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3269. Constructing Two Increasing Arrays 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minLargest(vector<int>& nums1, vector<int>& nums2) {
    const int m = nums1.size();
    const int n = nums2.size();
    // dp[i][j] := the minimum largest number for the first i nums1 and the
    // first j nums2
    vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MAX));
    dp[0][0] = 0;

    for (int i = 0; i <= m; ++i)
      for (int j = 0; j <= n; ++j) {
        if (i > 0)
          dp[i][j] = min(dp[i][j], f(dp[i - 1][j], nums1[i - 1]));
        if (j > 0)
          dp[i][j] = min(dp[i][j], f(dp[i][j - 1], nums2[j - 1]));
      }

    return dp[m][n];
  }

 private:
  // Returns the next number to fill in the array based on the previous number
  // and the current number.
  int f(int prev, int num) {
    return prev + (prev % 2 == num ? 2 : 1);
  }
};

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class Solution {
  public int minLargest(int[] nums1, int[] nums2) {
    final int m = nums1.length;
    final int n = nums2.length;
    // dp[i][j] := the minimum largest number for the first i nums1 and the
    // first j nums2
    int[][] dp = new int[m + 1][n + 1];

    Arrays.stream(dp).forEach(A -> Arrays.fill(A, Integer.MAX_VALUE));
    dp[0][0] = 0;

    for (int i = 0; i <= m; ++i)
      for (int j = 0; j <= n; ++j) {
        if (i > 0)
          dp[i][j] = Math.min(dp[i][j], f(dp[i - 1][j], nums1[i - 1]));
        if (j > 0)
          dp[i][j] = Math.min(dp[i][j], f(dp[i][j - 1], nums2[j - 1]));
      }

    return dp[m][n];
  }

  // Returns the next number to fill in the array based on the previous number
  // and the current number.
  private int f(int prev, int num) {
    return prev + (prev % 2 == num ? 2 : 1);
  }
}

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class Solution:
  def minLargest(self, nums1: list[int], nums2: list[int]) -> int:
    m = len(nums1)
    n = len(nums2)
    # dp[i][j] := the minimum largest number for the first i nums1 and the
    # first j nums2
    dp = [[math.inf] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = 0

    def f(prev: int, num: int) -> int:
      """
      Returns the next number to fill in the array based on the previous number
      and the current number.
      """
      return prev + (2 if prev % 2 == num else 1)

    for i in range(m + 1):
      for j in range(n + 1):
        if i > 0:
          dp[i][j] = min(dp[i][j], f(dp[i - 1][j], nums1[i - 1]))
        if j > 0:
          dp[i][j] = min(dp[i][j], f(dp[i][j - 1], nums2[j - 1]))

    return dp[m][n]