# 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 = -1, all values in range [1, M] are reachable as the 1st array element, therefore initialize dp[j] = 1 for all j in the range [1, M] otherwise initialize dp[arr] = 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 ``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 == -1) {``        ``// All integers in range [1, M]``        ``// are reachable``        ``for` `(``int` `j = 1; j <= M; j++) {``            ``dp[j] = 1;``        ``}``    ``}``    ``else` `{``        ``// Only reachable integer is arr``        ``dp[arr] = 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);``    ``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``            ``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``        ``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 == -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``            ``dp[0, arr] = 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

 ``
Output:
`5`

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

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