Count of arrays having consecutive element with different values (original) (raw)

Last Updated : 13 Sep, 2023

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 109 using namespace std;

// Return the number of arrays with given constraints. int countarray(int n, int k, int x) { int dp[MAXN] = { 0 };

// Initialising 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 Program int 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];

    // Initialising 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()

# Initialising 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 code

n = 4 k = 3 x = 2 print(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];

    // Initialising 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

JavaScript

PHP

k,k, k,x) { $dp = array( 0 ); // Initialising dp[0] and dp[1]. $dp[0] = 0; $dp[1] = 1; // Computing f(i) for // each 2 <= i <= n. for ( i=2;i = 2; i=2;i < n;n; n;i++) dp[dp[dp[i] = ($k - 2) * dp[dp[dp[i - 1] + ($k - 1) * dp[dp[dp[i - 2]; return ($x == 1 ? ($k - 1) * dp[dp[dp[n - 2] : dp[dp[dp[n - 1]); } // Driven Code n=4;n = 4; n=4;k = 3; $x = 2; echo countarray($n, k,k, k,x) ; // This code is contributed by anuj_67. ?>

**Time Complexity: O(n)
**Auxiliary Space: O(MAXN), here MAXN = 109

**Efficient approach: Space optimization O(1)

In the approach we have only used three variables , prev1 and prev2 to store the values of the previous two elements of the dp array and curr to store the current value. Therefore, the space complexity of the optimized code is O(1)

**Implementation Steps:

Create **2 variables **prev1 and **prev2 to keep track of the previous 2 values of **DP and curr to store the current value.
Initialize **prev1 and **prev2 with **0 and **1 as base cases.
Now iterate through loop and get the current value form previous **2 values.
after Every iteration assign **prev2 to **prev1 and **curr to **prev2 to iterate further;
At last return answer.

**Implementation:

C++ `

// CPP Program to find count of arrays. #include <bits/stdc++.h> #define MAXN 109 using namespace std;

// Return the number of arrays with given constraints. int countarray(int n, int k, int x) { // initialize variables to store previous values int prev1 = 0, prev2 = 1, curr;

// Computing f(i) for each 2 <= i <= n.
for (int i = 2; i < n; i++) {
    curr = (k - 2) * prev2 + (k - 1) * prev1;

    // assigning values to iterate further
    prev1 = prev2;
    prev2 = curr;
}

// return final answer
return (x == 1 ? (k - 1) * prev1 : prev2);

}

// Driven Program int main() { int n = 4, k = 3, x = 2;

// function call
cout << countarray(n, k, x) << endl;
return 0;

}

Java

import java.util.*;

public class Main {

// Return the number of arrays with given constants.
static int countArray(int n, int k, int x)
{
    // initialize variables to store previous values
    int prev1 = 0, prev2 = 1, curr;
    // Computing f(i) for each 2 <= i <= n.
    for (int i = 2; i < n; i++) {
        curr = (k - 2) * prev2 + (k - 1) * prev1;
        // assigning values to iterate further
        prev1 = prev2;
        prev2 = curr;
    }
    // return final answer
    return (x == 1 ? (k - 1) * prev1 : prev2);
}

// Driver Program
public static void main(String[] args)
{
    int n = 4, k = 3, x = 2;
    // function call
    System.out.println(countArray(n, k, x));
}

}

Python3

def countarray(n, k, x): # initialize variables to store previous values prev1 = 0 prev2 = 1

# Computing f(i) for each 2 <= i <= n.
for i in range(2, n):
    curr = (k - 2) * prev2 + (k - 1) * prev1

    # assigning values to iterate further
    prev1 = prev2
    prev2 = curr

# return final answer
return (k - 1) * prev1 if x == 1 else prev2

Driven Program

n = 4 k = 3 x = 2

function call

print(countarray(n, k, x))

C#

using System;

public class Program {

// Return the number of arrays with given constartints. public static int CountArray(int n, int k, int x) { // initialize variables to store previous values int prev1 = 0, prev2 = 1, curr;

// Computing f(i) for each 2 <= i <= n.
for (int i = 2; i < n; i++) {
  curr = (k - 2) * prev2 + (k - 1) * prev1;

  // assigning values to iterate further
  prev1 = prev2;
  prev2 = curr;
}

// return final answer
return (x == 1 ? (k - 1) * prev1 : prev2);

}

// Driven Program public static void Main() { int n = 4, k = 3, x = 2;

// function call
Console.WriteLine(CountArray(n, k, x));

} }

JavaScript

// Function to calculate the number of arrays with given constants. function countArray(n, k, x) { let prev1 = 0, prev2 = 1, curr;

// Computing f(i) for each 2 <= i < n.
for (let i = 2; i < n; i++) {
    // Calculate the current value using the given formula.
    curr = (k - 2) * prev2 + (k - 1) * prev1;
    
    // Update the previous values for the next iteration.
    prev1 = prev2;
    prev2 = curr;
}

// Return the final answer based on the value of x.
return (x === 1 ? (k - 1) * prev1 : prev2);

}

// Input values let n = 4, k = 3, x = 2;

// Calculate and output the result console.log(countArray(n, k, x));

**Time Complexity: O(n)
**Auxiliary Space: O(1)