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# Number of ways to form an array with distinct adjacent elements

Given three integers N, M and X, the task is to find the number of ways to form an array, such that all consecutive numbers of the array are distinct, and the value at any index of the array from 2 to N – 1(Considering 1 based indexing) lies between 1 and M, while the value at index 1 is X and the value at index N is 1.
Note: Value of X lies between 1 and M.
Examples:

Input: N = 4, M = 3, X = 2
Output:
The following arrays are possible:
1) 2, 1, 2, 1
2) 2, 1, 3, 1
3) 2, 3, 2, 1
Input: N = 2, M = 3, X = 2
Output:
The only possible array is: 2, 1

Approach: The problem can be solved using Dynamic Programming. Let, f(i) represent the number of ways to form the array till the ith index, such that every consecutive element of the array is distinct. Let f(i, One) represent the number of ways to form the array till the i-th index such that every consecutive element of the array is distinct and arri = 1.
Similarly, let f(i, Non-One) represent the number of ways to form the array till the ith index, such that every consecutive element of the array is distinct and arri is not equal to 1.
The following recurrence is formed:

`f(i, Non-One) = f(i - 1, One) * (M - 1) + f(i - 1, Non-One) * (M - 2)`

which means that the number of ways to form the array till the ith index with arrayi not equal to 1 is formed using two cases:

1. If the number at arrayi – 1 is 1, then opt one number out of (M – 1) options to place at the ith index, since arrayi is not equal to 1.
2. If the number at arrayi – 1 is not 1, then we need to opt one number out of (M – 2) options, since arrayi is not equal to 1 and arrayi ≠ arrayi – 1.

Similarly, f(i, One) = f(i – 1, Non-One), since the number of ways to form the array till the ith index with arrayi = 1, is same as number of ways to form the array till the (i – 1)th index with arrayi – 1 ≠ 1, thus at the ith index there is only one option. At the end the required answer if f(N, One) since arrayN needs to be equal to 1.
Below is the implementation of the above approach:

