Dynamic Programming

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Core of DP

mathematical induction

DP VS Recursion

recursion : big -> small
DP: small -> big

Longest increasing Subarray

Given an array of Integers A , find the length of Longest sub-array whose elements are in Non-Decreasing Order.

{9,7,2,4,1,5,8,9,5}

DP principle

Base case: M[0] = 1;
Induction rule: M[i] represents the length of the longest increasing subarray from A[0] to A[i].
- M[i] = M[i-1] + 1 if(A[i] > A[i-1])
- M[i] = 1;

Code

Public static int longestLength(int[] A){
    int length = 1;
    int maxLength = 0;
    for(int i = 1; i < A.length; i++){
        if(A[i] > A[i - 1]){
            length++;
            if(length > maxLength){
                maxLength = length;
            }
        }
        else{
            length = 1;
        }
    }
    return maxLength; 
}

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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1d DP solve pattern

1d original data
- if each smallest element is identical/similar

Linear scan and look back to the privous elements

Example

Longest increasing subarray

longest increasing subsequence

cut rope

Jump Game

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

For example:

A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

Index [0,1,2,3,4]

A = [2,3,1,1,4],

start goal

M = [x,x,x,x,T]

i j <--- scan direction

i + A[i]

Index [0,1,2,3,4]

A = [2,3,1,1,4],

start goal

M = [x,x,x,x,T]

i j <--- scan direction

i + A[i]

Base case: M[n-1] = true

Induction rule:

M[i] =

case 1: if(i + A[i] >= n-1) M[i] = true;

case 2: if(M[j] == true && j < i + A[i]) M[i] = true;

Index [0,1,2,3,4]

A = [2,3,1,1,4],

start goal

M = [F,F,F,F,T]

i j <--- scan direction

i + A[i]

Base case: M[n-1] = true

Induction rule:

M[i] =

case 1: if(i + A[i] >= n-1) M[i] = true;

case 2: if(M[j] == true && j < i + A[i]) M[i] = true;

    //DP
    bool canJump1(vector<int>& nums) {
        int n = (int)nums.size();
        vector<bool> dp(n,false);
        dp[n-1] = true;
        for(int i= n-2;i>=0;i--){
            dp[i] = false;
            for(int j=1;j<nums[i];j++){
                if(i+j>=n-1 || dp[i+j]){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[0];
    }

    //greedy
    bool canJump(vector<int>& nums) {
        int n = (int)nums.size();
        int res = nums[0];
        for(int i = 1; i < n; i++){
            --res;
            if(res<0) return false;
            res =max(res, nums[i]);
        }
        return true;
    }

Jump Game 2

same setup as Jump Game,

Return the minimum number of jumps needed to reach the end.

same setup as Jump Game,

Return the minimum number of jumps needed to reach the end.

1, base Case: M[n - 1] = 0

2 Induction Rule: M[i] ?????

same setup as Jump Game,

Return the minimum number of jumps needed to reach the end.

1, base Case: M[n - 1] = 0

2 Induction Rule: M[i] ?????

M[i] represent the least number of jumps to reach M[n - 1]

M[i] = 1 + min(M[j])

* j is all the elements that can be reached by 1 jump from i;

public static int minJumps(int A[]) {
    int n = A.length;
    int[] jumps = new int[n];
    int min = Integer.MAX_VALUE;
    jump[n-1] = 0;

    for(int i = n-2; i >= 0; i--){
        if(A[i] == 0){
            jumps[i] = Integer.MAX_VALUE;
        }
        else if(i + A[i] >= n - 1){ // jump over border
            jumps[i] = 1;
        }
        else{
            min = Integer.MAX_VALUE;
            for(int j = i + 1; j < n && j <= A[i] + i; j++){
                if(min > jumps[j]){
                    min = jumps[j];
                }
            }
            if(min != Integer.MAX_VALUE) {
                jumps[i] = min + 1;
            }
            else{
                jumps[i] = Integer.MAX_VALUE;
            }
        }
    } 
    return jumps[0];
}

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

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Example:

s1 = "asdf"

s2 = "hjkl"

s1 = c1 and s1r <- rest of s1

s2 = c2 and s2r <- rest of s2

1, replace: a -> h

h sdf

h jkl

editDistance(sdf, jkl) + 1

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Example:

s1 = "asdf"

s2 = "hjkl"

s1 = c1 and s1r <- rest of s1

s2 = c2 and s2r <- rest of s2

2, delete: a

a sdf

hjkl

editDistance(sdf, hjkl) + 1

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Example:

s1 = "asdf"

s2 = "hjkl"

s1 = c1 and s1r <- rest of s1

s2 = c2 and s2r <- rest of s2

3, insert: h

h asdf

h jkl

editDistance(asdf, jkl) + 1

x	x	x	x	x	x	x
x	x	x	x	x	x	x
x	x	x	x	x	x	x
x	x	x	x	x	x	x
x	x	x	x	x	x	x
x	x	x	x	x	x	x
x	x	x	x	x	x	x

M[i][j]

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]?

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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]?

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

edit Distance

Longest common Subsequence

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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

Core of DP

mathematical induction

DP VS Recursion

recursion : big -> small

DP: small -> big

Longest increasing Subarray

DP principle

Code

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

1d DP solve pattern

1d original data

Example

Longest increasing subarray

Jump Game

Jump Game 2

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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