# Count all N-length arrays made up of distinct consecutive elements whose first and last elements are equal

Given two integers M and N, the task is to find the number of N-length arrays possible having non-equal adjacent elements lying in the range [1, M] having elements at first and last indices equal.

Examples:

Input: N = 3, M = 3
Output: 6
Explanation:
The possible arrays are {1, 2, 1}, {1, 3, 1}, {2, 1, 2}, {2, 3, 2}, {3, 1, 3}, {3, 2, 3}.

Input: N = 5, M = 4
Output: 84

Approach: Follow the steps below to solve the problem:

• First fix arr[0] and arr[N-1] equal to 1.
• Now find the number of arrays possible of size i which ends with 1 (i.e., arr[i] = 1). Store this result into end_with_one[i].
• Now, find the number of arrays possible of size i which does not end with 1 (arr[i] ? 1). Store this result into end_not_with_one[i].
• Since, the number of ways to form the array till the ith index with arr[i] = 1, is same as the number of ways to form the array till the (i â€“ 1)th index with arr[i â€“ 1] ? 1, set end_with_one[i] = end_not_with_one[i – 1].
• Now, the number of ways to form the array till the ith index with arr[i] ? 1 is as follows:
• If arr[i – 1]= 1, there are (M â€“ 1) numbers to be placed at the ith index.
• If arr[i – 1] ? 1, then (M â€“ 2) numbers can be placed at index i, since arr[i] cannot be 1 and arr[i] cannot be equal to arr[i â€“ 1].
• Therefore, set end_not_with_one[i] = end_with_one[i-1] * (M â€“ 1) + end_not_with_one[i-1]* (M â€“ 2).
• Therefore, the number of ways to form arrays of size N with arr[0] and arr[N – 1] equal to 1 is end_with_one[N – 1].
• Similarly, arr[0] and arr[N – 1] can be set to any element from 1 to M.
• Therefore, the total number of arrays possible is M * end_with_one[N-1].

Below is the implementation of the above approach:

## C++14

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to print the count of` `// arrays satisfying given condition` `int` `totalArrays(``int` `N, ``int` `M)` `{`   `    ``int` `end_with_one[N + 1];` `    ``int` `end_not_with_one[N + 1];`   `    ``// First element of` `    ``// array is set as 1` `    ``end_with_one[0] = 1;` `    ``end_not_with_one[0] = 0;`   `    ``// Since the first element` `    ``// of arr[] is 1, the` `    ``// second element can't be 1` `    ``end_with_one[1] = 0;` `    ``end_not_with_one[1] = M - 1;`   `    ``// Traverse the remaining indices` `    ``for` `(``int` `i = 2; i < N; i++) {`   `        ``// If arr[i] = 1` `        ``end_with_one[i]` `            ``= end_not_with_one[i - 1];`   `        ``// If arr[i] ? 1` `        ``end_not_with_one[i]` `            ``= end_with_one[i - 1] * (M - 1)` `              ``+ end_not_with_one[i - 1] * (M - 2);` `    ``}`   `    ``// Since last element needs to be 1` `    ``return` `end_with_one[N - 1];` `}`   `// Driver Code` `int` `main()` `{`   `    ``int` `N = 3, M = 3;`   `    ``// Stores the count of arrays` `    ``// where arr[0] = arr[N - 1] = 1` `    ``int` `temp = totalArrays(N, M);`   `    ``// Since arr[0] and arr[N - 1]` `    ``// can be any number from 1 to M` `    ``int` `ans = M * temp;`   `    ``// Print answer` `    ``cout << ans << ``"\n"``;`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;` `class` `GFG` `{`   `// Function to print the count of` `// arrays satisfying given condition` `static` `int` `totalArrays(``int` `N, ``int` `M)` `{` `    ``int` `[]end_with_one = ``new` `int``[N + ``1``];` `    ``int` `[]end_not_with_one = ``new` `int``[N + ``1``];`   `    ``// First element of` `    ``// array is set as 1` `    ``end_with_one[``0``] = ``1``;` `    ``end_not_with_one[``0``] = ``0``;`   `    ``// Since the first element` `    ``// of arr[] is 1, the` `    ``// second element can't be 1` `    ``end_with_one[``1``] = ``0``;` `    ``end_not_with_one[``1``] = M - ``1``;`   `    ``// Traverse the remaining indices` `    ``for` `(``int` `i = ``2``; i < N; i++)` `    ``{`   `        ``// If arr[i] = 1` `        ``end_with_one[i]` `            ``= end_not_with_one[i - ``1``];`   `        ``// If arr[i] ? 1` `        ``end_not_with_one[i]` `            ``= end_with_one[i - ``1``] * (M - ``1``)` `              ``+ end_not_with_one[i - ``1``] * (M - ``2``);` `    ``}`   `    ``// Since last element needs to be 1` `    ``return` `end_with_one[N - ``1``];` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``3``, M = ``3``;`   `    ``// Stores the count of arrays` `    ``// where arr[0] = arr[N - 1] = 1` `    ``int` `temp = totalArrays(N, M);`   `    ``// Since arr[0] and arr[N - 1]` `    ``// can be any number from 1 to M` `    ``int` `ans = M * temp;`   `    ``// Print answer` `    ``System.out.print(ans+ ``"\n"``);` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python program for the above approach`   `# Function to print the count of` `# arrays satisfying given condition` `def` `totalArrays(N, M):` `    ``end_with_one ``=` `[``0``] ``*` `(N ``+` `1``);` `    ``end_not_with_one ``=` `[``0``] ``*` `(N ``+` `1``);`   `    ``# First element of` `    ``# array is set as 1` `    ``end_with_one[``0``] ``=` `1``;` `    ``end_not_with_one[``0``] ``=` `0``;`   `    ``# Since the first element` `    ``# of arr is 1, the` `    ``# second element can't be 1` `    ``end_with_one[``1``] ``=` `0``;` `    ``end_not_with_one[``1``] ``=` `M ``-` `1``;`   `    ``# Traverse the remaining indices` `    ``for` `i ``in` `range``(``2``, N):` `      `  `        ``# If arr[i] = 1` `        ``end_with_one[i] ``=` `end_not_with_one[i ``-` `1``];`   `        ``# If arr[i] ? 1` `        ``end_not_with_one[i] ``=` `end_with_one[i ``-` `1``] ``*` `(M ``-` `1``) ``+` `end_not_with_one[i ``-` `1``] ``*` `(M ``-` `2``);`   `    ``# Since last element needs to be 1` `    ``return` `end_with_one[N ``-` `1``];`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``N ``=` `3``;` `    ``M ``=` `3``;`   `    ``# Stores the count of arrays` `    ``# where arr[0] = arr[N - 1] = 1` `    ``temp ``=` `totalArrays(N, M);`   `    ``# Since arr[0] and arr[N - 1]` `    ``# can be any number from 1 to M` `    ``ans ``=` `M ``*` `temp;`   `    ``# Print answer` `    ``print``(ans);` `    `  `# This code is contributed by 29AjayKumar`

