This problem required very little memoization. I noticed a pattern; consider the number of arrays of length 2: [0,1,1] where index is the ending number x. Using a dynamic coding strategy, you might try to find the number of arrays of length 3 by summing up elements of this array: [2,1,1] You don't add the number of arrays of length 2 ending in the same x, since the numbers can't appear twice in a row. Therefore, index 1 is one greater, since it ignored the one element that was 1 less than the others. Using this array, the number of arrays of length 4 is: [2,3,3] This time, index 1 is one less since it ignored the one element that was one more than the rest. The pattern continues: [6,5,5] [10,11,11]

So we can ditch using arrays and just consider the number of arrays ending in x>1. The number N(n) of arrays of length n ending in x>1 is equal to (k-1)*N(n-1) + (1 if n is even, -1 if n is odd)

if x==1, subtract 1 if n is even, add 1 if n is odd

function countArray(n: number, k: number, x: number): number {
    // Return the number of ways to fill in the array.
    //ending in number col+1
    //array of length row+2
    //NOTE: for EVEN length, there is one less array ending in 1. for ODD, there is one extra.
    // Otherwise, the number of constructable arrays of length n ending in k is equal to the sum of all arrays of length n-1, + or - one depending on n's parity.
    let curr = 1;
    
    let mod = (Math.pow(10,9)+7)
    
    let even;
    let odd;
    //iterate over possible lengths
    for(let i=2; i<n; i++)
    {
        even = i%2
        odd = even==1?0:1
        //add even bonus, remove odd
        curr = (curr*(k-1) + even - odd) % mod
    }
    if(x==1) curr += (-even + odd)
    
    return curr
}