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97. Interleaving String 👍

Approach 1: 2D DP

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  bool isInterleave(string s1, string s2, string s3) {
    const int m = s1.length();
    const int n = s2.length();
    if (m + n != s3.length())
      return false;

    // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    //             s1[0..i) and s2[0..j)
    vector<vector<bool>> dp(m + 1, vector<bool>(n + 1));
    dp[0][0] = true;

    for (int i = 1; i <= m; ++i)
      dp[i][0] = dp[i - 1][0] && s1[i - 1] == s3[i - 1];

    for (int j = 1; j <= n; ++j)
      dp[0][j] = dp[0][j - 1] && s2[j - 1] == s3[j - 1];

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1] ||
                   dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];

    return dp[m][n];
  }
};
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class Solution {
  public boolean isInterleave(String s1, String s2, String s3) {
    final int m = s1.length();
    final int n = s2.length();
    if (m + n != s3.length())
      return false;

    // dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    //             s1[0..i) and s2[0..j)
    boolean[][] dp = new boolean[m + 1][n + 1];
    dp[0][0] = true;

    for (int i = 1; i <= m; ++i)
      dp[i][0] = dp[i - 1][0] && s1.charAt(i - 1) == s3.charAt(i - 1);

    for (int j = 1; j <= n; ++j)
      dp[0][j] = dp[0][j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) ||
                   dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);

    return dp[m][n];
  }
}
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class Solution:
  def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
    m = len(s1)
    n = len(s2)
    if m + n != len(s3):
      return False

    # dp[i][j] := true if s3[0..i + j) is formed by the interleaving of
    #             s1[0..i) and s2[0..j)
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True

    for i in range(1, m + 1):
      dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]

    for j in range(1, n + 1):
      dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]

    for i in range(1, m + 1):
      for j in range(1, n + 1):
        dp[i][j] = (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or \
            (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1])

    return dp[m][n]

Approach 2: 1D DP

  • Time: $O(mn)$
  • Space: $O(n)$
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class Solution {
 public:
  bool isInterleave(string s1, string s2, string s3) {
    const int m = s1.length();
    const int n = s2.length();
    if (m + n != s3.length())
      return false;

    vector<bool> dp(n + 1);

    for (int i = 0; i <= m; ++i)
      for (int j = 0; j <= n; ++j)
        if (i == 0 && j == 0)
          dp[j] = true;
        else if (i == 0)
          dp[j] = dp[j - 1] && s2[j - 1] == s3[j - 1];
        else if (j == 0)
          dp[j] = dp[j] && s1[i - 1] == s3[i - 1];
        else
          dp[j] = dp[j] && s1[i - 1] == s3[i + j - 1] ||
                  dp[j - 1] && s2[j - 1] == s3[i + j - 1];

    return dp[n];
  }
};
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class Solution {
  public boolean isInterleave(String s1, String s2, String s3) {
    final int m = s1.length();
    final int n = s2.length();
    if (m + n != s3.length())
      return false;

    boolean[] dp = new boolean[n + 1];

    for (int i = 0; i <= m; ++i)
      for (int j = 0; j <= n; ++j)
        if (i == 0 && j == 0)
          dp[j] = true;
        else if (i == 0)
          dp[j] = dp[j - 1] && s2.charAt(j - 1) == s3.charAt(j - 1);
        else if (j == 0)
          dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i - 1);
        else
          dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) ||
                  dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);

    return dp[n];
  }
}
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class Solution:
  def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
    m = len(s1)
    n = len(s2)
    if m + n != len(s3):
      return False

    dp = [False] * (n + 1)

    for i in range(m + 1):
      for j in range(n + 1):
        if i == 0 and j == 0:
          dp[j] = True
        elif i == 0:
          dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]
        elif j == 0:
          dp[j] = dp[j] and s1[i - 1] == s3[i - 1]
        else:
          dp[j] = (dp[j] and s1[i - 1] == s3[i + j - 1]) or \
              (dp[j - 1] and s2[j - 1] == s3[i + j - 1])

    return dp[n]
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