## C#

 `// C# program for the above approach ` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG ` `{` `    `  `    ``// Function to print the count of ` `    ``// arrays satisfying given condition ` `    ``static` `int` `totalArrays(``int` `N, ``int` `M) ` `    ``{ ` `      `  `        ``int``[] end_with_one = ``new` `int``[N + 1]; ` `        ``int``[] end_not_with_one = ``new` `int``[N + 1]; ` `      `  `        ``// First element of ` `        ``// array is set as 1 ` `        ``end_with_one[0] = 1; ` `        ``end_not_with_one[0] = 0; ` `      `  `        ``// Since the first element ` `        ``// of arr[] is 1, the ` `        ``// second element can't be 1 ` `        ``end_with_one[1] = 0; ` `        ``end_not_with_one[1] = M - 1; ` `      `  `        ``// Traverse the remaining indices ` `        ``for` `(``int` `i = 2; i < N; i++) { ` `      `  `            ``// If arr[i] = 1 ` `            ``end_with_one[i] ` `                ``= end_not_with_one[i - 1]; ` `      `  `            ``// If arr[i] ? 1 ` `            ``end_not_with_one[i] ` `                ``= end_with_one[i - 1] * (M - 1) ` `                  ``+ end_not_with_one[i - 1] * (M - 2); ` `        ``} ` `      `  `        ``// Since last element needs to be 1 ` `        ``return` `end_with_one[N - 1]; ` `    ``}  `   `  ``// Driver code` `  ``static` `void` `Main()` `  ``{` `      `  `    ``int` `N = 3, M = 3; ` `  `  `    ``// Stores the count of arrays ` `    ``// where arr[0] = arr[N - 1] = 1 ` `    ``int` `temp = totalArrays(N, M); ` `  `  `    ``// Since arr[0] and arr[N - 1] ` `    ``// can be any number from 1 to M ` `    ``int` `ans = M * temp; ` `  `  `    ``// Print answer ` `    ``Console.WriteLine(ans); ` `  ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output:

`6`

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

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