[N][M + 2];
    memset(dp, 0, sizeof dp);
 
    // Case where 1st array element is missing
    if (arr[0] == -1) {
        // All integers in range [1, M]
        // are reachable
        for (int j = 1; j <= M; j++) {
            dp[0][j] = 1;
        }
    }
    else {
        // Only reachable integer is arr[0]
        dp[0][arr[0]] = 1;
    }
 
    // Iterate through all values of i
    for (int i = 1; i < N; i++) {
 
        // If arr[i] is not missing
        if (arr[i] != -1) {
 
            // Only valid value of j is arr[i]
            int j = arr[i];
            dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j]
                        + dp[i - 1][j + 1];
        }
 
        // If arr[i] is missing
        if (arr[i] == -1) {
 
            // Iterate through all possible
            // values of j in range [1, M]
            for (int j = 1; j <= M; j++) {
                dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j]
                            + dp[i - 1][j + 1];
            }
        }
    }
 
    // Stores the count of valid arrays
    int arrCount = 0;
 
    // Calculate the total count of
    // valid arrays
    for (int j = 1; j <= M; j++) {
        arrCount += dp[N - 1][j];
    }
 
    // Return answer
    return arrCount;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, -1, 2, 1, -1, -1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int M = 10;
 
    // Function Call
    cout << countArray(arr, N, M);
 
    return 0;
}

Java




// Java program of the above approach
class GFG
{
   
    // Function to find the count of possible
    // arrays such that the absolute difference
    // between any adjacent elements is atmost 1
    public static int countArray(int arr[], int N, int M)
    {
       
        // Stores the dp states where dp[i][j]
        // represents count of arrays of length
        // i+1 having their last element as j
        int[][] dp = new int[N][M + 2];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M + 2; j++) {
                dp[i][j] = 0;
            }
        }
 
        // Case where 1st array element is missing
        if (arr[0] == -1)
        {
           
            // All integers in range [1, M]
            // are reachable
            for (int j = 1; j <= M; j++) {
                dp[0][j] = 1;
            }
        } else {
            // Only reachable integer is arr[0]
            dp[0][arr[0]] = 1;
        }
 
        // Iterate through all values of i
        for (int i = 1; i < N; i++) {
 
            // If arr[i] is not missing
            if (arr[i] != -1) {
 
                // Only valid value of j is arr[i]
                int j = arr[i];
                dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j] + dp[i - 1][j + 1];
            }
 
            // If arr[i] is missing
            if (arr[i] == -1) {
 
                // Iterate through all possible
                // values of j in range [1, M]
                for (int j = 1; j <= M; j++) {
                    dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j] + dp[i - 1][j + 1];
                }
            }
        }
 
        // Stores the count of valid arrays
        int arrCount = 0;
 
        // Calculate the total count of
        // valid arrays
        for (int j = 1; j <= M; j++) {
            arrCount += dp[N - 1][j];
        }
 
        // Return answer
        return arrCount;
    }
 
    // Driver Code
    public static void main(String args[]) {
        int arr[] = { 4, -1, 2, 1, -1, -1 };
        int N = arr.length;
        int M = 10;
 
        // Function Call
        System.out.println(countArray(arr, N, M));
 
    }
}
 
// This code is contributed by _saurabh_jaiswal.

Python3




# Python 3 program of the above approach
 
# Function to find the count of possible
# arrays such that the absolute difference
# between any adjacent elements is atmost 1
def countArray(arr, N, M):
   
    # Stores the dp states where dp[i][j]
    # represents count of arrays of length
    # i+1 having their last element as j
    dp = [[0 for i in range(M+2)] for j in range(N)]
 
    # Case where 1st array element is missing
    if (arr[0] == -1):
       
        # All integers in range [1, M]
        # are reachable
        for j in range(1,M+1,1):
            dp[0][j] = 1
 
    else:
        # Only reachable integer is arr[0]
        dp[0][arr[0]] = 1
 
    # Iterate through all values of i
    for i in range(1, N, 1):
       
        # If arr[i] is not missing
        if(arr[i] != -1):
           
            # Only valid value of j is arr[i]
            j = arr[i]
            dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j] + dp[i - 1][j + 1]
 
        # If arr[i] is missing
        if (arr[i] == -1):
           
            # Iterate through all possible
            # values of j in range [1, M]
            for j in range(1,M+1,1):
                dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j] + dp[i - 1][j + 1]
 
