Help debug dynamic programming problem!

#	User	Rating
1	tourist	4009
2	jiangly	3831
3	Radewoosh	3646
4	jqdai0815	3620
4	Benq	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	gamegame	3386
10	ksun48	3373

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
6	adamant	157
8	TheScrasse	154
8	nor	154
10	Dominater069	153

Recently, I have encounter this dynamic programming problem and I can not debug where I have problem, I used forward DP style and I have checked the transition and base case. Please help me to look at it, I'm so frustrating. Thank in advance.

Problem link: leetcode

My code:

code

#define ll long long
const ll INF = 1e12;
class Solution {
public:
    long long minimumTotalDistance(vector<int>& robot, vector<vector<int>>& factory) {
        sort(factory.begin(), factory.end());
        sort(robot.begin(), robot.end());
        vector<int> fact;
        for (auto& f : factory)
            for (int i = 0; i < f[1]; i++) fact.push_back(f[0]);

        int n = robot.size();
        int m = fact.size();
        // dp[i][j]: min dist if only fix first (i - 1) robots and use first (j - 1) factories (1-based index)
        vector<vector<ll>> dp(n + 2, vector<ll>(m + 2, INF));
        for (int j = 1; j <= m - n + 1; j++) {
            dp[1][j] = 0;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (dp[i][j] != INF) {
                    if (m - j > n - i) dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]); // not pick
                    dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + abs(robot[i - 1] - fact[j - 1])); // pick
                }
            }
        }
        return dp[n + 1][m + 1];
    }
};

Updated fix my problem: my for loop only run until i = n, but the final dp state which is dp[n + 1][m + 1] can be reached from dp[n + 1][m] => My for loop should run till n + 1 so this relaxation can happen.

fixed

long long minimumTotalDistance(vector<int>& robot, vector<vector<int>>& factory) {
        sort(factory.begin(), factory.end());
        sort(robot.begin(), robot.end());
        vector<int> fact;
        for (auto& f : factory)
            for (int i = 0; i < f[1]; i++) fact.push_back(f[0]);

        int n = robot.size();
        int m = fact.size();
        // dp[i][j]: min dist if only fix first (i - 1) robots and use first (j - 1) factories (1-based index)
        vector<vector<ll>> dp(n + 2, vector<ll>(m + 2, INF));
        for (int j = 1; j <= m - n + 1; j++) {
            dp[1][j] = 0;
        }
        for (int i = 1; i <= n + 1; i++) {
            for (int j = 1; j <= m + 1; j++) {
                if (dp[i][j] != INF) {
                    if (m - j > n - i) dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
                    if (i - 1 < n && j - 1 < m) dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + abs(robot[i - 1] - fact[j - 1]));  
                }
            }
        }
        return dp[n + 1][m + 1];
    }

Comments (5)

Write comment?

frissky9

5 hours ago, # |

← Rev. 2 →

Everything looks great to me but the initialization of dp[1][j] = 0, it should be dp[0][j] = 0 and if you are iterating from (m — n +i > j) you should be returning dp[n][m] try these once.

Edited code

#define ll long long
const ll INF = 1e12;
class Solution {
public:
    long long minimumTotalDistance(vector<int>& robot, vector<vector<int>>& factory) {
        sort(factory.begin(), factory.end());
        sort(robot.begin(), robot.end());
        vector<int> fact;
        for (auto& f : factory)
            for (int i = 0; i < f[1]; i++) fact.push_back(f[0]);

        int n = robot.size();
        int m = fact.size();
        // dp[i][j]: min dist if only fix first (i - 1) robots and use first (j - 1) factories (1-based index)
        vector<vector<ll>> dp(n + 10, vector<ll>(m + 10
        , INF));
        for (int j = 0; j <= m; j++) {
            dp[0][j] = 0;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                dp[i][j] = min(dp[i][j], dp[i][j - 1]);
                dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + abs(robot[i - 1] - fact[j - 1]));
            }
        }
        return dp[n][m];
    }
};

→ Reply

lvisbl_

110 minutes ago, # ^ |

This is actually backward DP, I'm intended to use forward DP, but for this problem backward DP seem more intuitive. Thank you! Btw I have fixed forward DP following phantrongnghia510 suggestion.

phantrongnghia510

3 hours ago, # |

← Rev. 3 →

You should consider all valid costs in all final valid states of $$$i$$$: $$$min_j(dp[n + 1][j])$$$ for all $$$j\in [n + 1, m + 1]$$$ since we have just advertised the cost upwards but yet to finalize any optimal cost at $$$dp[i][m + 1]$$$

2 hours ago, # ^ |

Thank this is the expected answer, but how do you visualize the problem to get the optimal cost. I think I have "visited" all the states that can reach dp[n + 1][m + 1], and use all those states to "optimize gradually" dp[n + 1][m + 1], hence my expectation that when I reach dp[n + 1][m + 1], it should have been optimized in its optimal cost. But turn out, it is not.

114 minutes ago, # ^ |

Ah, I finally got it, the state dp[n + 1][m + 1] can be reached from dp[n + 1][m] but my for loop run until i = n then dp[n + 1][m + 1] never be optimized by dp[n + 1][m]. Thank you so much for pointing it out.

lvisbl_'s blog