Dynamic Programming

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What is Dynamic Programming?

A very powerful and general tool for solving certain optimization problems

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Dynamic Programming

VS Recursion

Both are using sub-problems to solve the main problem
Recursion is much more general
If it can be solved by dynamic programming, DP is much faster
Usually DP requires more space
SPACE VS TIME

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When we use DP?

It can be resolved by recursion and not tail recursion
Recursion is making some redundant computation
It has to have optimal sub-structures:
optimal solutions to the original problem contains optimal solutions to sub-problems

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When we use DP?

It can be resolved by recursion and not tail recursion
Recursion is making some redundant computation

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It means:

1. If it cannot be solved by recursion, do not bother finding DP.

2. not all the recursion can use DP to solve. E.G. merge sort

What is Tail Recursion

The recursion call happens at last
It can always be rewritten into a loop

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int A() {
    ...
    ...
    A;
}

int Factorial(int n) {
    if(n == 1) return 1;
    return n * Factorial(n-1);
}

int Factorial(int n) {
    int sum = 1;
    for(int i = 1; i <= n; i ++) {
        sum = sum * i;
    }
    return sum;
}

How to use DP?

Think about the recursion solution
Find sub problems that have been computed multiple times
See if the sub problems are optimal sub-structure
Try to memorize them to avoid the redundant computing

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Fibonacci Numbers

int Fabonacci(int n) {
    if(n == 0 || n == 1) return 1;
    return F(n-1) + F(n-2);
}

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F(6)

F(5)

F(4)

F(3)

F(2)

F(1)

F(0)

F(1)

F(0)

F(1)

F(0)

F(1)

F(0)

F(2)

F(1)

F(0)

F(2)

Need some space to avoid the computation

int Fabonacci(int n) {
    if(n <= 1) return 1;
    int[] result = new int[n + 1];
    result[0] = 1;
    result[1] = 1;
    for(int i = 2; i < n + 1; i ++) {
        result[i] = result[i-1] + result[i-2];
    }
    return result[n];
}

We use an Array to store the temp result

What is the optimal sub-structure?

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Unique Paths

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Unique Paths

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int uniquePaths(int m, int n) {
      if(m==1 || n==1) return 1;
      return uniquePaths(m-1, n) + uniquePaths(m, n-1);
}

Where is the redundancy?

What is the optimal sub-structure?

Unique Paths

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Optimal Sub-structure

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

Unique Paths

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public int uniquePaths(int m, int n) {  
    int[][] a = new int[m][n];  
    for (int i = 0; i < m; i++) {  
        a[i][0] = 1;  
    }  
    for (int i = 0; i < n; i++) {  
        a[0][i] = 1;  
    }  
    for (int i = 1; i < m; i++) {  
        for (int j = 1; j < n; j++) {  
            a[i][j] = a[i-1][j] + a[i][j-1];  
        }  
    }  
    return a[m-1][n-1];  
}

Minimum Path Sum

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1	3	4	2
3	5	2	3
2	1	2	3
2	2	4	2

Best optimal sub-structure?

Minimum Path Sum

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1	3	4	2
3	5	2	3
2	1	2	3
2	2	4	2

PathSum(m,n) = MIN(PathSum(m,n-1),PathSum(m-1,n)) + matrix(m,n)

Minimum Path Sum

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public int minPathSum(int[][] grid) {
    if(grid == null || grid.length==0)
        return 0;
    int m = grid.length;
    int n = grid[0].length;
    int[][] dp = new int[m][n];
    dp[0][0] = grid[0][0];    
    for(int i=1; i<n; i++){
        dp[0][i] = dp[0][i-1] + grid[0][i];
    }
    for(int j=1; j<m; j++){
        dp[j][0] = dp[j-1][0] + grid[j][0];
    }
    for(int i=1; i<m; i++){
        for(int j=1; j<n; j++){
            if(dp[i-1][j] > dp[i][j-1]){
                dp[i][j] = dp[i][j-1] + grid[i][j];
            }else{
                dp[i][j] = dp[i-1][j] + grid[i][j];
            }
        }
    }
    return dp[m-1][n-1];
}

Minimum Path Sum

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What is time complexity?

Can we do better?

What is space complexity?

Can we do better?

Any improvement?

Minimum Path Sum

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Any improvement?

