Construct the Array


Problem Statement :


Your goal is to find the number of ways to construct an array such that consecutive positions contain different values.

Specifically, we want to construct an array with n elements such that each element between 1 and k, inclusive. We also want the first and last elements of the array to be 1 and x.

Given n, k and x, find the number of ways to construct such an array. Since the answer may be large, only find it modulo 10^9 + 7.

For example, for n=4, k=3, x=2, there are 3 ways, as shown here:

image

Complete the function countArray which takes input n, k and x. Return the number of ways to construct the array such that consecutive elements are distinct.

Constraints

3 <= n <=10^5
2 <= k <= 10^5
1 <= x <= k



Solution :



title-img


                            Solution in C :

In C++ :





#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int mod = 1000000007;

int Inv(int a)
{
    int res = 1;
    int p = mod - 2;
    while (p) {
        if (p & 1) res = ll(res) * a % mod;
        p >>= 1; a = ll(a) * a % mod;
    }
    return res;
}

int main() {
    int n;
    int k;
    int x;
    cin >> n >> k >> x;
    int res1 = 1, res0 = 0;
    for (int i = 1; i < n; i++) {
        int nres1 = res0;
        int nres0 = (ll(res1) * (k - 1) + ll(res0) * (k - 2)) % mod;
        res1 = nres1; res0 = nres0;
    }
    if (x == 1) printf("%d\n", res1);
    else {
        int res = ll(res0) * Inv(k - 1) % mod;
        printf("%d\n", res);
    }
    return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static long countArray(int n, int k, int x) {
        // Return the number of ways to fill in the array.
        long max = 1000000007;
        long[] f = new long[n + 1]; //[1,...,1]
        long[] g = new long[n + 1]; //[1,...,2]
        f[3] = k - 1;
        g[2] = 1;
        g[3] = k - 2;
        for (int i = 4; i <= n; i++) {
            f[i] = (k - 1) * g[i - 1] % max;
            g[i] = (f[i - 1] + (k - 2) * g[i - 1]) % max;
        }
        return x == 1 ? f[n] : g[n];
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int x = in.nextInt();
        long answer = countArray(n, k, x);
        System.out.println(answer);
        in.close();
    }
}








In C :





#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#include <stdint.h>
#include <inttypes.h>

long int countArray(int n, int k, int x) {
    // Return the number of ways to fill in the array.
    int64_t MOD=1e9+7;
    int64_t eq_x = 0;
    int64_t neq_x = 0;
    if (x == 1) {
        eq_x = 1;
        neq_x = 0;
    } else {
        eq_x = 0;
        neq_x = 1;
    }
    for (int i = 1; i < n; i++) {
        int64_t eq_x1 = neq_x;
        int64_t neq_x1 = (k-1) * eq_x + (k-2) * neq_x;

        eq_x = eq_x1 % MOD;
        neq_x = neq_x1 % MOD;
    }
    return eq_x;
}

int main() {
    int n; 
    int k; 
    int x; 
    scanf("%i %i %i", &n, &k, &x);
    long int answer = countArray(n, k, x);
    printf("%ld\n", answer);
    return 0;
}








In Python3 :





#!/bin/python3

import sys

def countArray(n, k, x):
    # Return the number of ways to fill in the array.
    d, s = 1, 0
    for i in range(2, n):
        d, s = (k - 2) * d + s, (k - 1) * d
    return d if x != 1 else s

if __name__ == "__main__":
    n, k, x = input().strip().split(' ')
    n, k, x = [int(n), int(k), int(x)]
    answer = countArray(n, k, x)
    print(answer % (10 ** 9 + 7))
                        








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