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Commit 674c364

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feat: add solutions to lc problem: No.3339 (doocs#3698)
No.3339.Find the Number of K-Even Arrays
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‎solution/3300-3399/3339.Find the Number of K-Even Arrays/README.md‎

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‎solution/3300-3399/3339.Find the Number of K-Even Arrays/README_EN.md‎

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class Solution {
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public:
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int countOfArrays(int n, int m, int k) {
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int f[n][k + 1][2];
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memset(f, -1, sizeof(f));
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const int mod = 1e9 + 7;
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int cnt0 = m / 2;
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int cnt1 = m - cnt0;
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auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
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if (j < 0) {
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return 0;
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}
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if (i >= n) {
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return j == 0 ? 1 : 0;
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}
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if (f[i][j][k] != -1) {
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return f[i][j][k];
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}
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int a = 1LL * cnt1 * dfs(dfs, i + 1, j, 1) % mod;
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int b = 1LL * cnt0 * dfs(dfs, i + 1, j - (k & 1 ^ 1), 0) % mod;
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return f[i][j][k] = (a + b) % mod;
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};
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return dfs(dfs, 0, k, 1);
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}
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};
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func countOfArrays(n int, m int, k int) int {
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f := make([][][2]int, n)
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for i := range f {
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f[i] = make([][2]int, k+1)
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for j := range f[i] {
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f[i][j] = [2]int{-1, -1}
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}
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}
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const mod int = 1e9 + 7
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cnt0 := m / 2
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cnt1 := m - cnt0
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var dfs func(int, int, int) int
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dfs = func(i, j, k int) int {
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if j < 0 {
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return 0
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}
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if i >= n {
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if j == 0 {
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return 1
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}
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return 0
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}
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if f[i][j][k] != -1 {
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return f[i][j][k]
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}
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a := cnt1 * dfs(i+1, j, 1) % mod
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b := cnt0 * dfs(i+1, j-(k&1^1), 0) % mod
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f[i][j][k] = (a + b) % mod
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return f[i][j][k]
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}
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return dfs(0, k, 1)
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}
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class Solution {
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private Integer[][][] f;
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private long cnt0, cnt1;
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private final int mod = (int) 1e9 + 7;
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public int countOfArrays(int n, int m, int k) {
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f = new Integer[n][k + 1][2];
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cnt0 = m / 2;
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cnt1 = m - cnt0;
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return dfs(0, k, 1);
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}
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private int dfs(int i, int j, int k) {
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if (j < 0) {
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return 0;
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}
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if (i >= f.length) {
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return j == 0 ? 1 : 0;
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}
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if (f[i][j][k] != null) {
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return f[i][j][k];
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}
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int a = (int) (cnt1 * dfs(i + 1, j, 1) % mod);
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int b = (int) (cnt0 * dfs(i + 1, j - (k & 1 ^ 1), 0) % mod);
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return f[i][j][k] = (a + b) % mod;
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}
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}
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class Solution:
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def countOfArrays(self, n: int, m: int, k: int) -> int:
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@cache
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def dfs(i: int, j: int, k: int) -> int:
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if j < 0:
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return 0
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if i >= n:
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return int(j == 0)
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return (
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cnt1 * dfs(i + 1, j, 1) + cnt0 * dfs(i + 1, j - (k & 1 ^ 1), 0)
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) % mod
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cnt0 = m // 2
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cnt1 = m - cnt0
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mod = 10**9 + 7
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ans = dfs(0, k, 1)
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dfs.cache_clear()
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return ans
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function countOfArrays(n: number, m: number, k: number): number {
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const f = Array.from({ length: n }, () =>
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Array.from({ length: k + 1 }, () => Array(2).fill(-1)),
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);
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const mod = 1e9 + 7;
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const cnt0 = Math.floor(m / 2);
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const cnt1 = m - cnt0;
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const dfs = (i: number, j: number, k: number): number => {
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if (j < 0) {
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return 0;
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}
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if (i >= n) {
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return j === 0 ? 1 : 0;
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}
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if (f[i][j][k] !== -1) {
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return f[i][j][k];
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}
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const a = (cnt1 * dfs(i + 1, j, 1)) % mod;
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const b = (cnt0 * dfs(i + 1, j - ((k & 1) ^ 1), 0)) % mod;
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return (f[i][j][k] = (a + b) % mod);
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};
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return dfs(0, k, 1);
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}
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class Solution {
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public:
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int countOfArrays(int n, int m, int k) {
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int f[n + 1][k + 1][2];
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memset(f, 0, sizeof(f));
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f[0][0][1] = 1;
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const int mod = 1e9 + 7;
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int cnt0 = m / 2;
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int cnt1 = m - cnt0;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j][0] = 1LL * (f[i - 1][j][1] + (j ? f[i - 1][j - 1][0] : 0)) * cnt0 % mod;
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f[i][j][1] = 1LL * (f[i - 1][j][0] + f[i - 1][j][1]) * cnt1 % mod;
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}
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}
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return (f[n][k][0] + f[n][k][1]) % mod;
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}
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};
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func countOfArrays(n int, m int, k int) int {
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f := make([][][2]int, n+1)
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for i := range f {
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f[i] = make([][2]int, k+1)
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}
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f[0][0][1] = 1
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cnt0 := m / 2
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cnt1 := m - cnt0
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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for j := 0; j <= k; j++ {
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f[i][j][0] = cnt0 * f[i-1][j][1] % mod
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if j > 0 {
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f[i][j][0] = (f[i][j][0] + cnt0*f[i-1][j-1][0]%mod) % mod
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}
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f[i][j][1] = cnt1 * (f[i-1][j][0] + f[i-1][j][1]) % mod
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}
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}
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return (f[n][k][0] + f[n][k][1]) % mod
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}
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class Solution {
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public int countOfArrays(int n, int m, int k) {
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int[][][] f = new int[n + 1][k + 1][2];
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int cnt0 = m / 2;
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int cnt1 = m - cnt0;
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final int mod = (int) 1e9 + 7;
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f[0][0][1] = 1;
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for (int i = 1; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j][0]
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= (int) (1L * cnt0 * (f[i - 1][j][1] + (j > 0 ? f[i - 1][j - 1][0] : 0)) % mod);
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f[i][j][1] = (int) (1L * cnt1 * (f[i - 1][j][0] + f[i - 1][j][1]) % mod);
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}
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}
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return (f[n][k][0] + f[n][k][1]) % mod;
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}
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}

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