We can use less space to get the same result without hurting time complexity

Minimum Path Sum II

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public static int minPathSum(int[][] grid) {
    if(grid == null || grid.length==0)
        return 0;
    int m = grid.length;
    int n = grid[0].length;
     int[] newline = new int[n];
     int[] oldline = new int[n];
     oldline[0] = grid[0][0];    
     for(int i=1; i<n; i++){
    	 oldline[i] = oldline[i-1] + grid[0][i];
     }
     for(int i=1; i<m; i++){
    	 newline[0] = grid[i][0] + oldline[0];
         for(int j=1; j<n; j++){
             if(oldline[j] > newline[j-1]){
                 newline[j] = newline[j-1] + grid[i][j];
             }else{
                 newline[j] = oldline[j] + grid[i][j];
             }
         }
         oldline = newline;
     }
     return newline[n-1];
}

0-1 Knapsack

Given a knapsack which can hold w pounds of items, and a set of items with weight w1, w2, ... wn. Each item has its value s1,s2,...,sn. Try to select the items that could put in knapsack and contains most value.

What is the optimal sub-structure?

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0-1 Knapsack

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w[i][j]: for the previous total i items, the max value it can have for capacity j

Which two we need to use to compare?

w[i][j]: for the previous total i items, the max value it can have for capacity j

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When you iterate i, and j, you need to try:

If the new i could be added into j
if it could and added, could it be better

	1	2	3	4	5	6	7
0	0	0	0	0	0	0	0
1	3	3	3	3	3	3	3
2	3	3	8	11	11	11	11
3	3	3	8	11	11	11	12
4	3	3	8	11	11	11	12

w[i][j]: for the previous total i items, the max value it can have for capacity j

Example: weights{1,3,4,5} values{3,8,4,7}

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public int knapsack(int capacity, int[] weights, int[] values) {
	int length = weights.length;
	if (capacity == 0 || length == 0)
		return 0;
	int[][] w = new int[length + 1][capacity + 1];
	for (int i = 1; i <= length; i++) {
		int index = i - 1;
		for (int j = 1; j <= capacity; j++) {
			if (j < weights[index]) {
				w[i][j] = w[i - 1][j];
			} else if (w[i - 1][j - weights[index]] + values[index] > w[i - 1][j]) {
				w[i][j] = w[i - 1][j - weights[index]] + values[index];
			} else {
				w[i][j] = w[i - 1][j];
			}
		}
	}
	return w[length][capacity];
}

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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];
}

Matrix Multiplication

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Matrix A m*n

Matrix B n*p

C = A * B will need m*n*p times multiplication

A 100 * 10, B 10 * 100, C 100 * 5

D = A * B * C

If we do (AB)C, need 100*10*100 + 100*100*5 = 150000 times

If we do A(BC), need 100*10*5 + 10*100*5 = 10000 times

Matrix Multiplication

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Give A0,A1,.....,An-1 n different matrixs

Find minimum of multiplication it needs to get the result

input: An array P with n+1 numbers

A0 = p0*p1, An-1 = Pn-1*Pn

Matrix Multiplication

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What is the optimal sub-structure?

For the result A1*...*An, if we split from Ak

It becomes (A1*...Ak)(Ak+1*...An)

T(1,n) = T(1,k) + T(k+1,n) + p0*pk*pn

So if T(1,n) is the best, T(1,k) and T(k+1,n) must be the best

Matrix Multiplication

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So how do we get the T(1,k)?

We need to start from the chain with length 1 to finally get to length n

public static int MatrixChain(int[] p)  
{  
    int n = p.length;
    n --;
    int[][] m = new int[n][n];
    for(int i = 0; i < n; i++)  
        m[i][i] = 0; 
    for(int r = 2; r <= n; r++)  
    { 
        for(int i = 0; i < n - r + 1; i ++)  
        { 
            int j = i + r - 1; 
            m[i][j] = m[i + 1][j]  + p[i] * p[i+1] * p[j + 1];  
            for(int k = i + 1; k < j; k++)  
            {  
                int t = m[i][k] + m[k + 1][j] + p[i] * p[k+1] * p[j+1];  
                if( t < m[i][j])  
                    m[i][j] = t;  
            }  
        }  
    }
    return m[0][n-1];
}

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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