Number of ways to cut pizza

MOST COMMON PROBLEM FOR TIKTOK

i,j,k = coords and k cuts prefix sum

const M = int(1e9 + 7)

var dp [][][]int 
var a [][]int 
var n int
var m int 

func getSum(i, j, u, v int) int {

    // get sum of tomatoes inside rec (i, j) -> (u, v)
    return a[u][v] - a[u][j - 1] - a[i - 1][v] + a[i - 1][j - 1]
}

func DP(last_row, last_col, k int) int {

    // use enough slice
    if (k == 0) {
        // if the rest having larger than 1 tomato
        if getSum(last_row, last_col, n, m) > 0 {
            return 1
        }
        return 0
    }

    // if the rest is not a sub-rec  
    if last_col == m + 1 || last_row == n + 1 {
        return 0
    }

    // if solution have been already exist
    if dp[last_row][last_col][k] != -1 {
        return dp[last_row][last_col][k]
    }

    ans := 0
    
    // slice row
    for i := last_row; i <= n; i++ {
        if getSum(last_row, last_col, i, m) > 0 {
            ans = (ans + DP(i + 1, last_col, k - 1)) % M
        }
    }

    // slice col
    for i := last_col; i <= m; i++ {
        if getSum(last_row, last_col, n, i) > 0 {
            ans = (ans + DP(last_row, i + 1, k - 1)) % M
        }
    }

    dp[last_row][last_col][k] = ans
    return ans

}

func ways(pizza []string, k int) int {

    // number of row
    n = len(pizza)
    // number of col
    m = len(pizza[0])

    // init a
    a = make([][]int, n + 1)
    for i := 0; i < n + 1; i++ {
        a[i] = make([]int, m + 1)
    }

    // init dp
    dp = make([][][]int, n + 1)
    for i := 0; i < n + 1; i++ {
        dp[i] = make([][]int, m + 1)
        for j := 0; j < m + 1; j++ {
            dp[i][j] = make([]int, k + 1)
            for x := 0; x < k + 1; x++ {
                dp[i][j][x] = -1
            }
        }
    }

    // calc prefix sum of matrix 2D a ('A' is 1, '.' is 0)
    for i := 1; i <= n; i++ {
        for j := 1; j <= m; j++ {
            if pizza[i - 1][j - 1] == 'A' {
                a[i][j] = a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1] + 1
            } else {
                a[i][j] = a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1]
            }
        }
    }

    return DP(1, 1, k - 1)

}