## C++

 `// C++ program to count the number of ways``// to form arrays of N numbers such that the``// first and last numbers are fixed``// and all consecutive numbers are distinct``#include ``using` `namespace` `std;` `// Returns the total ways to form arrays such that``// every consecutive element is different and each``// element except the first and last can take values``// from 1 to M``int` `totalWays(``int` `N, ``int` `M, ``int` `X)``{` `    ``// define the dp[][] array``    ``int` `dp[N + 1];` `    ``// if the first element is 1``    ``if` `(X == 1) {` `        ``// there is only one way to place``        ``// a 1 at the first index``        ``dp = 1;``    ``}``    ``else` `{` `        ``// the value at first index needs to be 1,``        ``// thus there is no way to place a non-one integer``        ``dp = 0;``    ``}` `    ``// if the first element was 1 then at index 1,``    ``// only non one integer can be placed thus``    ``// there are M - 1 ways to place a non one integer``    ``// at index 2 and 0 ways to place a 1 at the 2nd index``    ``if` `(X == 1) {``        ``dp = 0;``        ``dp = M - 1;``    ``}` `    ``// Else there is one way to place a one at``    ``// index 2 and if a non one needs to be placed here,``    ``// there are (M - 2) options, i.e``    ``// neither the element at this index``    ``// should be 1, neither should it be``    ``// equal to the previous element``    ``else` `{``        ``dp = 1;``        ``dp = (M - 2);``    ``}` `    ``// Build the dp array in bottom up manner``    ``for` `(``int` `i = 2; i < N; i++) {` `        ``// f(i, one) = f(i - 1, non-one)``        ``dp[i] = dp[i - 1];` `        ``// f(i, non-one) = f(i - 1, one) * (M - 1) +``        ``// f(i - 1, non-one) * (M - 2)``        ``dp[i] = dp[i - 1] * (M - 1) + dp[i - 1] * (M - 2);``    ``}` `    ``// last element needs to be one, so return dp[n - 1]``    ``return` `dp[N - 1];``}` `// Driver Code``int` `main()``{` `    ``int` `N = 4, M = 3, X = 2;``    ``cout << totalWays(N, M, X) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to count the``// number of ways to form``// arrays of N numbers such``// that the first and last``// numbers are fixed and all``// consecutive numbers are``// distinct``import` `java.io.*;` `class` `GFG``{` `// Returns the total ways to``// form arrays such that every``// consecutive element is``// different and each element``// except the first and last``// can take values from 1 to M``static` `int` `totalWays(``int` `N,``                     ``int` `M, ``int` `X)``{` `    ``// define the dp[][] array``    ``int` `dp[][] = ``new` `int``[N + ``1``][``2``];` `    ``// if the first element is 1``    ``if` `(X == ``1``)``    ``{` `        ``// there is only one``        ``// way to place a 1``        ``// at the first index``        ``dp[``0``][``0``] = ``1``;``    ``}``    ``else``    ``{` `        ``// the value at first index``        ``// needs to be 1, thus there``        ``// is no way to place a``        ``// non-one integer``        ``dp[``0``][``1``] = ``0``;``    ``}` `    ``// if the first element was 1``    ``// then at index 1, only non``    ``// one integer can be placed``    ``// thus there are M - 1 ways``    ``// to place a non one integer``    ``// at index 2 and 0 ways to``    ``// place a 1 at the 2nd index``    ``if` `(X == ``1``)``    ``{``        ``dp[``1``][``0``] = ``0``;``        ``dp[``1``][``1``] = M - ``1``;``    ``}` `    ``// Else there is one way to``    ``// place a one at index 2``    ``// and if a non one needs to``    ``// be placed here, there are``    ``// (M - 2) options, i.e neither``    ``// the element at this index``    ``// should be 1, neither should``    ``// it be equal to the previous``    ``// element``    ``else``    ``{``        ``dp[``1``][``0``] = ``1``;``        ``dp[``1``][``1``] = (M - ``2``);``    ``}` `    ``// Build the dp array``    ``// in bottom up manner``    ``for` `(``int` `i = ``2``; i < N; i++)``    ``{` `        ``// f(i, one) = f(i - 1,``        ``// non-one)``        ``dp[i][``0``] = dp[i - ``1``][``1``];` `        ``// f(i, non-one) =``        ``// f(i - 1, one) * (M - 1) +``        ``// f(i - 1, non-one) * (M - 2)``        ``dp[i][``1``] = dp[i - ``1``][``0``] * (M - ``1``) +``                   ``dp[i - ``1``][``1``] * (M - ``2``);``    ``}` `    ``// last element needs to be``    ``// one, so return dp[n - 1]``    ``return` `dp[N - ``1``][``0``];``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``4``, M = ``3``, X = ``2``;``    ``System.out.println(totalWays(N, M, X));``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Python 3 program to count the number of ways``# to form arrays of N numbers such that the``# first and last numbers are fixed``# and all consecutive numbers are distinct` `# Returns the total ways to form arrays such that``# every consecutive element is different and each``# element except the first and last can take values``# from 1 to M``def` `totalWays(N,M,X):``    ` `    ``# define the dp[][] array``    ``dp ``=` `[[``0` `for` `i ``in` `range``(``2``)] ``for` `j ``in` `range``(N``+``1``)]` `    ``# if the first element is 1``    ``if` `(X ``=``=` `1``):``        ` `        ``# there is only one way to place``        ``# a 1 at the first index``        ``dp[``0``][``0``] ``=` `1` `    ``else``:``        ``# the value at first index needs to be 1,``        ``# thus there is no way to place a non-one integer``        ``dp[``0``][``1``] ``=` `0` `    ``# if the first element was 1 then at index 1,``    ``# only non one integer can be placed thus``    ``# there are M - 1 ways to place a non one integer``    ``# at index 2 and 0 ways to place a 1 at the 2nd index``    ``if` `(X ``=``=` `1``):``        ``dp[``1``][``0``] ``=` `0``        ``dp[``1``][``1``] ``=` `M ``-` `1` `    ``# Else there is one way to place a one at``    ``# index 2 and if a non one needs to be placed here,``    ``# there are (M - 2) options, i.e``    ``# neither the element at this index``    ``# should be 1, neither should it be``    ``# equal to the previous element``    ``else``:``        ``dp[``1``][``0``] ``=` `1``        ``dp[``1``][``1``] ``=` `(M ``-` `2``)` `    ``# Build the dp array in bottom up manner``    ``for` `i ``in` `range``(``2``,N):``        ``# f(i, one) = f(i - 1, non-one)``        ``dp[i][``0``] ``=` `dp[i ``-` `1``][``1``]` `        ``# f(i, non-one) = f(i - 1, one) * (M - 1) +``        ``# f(i - 1, non-one) * (M - 2)``        ``dp[i][``1``] ``=` `dp[i ``-` `1``][``0``] ``*` `(M ``-` `1``) ``+` `dp[i ``-` `1``][``1``] ``*` `(M ``-` `2``)` `    ``# last element needs to be one, so return dp[n - 1]``    ``return` `dp[N ``-` `1``][``0``]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `4``    ``M ``=` `3``    ``X ``=` `2``    ``print``(totalWays(N, M, X))``    ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to count the``// number of ways to form``// arrays of N numbers such``// that the first and last``// numbers are fixed and all``// consecutive numbers are``// distinct``using` `System;` `class` `GFG``{` `// Returns the total ways to``// form arrays such that every``// consecutive element is``// different and each element``// except the first and last``// can take values from 1 to M``static` `int` `totalWays(``int` `N,``                     ``int` `M, ``int` `X)``{` `    ``// define the dp[][] array``    ``int` `[,]dp = ``new` `int``[N + 1, 2];` `    ``// if the first element is 1``    ``if` `(X == 1)``    ``{` `        ``// there is only one``        ``// way to place a 1``        ``// at the first index``        ``dp[0, 0] = 1;``    ``}``    ``else``    ``{` `        ``// the value at first index``        ``// needs to be 1, thus there``        ``// is no way to place a``        ``// non-one integer``        ``dp[0, 1] = 0;``    ``}` `    ``// if the first element was 1``    ``// then at index 1, only non``    ``// one integer can be placed``    ``// thus there are M - 1 ways``    ``// to place a non one integer``    ``// at index 2 and 0 ways to``    ``// place a 1 at the 2nd index``    ``if` `(X == 1)``    ``{``        ``dp[1, 0] = 0;``        ``dp[1, 1] = M - 1;``    ``}` `    ``// Else there is one way to``    ``// place a one at index 2``    ``// and if a non one needs to``    ``// be placed here, there are``    ``// (M - 2) options, i.e neither``    ``// the element at this index``    ``// should be 1, neither should``    ``// it be equal to the previous``    ``// element``    ``else``    ``{``        ``dp[1, 0] = 1;``        ``dp[1, 1] = (M - 2);``    ``}` `    ``// Build the dp array``    ``// in bottom up manner``    ``for` `(``int` `i = 2; i < N; i++)``    ``{` `        ``// f(i, one) = f(i - 1,``        ``// non-one)``        ``dp[i, 0] = dp[i - 1, 1];` `        ``// f(i, non-one) =``        ``// f(i - 1, one) * (M - 1) +``        ``// f(i - 1, non-one) * (M - 2)``        ``dp[i, 1] = dp[i - 1, 0] * (M - 1) +``                   ``dp[i - 1, 1] * (M - 2);``    ``}` `    ``// last element needs to be``    ``// one, so return dp[n - 1]``    ``return` `dp[N - 1, 0];``}` `// Driver Code``public` `static` `void` `Main ()``{``    ``int` `N = 4, M = 3, X = 2;``    ``Console.WriteLine(totalWays(N, M, X));``}``}` `// This code is contributed``// by anuj_67.`

