SE103 - Week 9, Session 2

Transcript

1. Week 9:
   Dynamic Programming,
   Continued!
   

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2. - Fibonacci
   - Knapsack
   - Unbounded knapsack
   - Palindromic Sequences
   - Common Substrings
   Common Patterns in DP
   

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3. - Fibonacci
   - Knapsack
   - Unbounded knapsack
   - Palindromic Sequences
   - Common Substrings
   Common Patterns in DP
   

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4. Given a string, find the length of its Longest
   Palindromic Subsequence (LPS). In a palindromic
   subsequence, elements read the same backward
   and forward.
   A subsequence is a sequence that can be derived
   from another sequence by deleting some or no
   elements without changing the order of the remaining
   elements.
   Longest Palindromic Subsequence
   

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5. Given a string, find the length of its Longest Palindromic Subsequence
   (LPS). In a palindromic subsequence, elements read the same
   backward and forward.
   Example 1

   Given: abdbca

   Return: ??
   Longest Palindromic Subsequence
   

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6. Given a string, find the length of its Longest Palindromic Subsequence
   (LPS). In a palindromic subsequence, elements read the same
   backward and forward.
   Example 1

   Given: abdbca

   Return: 5 (abdba)
   Longest Palindromic Subsequence
   

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7. Given a string, find the length of its Longest Palindromic Subsequence
   (LPS). In a palindromic subsequence, elements read the same
   backward and forward.
   Example 1

   Given: abdbca

   Return: 5 (abdba)
   Example 2

   Given: cddpd

   Return: ??
   Longest Palindromic Subsequence
   

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8. Given a string, find the length of its Longest Palindromic Subsequence
   (LPS). In a palindromic subsequence, elements read the same
   backward and forward.
   Example 1

   Given: abdbca

   Return: 5 (abdba)
   Example 2

   Given: cddpd

   Return: 3 (ddd or dpd)
   Longest Palindromic Subsequence
   

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9. Given a string, find the length of its Longest Palindromic Subsequence
   (LPS). In a palindromic subsequence, elements read the same
   backward and forward.
   Example 1

   Given: abdbca

   Return: 5 (abdba)
   Example 2

   Given: cddpd

   Return: 3 (ddd or dpd)

   

   Example 3

   Given: abc

   Return: ??

   Longest Palindromic Subsequence
   

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10. Given a string, find the length of its Longest Palindromic Subsequence
    (LPS). In a palindromic subsequence, elements read the same
    backward and forward.
    Example 1

    Given: abdbca

    Return: 5 (abdba)
    Example 2

    Given: cddpd

    Return: 3 (ddd or dpd)

    

    Example 3

    Given: abc

    Return: 1 (a or b or c)

    Longest Palindromic Subsequence
    

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11. Given: cddpd

    Return: 3 (ddd or dpd)
    Intuitively…

    Let’s have two pointers, one starting at the beginning and one starting
    at the end

    cddpd

    

    

    Longest Palindromic Subsequence
    

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12. Given: cddpd

    Return: 3 (ddd or dpd)
    Intuitively…

    Let’s have two pointers, one starting at the beginning and one starting
    at the end

    cddpd

    

    

    If the letters are the same

    - count of palindromic subsequence = 2

    - recurively do this for the rest of the substring

    

    Longest Palindromic Subsequence
    

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13. Given: cddpd

    Return: 3 (ddd or dpd)
    Intuitively…

    Let’s have two pointers, one starting at the beginning and one starting
    at the end

    cddpd

    

    

    If the letters are the same

    - count of palindromic subsequence = 2

    if letters are NOT the same

    - try advancing starting pointer cddpd

    

    - try advancing ending pointer cddpd
    Longest Palindromic Subsequence
    

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14. Longest Palindromic Subsequence
    cddpd
    

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15. Longest Palindromic Subsequence
    c != d
    cddpd
    

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16. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    