    # Stores the count of valid arrays
    arrCount = 0
 
    # Calculate the total count of
    # valid arrays
    for j in range(1,M+1,1):
        arrCount += dp[N - 1][j]
 
    # Return answer
    return arrCount
 
# Driver Code
if __name__ == '__main__':
    arr = [4, -1, 2, 1, -1, -1]
    N = len(arr)
    M = 10
 
    # Function Call
    print(countArray(arr, N, M))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program of the above approach
using System;
public class GFG
{
   
    // Function to find the count of possible
    // arrays such that the absolute difference
    // between any adjacent elements is atmost 1
    public static int countArray(int []arr, int N, int M)
    {
       
        // Stores the dp states where dp[i][j]
        // represents count of arrays of length
        // i+1 having their last element as j
        int[,] dp = new int[N, M + 2];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < M + 2; j++) {
                dp[i, j] = 0;
            }
        }
 
        // Case where 1st array element is missing
        if (arr[0] == -1)
        {
           
            // All integers in range [1, M]
            // are reachable
            for (int j = 1; j <= M; j++) {
                dp[0, j] = 1;
            }
        } else {
            // Only reachable integer is arr[0]
            dp[0, arr[0]] = 1;
        }
 
        // Iterate through all values of i
        for (int i = 1; i < N; i++) {
 
            // If arr[i] is not missing
            if (arr[i] != -1) {
 
                // Only valid value of j is arr[i]
                int j = arr[i];
                dp[i, j] += dp[i - 1, j - 1] + dp[i - 1, j] + dp[i - 1, j + 1];
            }
 
            // If arr[i] is missing
            if (arr[i] == -1) {
 
                // Iterate through all possible
                // values of j in range [1, M]
                for (int j = 1; j <= M; j++) {
                    dp[i, j] += dp[i - 1, j - 1] + dp[i - 1, j] + dp[i - 1, j + 1];
                }
            }
        }
 
        // Stores the count of valid arrays
        int arrCount = 0;
 
        // Calculate the total count of
        // valid arrays
        for (int j = 1; j <= M; j++) {
            arrCount += dp[N - 1, j];
        }
 
        // Return answer
        return arrCount;
    }
 
    // Driver Code
    public static void Main(String[] args) {
        int []arr = { 4, -1, 2, 1, -1, -1 };
        int N = arr.Length;
        int M = 10;
 
        // Function Call
        Console.WriteLine(countArray(arr, N, M));
    }
}
 
// This code is contributed by AnkThon.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find the count of possible
        // arrays such that the absolute difference
        // between any adjacent elements is atmost 1
        function countArray(arr, N, M) {
            // Stores the dp states where dp[i][j]
            // represents count of arrays of length
            // i+1 having their last element as j
            let dp = new Array(N);
 
 
 
            // Loop to create 2D array using 1D array
            for (let i = 0; i < dp.length; i++) {
                dp[i] = new Array(M + 2).fill(0);
            }
 
 
            // Case where 1st array element is missing
            if (arr[0] == -1) {
                // All integers in range [1, M]
                // are reachable
                for (let j = 1; j <= M; j++) {
                    dp[0][j] = 1;
                }
            }
            else {
                // Only reachable integer is arr[0]
                dp[0][arr[0]] = 1;
            }
 
            // Iterate through all values of i
            for (let i = 1; i < N; i++) {
 
                // If arr[i] is not missing
                if (arr[i] != -1) {
 
                    // Only valid value of j is arr[i]
                    let j = arr[i];
                    dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j]
                        + dp[i - 1][j + 1];
                }
 
                // If arr[i] is missing
                if (arr[i] == -1) {
 
                    // Iterate through all possible
                    // values of j in range [1, M]
                    for (let j = 1; j <= M; j++) {
                        dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j]
                            + dp[i - 1][j + 1];
                    }
                }
            }
 
            // Stores the count of valid arrays
            let arrCount = 0;
 
            // Calculate the total count of
            // valid arrays
            for (let j = 1; j <= M; j++) {
                arrCount += dp[N - 1][j];
            }
 
            // Return answer
            return arrCount;
        }
 
        // Driver Code
 
        let arr = [4, -1, 2, 1, -1, -1];
        let N = arr.length;
        let M = 10;
 
        // Function Call
        document.write(countArray(arr, N, M));
 
     // This code is contributed by Potta Lokesh
 
    </script>
Output: 
5

 

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




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