## PHP

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## Javascript

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Output:

`3`

Time Complexity: O(N), where N is the size of the array
Auxiliary Space: O(N), for dp array

Efficient approach : Space optimization O(1)

In previous approach we the current value dp[i] is only depend upon the previous value i.e. dp[i-1]. So to optimize the space we can keep track of previous and current values by the help of three variables prev and curr which will reduce the space complexity from O(N) to O(1).

Implementation Steps:

• Create 2 variables prev and curr to keep track of previous and current values of DP.
• Initialize base case for X == 1.
• Iterate over subproblem using loop and update curr.
• After every iteration update prev and curr for further iterations.
• At last return prev.

Implementation:

## C++

 `// C++ program to count the number of ways``// to form arrays of N numbers such that the``// first and last numbers are fixed``// and all consecutive numbers are distinct``#include ``using` `namespace` `std;`  `// Returns the total ways to form arrays such that``// every consecutive element is different and each``// element except the first and last can take values``// from 1 to M``int` `totalWays(``int` `N, ``int` `M, ``int` `X)``{``    ``// initialize variable to keep track``    ``// of previous and current value``    ``int` `prev, curr;` `    ` `    ``// Base case``    ``if` `(X == 1) {``        ``prev = 1;``    ``} ``else` `{``        ``curr = 0;``    ``}` `    ``if` `(X == 1) {``        ``curr = M - 1;``    ``} ``else` `{``        ``prev = 1;``        ``curr = M - 2;``    ``}``    ` `    ``// iterating over subproblems to get the computation of``    ``// current value from previous computation``    ``for` `(``int` `i = 2; i < N; i++) {``        ``int` `temp = prev;``        ``prev = curr;``        ``curr = temp * (M - 1) + curr * (M - 2);``    ``}` `    ``// return answer``    ``return` `prev;``}` `// Driver Code``int` `main()``{``    ``int` `N = 4, M = 3, X = 2;``    ``cout << totalWays(N, M, X) << endl;` `    ``return` `0;``}`

Output:

`3`

Time Complexity: O(N), where N is the size of the array
Auxiliary Space: O(1), only use variables

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