Dynamic Programming

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Longest Increasing Subsequence

3, 1, 4, 5, 7, 6, 8, 2

1, 4, 5, 6, 8 (Or 1, 4, 5, 7, 8)

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Longest Increasing Subsequence

What is the optimal sub-structure?

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Longest Increasing Subsequence

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We store lis[i] for the LIS by i?

We store lis[i] for the LIS using sequence[i]

Longest Increasing Subsequence

public int longestIncreasingSubsequence(int[] nums) {
    if(nums.length == 0){
        return 0;
    }
    int[] lis = new int[nums.length];
    int max = 0;
    for (int i = 0; i < nums.length; i++){
        int localMax = 0;
        for (int j = 0; j < i; j++){
            if (lis[j] > localMax && nums[j] <= nums[i]){
                localMax = lis[j];
            }
        }
        lis[i] = localMax + 1;
        max = Math.max(max, lis[i]);
    }
    return max;
}

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What is the Time Complexity?

What is the Space Complexity?

Can we do better?

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Patient Sort

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1, 3, 5, 2, 8, 4, 7, 6, 0, 9, 10

1 -> 0

1,3 -> 1,2

1,3,5 -> 1,3,4

1,3,5,8 -> 1,3,5,7 -> 1,3,5,6

1,3,5,6,9

1,3,5,6,9,10

public int longestIncreasingSubsequence(int[] nums) {
    if(nums.length == 0){
        return 0;
    }
    int len = 0;
    int[] tails = new int[nums.length];
    tails[0] = nums[0];
    for(int i = 1; i < nums.length; i++){
        if(nums[i] < tails[0]){
            tails[0] = nums[i];
        } else if (nums[i] >= tails[len]){
            tails[++len] = nums[i];
        } else {
            tails[binarySearch(tails, 0, len, nums[i])] = nums[i];
        }
    }
    return len + 1;
}
private int binarySearch(int[] tails, int min, int max, int target){
    while(min < max){
        int mid = min + (max - min) / 2;
        if(tails[mid] == target){
            return mid;
        }
        else if(tails[mid] < target){
            min = mid + 1;
        }
        else max = mid;
    }
    return min;
}

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Longest Common Sequence

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Longest Common Sequence

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Example: abcfbc abfcab

return 4 (abcb)

Longest Common Sequence

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Example: abcfbc abfcab

return 4 (abcb)

Longest Common Sequence

What is the optimal sub-structure?

maxCommon(i,j): longest common string for String A(0,i) and String B(0,j)

We finally need to get maxCommon(stringA.length, stringB.length)

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Longest Common Sequence

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What is the relationship between maxCommon(i,j) and maxCommon(i-1,j-1)?

If(A[i-1] = B[j-1]) ?

If(A[i-1] != B[j-1])?

Longest Common Sequence

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What is the relationship between maxCommon(i,j) and maxCommon(i-1,j-1)?

If(A[i-1] = B[j-1]) ?

If(A[i-1] != B[j-1])?

maxCommon(i,j) = maxCommon(i-1,j-1) + 1

maxCommon(i,j) = max(maxCommon(i-1,j), maxCommon(i,j-1))

Longest Common Sequence

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public static int longestCommonString(String a, String b) {
	int m = a.length();
	int n = b.length();
	int[][] maxCommon = new int[m+1][n+1];
	for(int i = 0; i <= m; i ++) {
		maxCommon[i][0] = 0;
	}
	for(int j = 0; j <= n; j ++) {
		maxCommon[0][j] = 0;
	}
	for(int i = 1; i <= m; i ++) {
		for(int j = 1; j <= n; j ++) {
		  if(a.charAt(i-1) == b.charAt(j-1)) {
		    maxCommon[i][j] = maxCommon[i-1][j-1] + 1;
		  }
		  else {
		    maxCommon[i][j] = Math.max(maxCommon[i][j-1], maxCommon[i-1][j]);
		  }
		}
	}
	return maxCommon[m][n];
}

Edit Distance

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Replace: abc -> abd

Remove: abc -> ab

Add: abc -> abcd

How many steps you need from String A to B?

e.g. abca -> eeba 3 steps

abcdf -> eecf 3 steps

Edit Distance

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What is the optimal sub-structure?

what is the connection between EDIT[i,j]

and EDIT[i-1,j], EDIT[i,j-1] and EDIT[i-1,j-1]?

Edit Distance

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what is the connection between EDIT[i,j]

and EDIT[i-1,j], EDIT[i,j-1] and EDIT[i-1,j-1]?

if(A[i-1]!=B[j-1])?

if(A[i-1]=B[j-1])?

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public static int editDistance(String a, String b) {
    int m = a.length() + 1; 
    int n = b.length() + 1;
    int[][] f = new int[m][n];
    for(int i = 0; i < m; i ++) {
        f[i][0] = i;
    }
    for(int j = 0; j < n; j ++) {
        f[0][j] = j;
    }
	
    for(int i = 1; i < m; i ++) {
        for(int j = 1; j < n; j ++) {
            if(a.charAt(i-1) == b.charAt(j-1)) {
                f[i][j] = f[i-1][j-1];
            }
            else {
                f[i][j] = f[i-1][j-1] + 1;
            }
            f[i][j] = Math.min(f[i][j], Math.min(f[i-1][j]+1, f[i][j-1]+1));
        }
    }
    return f[m-1][n-1];
}

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Homework

Unique Path II

Climbing Stairs

Maximum Subarray

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Homework (Optional)

Longest Valid Parentheses

Triangle

Best Time to Buy and Sell Stock

Distinct Subsequences

[GoValley-201612] Dynamic Programming 2

By govalley201612

[GoValley-201612] Dynamic Programming 2

Dynamic Programming

Longest Increasing Subsequence

Longest Increasing Subsequence

Longest Increasing Subsequence

Longest Increasing Subsequence

What is the Time Complexity?

What is the Space Complexity?

Can we do better?

Patient Sort

Longest Common Sequence

Longest Common Sequence

Longest Common Sequence

Longest Common Sequence

Longest Common Sequence

Longest Common Sequence

Longest Common Sequence

Edit Distance

Edit Distance

Edit Distance

Homework

Homework (Optional)

[GoValley-201612] Dynamic Programming 2

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