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629. K Inverse Pairs Array 👍

  • Time: $O(nk)$
  • Space: $O(nk)$
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class Solution {
 public:
  int kInversePairs(int n, int k) {
    constexpr int kMod = 1'000'000'007;
    // dp[i][j] := the number of permutations of numbers 1..i with j inverse
    // pairs
    vector<vector<int>> dp(n + 1, vector<int>(k + 1));

    // If there's no inverse pair, the permutation is unique "123..i".
    for (int i = 0; i <= n; ++i)
      dp[i][0] = 1;

    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= k; ++j) {
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % kMod;
        if (j - i >= 0)
          dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + kMod) % kMod;
      }

    return dp[n][k];
  }
};

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class Solution {
  public int kInversePairs(int n, int k) {
    final int kMod = 1_000_000_007;
    // dp[i][j] := the number of permutations of numbers 1..i with j inverse pairs
    int[][] dp = new int[n + 1][k + 1];

    // If there's no inverse pair, the permutation is unique "123..i".
    for (int i = 0; i <= n; ++i)
      dp[i][0] = 1;

    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= k; ++j) {
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % kMod;
        if (j - i >= 0)
          dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + kMod) % kMod;
      }

    return dp[n][k];
  }
}

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class Solution:
  def kInversePairs(self, n: int, k: int) -> int:
    kMod = 1_000_000_007
    # dp[i][j] := the number of permutations of numbers 1..i with j inverse pairs
    dp = [[0] * (k + 1) for _ in range(n + 1)]

    # If there's no inverse pair, the permutation is unique '123..i'
    for i in range(n + 1):
      dp[i][0] = 1

    for i in range(1, n + 1):
      for j in range(1, k + 1):
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % kMod
        if j - i >= 0:
          dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + kMod) % kMod

    return dp[n][k]