动态规划学习指南

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Author：aatalyk
Origin：Link

Before starting the topic let me introduce myself. I am a Mobile Developer currently working in Warsaw and spending my free time for interview preparations. I started to prepare for interviews two years ago. At that time I should say I could not solve the two sum problem. Easy problems seemed to me like hard ones so most of the time I had to look at editorials and discuss section. Currently, I have solved ~800 problems and time to time participate in contests. I usually solve 3 problems in a contest and sometimes 4 problems. Ok, lets come back to the topic.

Recently I have concentrated my attention on Dynamic Programming cause its one of the hardest topics in an interview prep. After solving ~140 problems in DP I have noticed that there are few patterns that can be found in different problems. So I did a research on that and find the following topics. I will not give complete ways how to solve problems but these patterns may be helpful in solving DP.

Patterns

  1. Minimum (Maximum) Path to Reach a Target
  2. Distinct Ways
  3. Merging Intervals
  4. DP on Strings
  5. Decision Making

1.0 Minimum (Maximum) Path to Reach a Target

最大最小的条件路径

Generate problem statement for this pattern

Statement

Given a target find minimum (maximum) cost / path / sum to reach the target.

Approach

Choose minimum (maximum) path among all possible paths before the current state, then add value for the current state.

routes[i] = min(routes[i-1], routes[i-2], ... , routes[i-k]) + cost[i]

Generate optimal solutions for all values in the target and return the value for the target.

for (int i = 1; i <= target; ++i) {
   for (int j = 0; j < ways.size(); ++j) {
       if (ways[j] <= i) {
           dp[i] = min(dp[i], dp[i - ways[j]]) + cost / path / sum;
       }
   }
}
 
return dp[target]

Similar Problems

746. Min Cost Climbing Stairs Easy

for (int i = 2; i <= n; ++i) {
   dp[i] = min(dp[i-1], dp[i-2]) + (i == n ? 0 : cost[i]);
}
 
return dp[n]

64. Minimum Path Sum Medium

    for (int i = 1; i < n; ++i) {
   for (int j = 1; j < m; ++j) {
       grid[i][j] = min(grid[i-1][j], grid[i][j-1]) + grid[i][j];
   }
}
 
return grid[n-1][m-1]

322. Coin Change Medium

for (int j = 1; j <= amount; ++j) {
   for (int i = 0; i < coins.size(); ++i) {
       if (coins[i] <= j) {
           dp[j] = min(dp[j], dp[j - coins[i]] + 1);
       }
   }
}

931. Minimum Falling Path Sum Medium
983. Minimum Cost For Tickets Medium
650. 2 Keys Keyboard Medium
279. Perfect Squares Medium
1049. Last Stone Weight II Medium
120. Triangle Medium
474. Ones and Zeroes Medium
221. Maximal Square Medium
322. Coin Change Medium
1240. Tiling a Rectangle with the Fewest Squares Hard
174. Dungeon Game Hard
871. Minimum Number of Refueling Stops Hard

2.0 Distinct Ways

达到目标的不同方式总数

Generate problem statement for this pattern

Statement，问题描述

Given a target find a number of distinct ways to reach the target.

Approach，解题方法

Sum all possible ways to reach the current state.

routes[i] = routes[i-1] + routes[i-2], ... , + routes[i-k]

Generate sum for all values in the target and return the value for the target.

for (int i = 1; i <= target; ++i) {
   for (int j = 0; j < ways.size(); ++j) {
       if (ways[j] <= i) {
           dp[i] += dp[i - ways[j]];
       }
   }
}
 
return dp[target]

Similar Problems

70. Climbing Stairs easy

for (int stair = 2; stair <= n; ++stair) {
   for (int step = 1; step <= 2; ++step) {
       dp[stair] += dp[stair-step];   
   }
}

62. Unique Paths Medium

for (int i = 1; i < m; ++i) {
   for (int j = 1; j < n; ++j) {
       dp[i][j] = dp[i][j-1] + dp[i-1][j];
   }
}

1155. Number of Dice Rolls With Target Sum Medium

for (int rep = 1; rep <= d; ++rep) {
   vector<int> new_ways(target+1);
   for (int already = 0; already <= target; ++already) {
       for (int pipe = 1; pipe <= f; ++pipe) {
           if (already - pipe >= 0) {
               new_ways[already] += ways[already - pipe];
               new_ways[already] %= mod;
           }
       }
   }
   ways = new_ways;
}

Note，备注
Some questions point out the number of repetitions, in that case, add one more loop to simulate every repetition.
688. Knight Probability in Chessboard Medium
494. Target Sum Medium
377. Combination Sum IV Medium
935. Knight Dialer Medium
1223. Dice Roll Simulation Medium
416. Partition Equal Subset Sum Medium
808. Soup Servings Medium
790. Domino and Tromino Tiling Medium
801. Minimum Swaps To Make Sequences Increasing
673. Number of Longest Increasing Subsequence Medium
63. Unique Paths II Medium
576. Out of Boundary Paths Medium
1269. Number of Ways to Stay in the Same Place After Some Steps Hard
1220. Count Vowels Permutation Hard

3.0 Merging Intervals

区间合并

Generate problem statement for this pattern

Statement，问题描述

Given a set of numbers find an optimal solution for a problem considering the current number and the best you can get from the left and right sides.