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17. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    d == d

    count = 2
    

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18. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    

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19. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    

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20. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    d != p
    p d
    

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21. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    d != p
    p d
    p == p
    count = 1
    d == d
    count = 1
    

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22. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    d != p
    p d
    p == p
    count = 1
    d == d
    count = 1
    2 + Max(1, 1) = 3
    

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23. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    d != p
    p d
    p == p
    count = 1
    d == d
    count = 1
    2 + 1 = 3
    if(startIndex == endIndex)
    return 1;
    // case 1: beginning and the end are the same
    if(st.charAt(startIndex) == st.charAt(endIndex))
    return 2 + lpsLengthRecursive(st, startIndex+1, endIndex-1);
    // case 2: skip one element either from the beginning or the end
    int r1 = lpsLengthRecursive(startIndex+1, endIndex);
    int r2 = lpsLengthRecursive(startIndex, endIndex-1);
    return Math.max(r1, r2);
    Run time: ??
    

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24. Longest Palindromic Subsequence
    c != d
    cddpd
    ddpd cddp
    c != p
    d == d

    count = 2
    dp
    d != p
    p d
    p == p
    count = 1
    d == d
    count = 1
    2 + 1 = 3
    if(startIndex == endIndex)
    return 1;
    // case 1: beginning and the end are the same
    if(st.charAt(startIndex) == st.charAt(endIndex))
    return 2 + lpsLengthRecursive(st, startIndex+1, endIndex-1);
    // case 2: skip one element either from the beginning or the end
    int r1 = lpsLengthRecursive(startIndex+1, endIndex);
    int r2 = lpsLengthRecursive(startIndex, endIndex-1);
    return Math.max(r1, r2);
    Run time: 2 ^ n
    if(startIndex == endIndex)
    return 1;
    // case 1: beginning and the end are the same
    if(st.charAt(startIndex) == st.charAt(endIndex))
    return 2 + lpsLengthRecursive(st, startIndex+1, endIndex-1);
    // case 2: skip one element either from the beginning or the end
    int r1 = lpsLengthRecursive(startIndex+1, endIndex);
    int r2 = lpsLengthRecursive(startIndex, endIndex-1);
    return Math.max(r1, r2);
    

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25. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    c d d p d
    0 1 2 3 4
    ending index
    

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26. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 1
    

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27. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 1
    

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28. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    

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29. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    what substring is this?
    

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30. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “cd”
    

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31. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “cd”
    

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32. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “dd”
    

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33. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1 2
    1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “dd”
    

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34. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1 2
    1 1
    1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “dp”
    

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35. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 2
    “pd”
    

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36. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    length = 3
    

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37. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1
    1 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “cdd”
    max of “cd,”
    and “dd”
    

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38. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1 2
    1 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “cdd”
    max of “cd,”
    and “dd”
    

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39. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1 2
    1 2 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “ddp”
    max of “dd,”
    and “dp”
    

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40. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1 2
    1 2 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “dpd”
    max of “dp,”
    and “pd”
    

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41. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1 2
    1 2 2
    1 1
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “dpd”
    max of “dp,”
    and “pd”
    … 1?!?
    

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42. Given: cddpd

    Return: 3 (ddd or dpd)
    Longest Palindromic Subsequence
    1 1 2
    1 2 2
    1 1 3
    1 1
    1
    1 2 3 4
    0
    0
    1
    2
    3
    4
    starting index
    ending index
    c d d p d
    0 1 2 3 4
    “dpd”
    If start == end,
    2 + start[+1]end[-1]

    (“p”)
    

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43. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    

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44. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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45. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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46. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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47. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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48. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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49. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    start
    end
    

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50. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    

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51. // create dp[][] array
    dp[i][i] = 1; // every length 1 substring
    for (int startIndex = s.length() - 1; startIndex >= 0; startIndex--) {
    for (int endIndex = startIndex + 1; endIndex < s.length(); endIndex++) {
    // case 1: letter at the beginning and the end are the same
    if (s.charAt(startIndex) == s.charAt(endIndex)) {
    dp[startIndex][endIndex] = 2 + dp[startIndex + 1][endIndex - 1];
    } else {
    // case 2: skip one element either from the beginning or the end
    dp[startIndex][endIndex] = Math.max(
    dp[startIndex + 1][endIndex],
    dp[startIndex][endIndex - 1]);
    }
    }
    }
    return dp[0][st.length() - 1];
    

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52. What did we learn?
    

