**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 :

` ````
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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