Approach，解法方法

Find all optimal solutions for every interval and return the best possible answer.

// from i to j
dp[i][j] = dp[i][k] + result[k] + dp[k+1][j]

Get the best from the left and right sides and add a solution for the current position.

for(int l = 1; l<n; l++) {
   for(int i = 0; i<n-l; i++) {
       int j = i+l;
       for(int k = i; k<j; k++) {
           dp[i][j] = max(dp[i][j], dp[i][k] + result[k] + dp[k+1][j]);
       }
   }
}
 
return dp[0][n-1]

Similar Problems，类似问题

1130. Minimum Cost Tree From Leaf Values Medium

for (int l = 1; l < n; ++l) {
   for (int i = 0; i < n - l; ++i) {
       int j = i + l;
       dp[i][j] = INT_MAX;
       for (int k = i; k < j; ++k) {
           dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + maxs[i][k] * maxs[k+1][j]);
       }
   }
}

96. Unique Binary Search Trees Medium
1039. Minimum Score Triangulation of Polygon Medium
546. Remove Boxes Medium
1000. Minimum Cost to Merge Stones Medium
312. Burst Balloons Hard
375. Guess Number Higher or Lower II Medium

4.0 DP on Strings

字符串上的DP

General problem statement for this pattern can vary but most of the time you are given two strings where lengths of those strings are not big

Statement，问题描述

Given two strings s1 and s2, return some result.

Approach，解决方法

Most of the problems on this pattern requires a solution that can be accepted in O(n^2) complexity.

// i - indexing string s1
// j - indexing string s2
for (int i = 1; i <= n; ++i) {
   for (int j = 1; j <= m; ++j) {
       if (s1[i-1] == s2[j-1]) {
           dp[i][j] = /*code*/;
       } else {
           dp[i][j] = /*code*/;
       }
   }
}

If you are given one string s the approach may little vary

for (int l = 1; l < n; ++l) {
   for (int i = 0; i < n-l; ++i) {
       int j = i + l;
       if (s[i] == s[j]) {
           dp[i][j] = /*code*/;
       } else {
           dp[i][j] = /*code*/;
       }
   }
}

1143. Longest Common Subsequence Medium

for (int i = 1; i <= n; ++i) {
   for (int j = 1; j <= m; ++j) {
       if (text1[i-1] == text2[j-1]) {
           dp[i][j] = dp[i-1][j-1] + 1;
       } else {
           dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
       }
   }
}

647. Palindromic Substrings Medium

for (int l = 1; l < n; ++l) {
   for (int i = 0; i < n-l; ++i) {
       int j = i + l;
       if (s[i] == s[j] && dp[i+1][j-1] == j-i-1) {
           dp[i][j] = dp[i+1][j-1] + 2;
       } else {
           dp[i][j] = 0;
       }
   }
}

516. Longest Palindromic Subsequence Medium
1092. Shortest Common Supersequence Medium
72. Edit Distance Hard
115. Distinct Subsequences Hard
712. Minimum ASCII Delete Sum for Two Strings Medium
5. Longest Palindromic Substring Medium

5.0 Decision Making

决策，取不取当前元素

The general problem statement for this pattern is forgiven situation decide whether to use or not to use the current state. So, the problem requires you to make a decision at a current state.

Statement，问题描述

Given a set of values find an answer with an option to choose or ignore the current value.

Approach，解决方法

If you decide to choose the current value use the previous result where the value was ignored; vice-versa, if you decide to ignore the current value use previous result where value was used.

// i - indexing a set of values
// j - options to ignore j values
for (int i = 1; i < n; ++i) {
   for (int j = 1; j <= k; ++j) {
       dp[i][j] = max({dp[i][j], dp[i-1][j] + arr[i], dp[i-1][j-1]});
       dp[i][j-1] = max({dp[i][j-1], dp[i-1][j-1] + arr[i], arr[i]});
   }
}

198. House Robber Easy

for (int i = 1; i < n; ++i) {
   dp[i][1] = max(dp[i-1][0] + nums[i], dp[i-1][1]);
   dp[i][0] = dp[i-1][1];
}

121. Best Time to Buy and Sell Stock Easy
714. Best Time to Buy and Sell Stock with Transaction Fee Medium
309. Best Time to Buy and Sell Stock with Cooldown Medium
123. Best Time to Buy and Sell Stock III Hard
188. Best Time to Buy and Sell Stock IV Hard

I hope these tips will be helpful.