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53. - Solving it ‘naively’ first helps come up with DP
    approach
    - Recursive -> bottom up approach
    - Palindromic questions -> two pointers, one at start
    and one at end
    What did we learn?
    

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54. 2 problems
    30 minutes for each problem
    Pick the person that has their birthday coming up
    next!
    Mock Interviews
    

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55. How did it go??
    

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56. - DP problems are hard, practice is key

    - We didn’t cover DP + Binary Trees, would
    recommend spending some time practicing that
    - HackerRank
    - More practice next week!
    Before next session
    

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57. Journey of Caren’s career
    

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58. Interned at Dreamworks for two semesters

    

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59. Interned at Dreamworks for two semesters

    -> realized I didn’t like big companies
    

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60. Interned at Dreamworks 

    -> realized I didn’t like big companies
    Recruited a bunch senior year, failed a bunch of interviews
    

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61. Interned at Dreamworks 

    -> realized I didn’t like big companies
    Recruited a bunch senior year, failed a bunch of interviews
    Got a full time offer at a financial startup (75k) -> wasn’t really excited, but
    wanted a job
    

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62. Interned at Dreamworks 

    -> realized I didn’t like big companies
    Recruited a bunch senior year, failed a bunch of interviews
    Got a full time offer at a financial startup (75k) -> wasn’t really excited, but
    wanted a job
    Found an internship at a startup (7k / month)

    Converted to full time after 3 months (90k) -> got a raise after 6 months
    (100k)
    

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63. Interned at Dreamworks 

    -> realized I didn’t like big companies
    Recruited a bunch senior year, failed a bunch of interviews
    Got a full time offer at a financial startup (75k) -> wasn’t really excited, but
    wanted a job
    Found an internship at a startup (7k / month)

    Converted to full time after 3 months (90k) -> got a raise after 6 months
    (100k)
    One year later… Wanted to feel more challenged at my job -> took a
    CodePath Android class (8 weeks)
    

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64. Interned at Dreamworks 

    -> realized I didn’t like big companies
    Recruited a bunch senior year, failed a bunch of interviews
    Got a full time offer at a financial startup (75k) -> wasn’t really excited, but
    wanted a job
    Found an internship at a startup (7k / month)

    Converted to full time after 3 months (90k) -> got a raise after 6 months
    (100k)
    One year later… Wanted to feel more challenged at my job -> took a
    CodePath Android class (8 weeks)
    Recruited for Android jobs -> failed a bunch
    

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65. Got two full time offers at two different startups 

    - company A (110k)

    - company B (105k)

    - negotiated a bit: company A -> 110k, company B -> 115k + 15k sign
    on bonus
    

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66. Got two full time offers at two different startups 

    - company A (110k)

    - company B (105k)

    - negotiated a bit: company A -> 110k, company B -> 115k + 15k sign
    on bonus
    1.5 years later -> Felt like I wasn’t learning much anymore -> looked for
    jobs at hardware companies
    

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67. Got two full time offers at two different startups 

    - company A (110k)

    - company B (105k)

    - negotiated a bit: company A -> 110k, company B -> 115k + 15k sign
    on bonus
    1.5 years later -> Felt like I wasn’t learning much anymore -> looked for
    jobs at hardware companies
    2 years later -> Thought about starting to save more money -> Recruited
    at big companies
    

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