Given three positive integers n, k and x. The task is to count the number of different array that can be formed of size n such that each element is between 1 to k and two consecutive element are different. Also, the first and last elements of each array should be 1 and x respectively.
Examples :
Input : n = 4, k = 3, x = 2 Output : 3
The idea is to use Dynamic Programming and combinatorics to solve the problem.
First of all, notice that the answer is same for all x from 2 to k. It can easily be proved. This will be useful later on.
Let the state f(i) denote the number of ways to fill the range [1, i] of array A such that A1 = 1 and Ai ≠ 1.
Therefore, if x ≠ 1, the answer to the problem is f(n)/(k – 1), because f(n) is the number of way where An is filled with a number from 2 to k, and the answer are equal for all such values An, so the answer for an individual value is f(n)/(k – 1).
Otherwise, if x = 1, the answer is f(n – 1), because An – 1 ≠ 1, and the only number we can fill An with is x = 1.
Now, the main problem is how to calculate f(i). Consider all numbers that Ai – 1 can be. We know that it must lie in [1, k].
- If Ai – 1 ≠ 1, then there are (k – 2)f(i – 1) ways to fill in the rest of the array, because Ai cannot be 1 or Ai – 1 (so we multiply with (k – 2)), and for the range [1, i – 1], there are, recursively, f(i – 1) ways.
- If Ai – 1 = 1, then there are (k – 1)f(i – 2) ways to fill in the rest of the array, because Ai – 1 = 1 means Ai – 2 ≠ 1 which means there are f(i – 2)ways to fill in the range [1, i – 2] and the only value that Ai cannot be 1, so we have (k – 1) choices for Ai.
By combining the above, we get
f(i) = (k - 1) * f(i - 2) + (k - 2) * f(i - 1)
This will help us to use dynamic programming using f(i).
Below is the implementation of this approach:
C++
// CPP Program to find count of arrays.#include <bits/stdc++.h>#define MAXN 109using namespace std;// Return the number of arrays with given constartints.int countarray(int n, int k, int x){ int dp[MAXN] = { 0 }; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for (int i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]);}// Driven Programint main(){ int n = 4, k = 3, x = 2; cout << countarray(n, k, x) << endl; return 0;} |
Java
// Java program to find count of arrays.import java.util.*;class Counting{ static int MAXN = 109; public static int countarray(int n, int k, int x) { int[] dp = new int[109]; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for (int i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]); } // driver code public static void main(String[] args) { int n = 4, k = 3, x = 2; System.out.println(countarray(n, k, x)); }}// This code is contributed by rishabh_jain |
Python3
# Python3 code to find count of arrays.# Return the number of lists with# given constraints.def countarray( n , k , x ): dp = list() # Initalising dp[0] and dp[1] dp.append(0) dp.append(1) # Computing f(i) for each 2 <= i <= n. i = 2 while i < n: dp.append( (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]) i = i + 1 return ( (k - 1) * dp[n - 2] if x == 1 else dp[n - 1])# Driven coden = 4k = 3x = 2print(countarray(n, k, x))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to find count of arrays.using System;class GFG{// static int MAXN = 109; public static int countarray(int n, int k, int x) { int[] dp = new int[109]; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for (int i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]); } // Driver code public static void Main() { int n = 4, k = 3, x = 2; Console.WriteLine(countarray(n, k, x)); }}// This code is contributed by vt_m |
PHP
<?php// PHP Program to find// count of arrays.$MAXN = 109;// Return the number of arrays// with given constartints.function countarray($n, $k, $x){ $dp = array( 0 ); // Initalising dp[0] and dp[1]. $dp[0] = 0; $dp[1] = 1; // Computing f(i) for // each 2 <= i <= n. for ( $i = 2; $i < $n; $i++) $dp[$i] = ($k - 2) * $dp[$i - 1] + ($k - 1) * $dp[$i - 2]; return ($x == 1 ? ($k - 1) * $dp[$n - 2] : $dp[$n - 1]);}// Driven Code$n = 4; $k = 3; $x = 2;echo countarray($n, $k, $x) ;// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript program to find count of arrays.let MAXN = 109; function countarray(n, k, x){ let dp = []; // Initalising dp[0] and dp[1]. dp[0] = 0; dp[1] = 1; // Computing f(i) for each 2 <= i <= n. for(let i = 2; i < n; i++) dp[i] = (k - 2) * dp[i - 1] + (k - 1) * dp[i - 2]; return (x == 1 ? (k - 1) * dp[n - 2] : dp[n - 1]);}// Driver codelet n = 4, k = 3, x = 2;document.write(countarray(n, k, x));// This code is contributed by sanjoy_62</script> |
Output :
3
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