Help with a Leetcode problem

Revision en1, by jobin491, 2025-05-02 11:57:12

Problem

You have n tasks and m workers. Each task has a strength requirement stored in a 0-indexed integer array tasks, with the ith task requiring tasks[i] strength to complete. The strength of each worker is stored in a 0-indexed integer array workers, with the jth worker having workers[j] strength. Each worker can only be assigned to a single task and must have a strength greater than or equal to the task's strength requirement (i.e., workers[j] >= tasks[i]). Additionally, you have pills magical pills that will increase a worker's strength by strength. You can decide which workers receive the magical pills, however, you may only give each worker at most one magical pill. Given the 0-indexed integer arrays tasks and workers and the integers pills and strength, return the maximum number of tasks that can be completed.

below is the code that i tried with a complexity of N(logN)(logN). N=max(n,m); approach: binary search on answer, check the mid smallest tasks: start from the largest task and worker if(task<worker) this worker completes task without pill. if(task>worker) we check for the smallest worker who can complete this task with a pill.

class Solution {
public:
    int maxTaskAssign(vector<int>& tasks, vector<int>& workers, int p, int strength) {
        int ans = 0;
        int n=tasks.size();
        int m=workers.size();
        sort(workers.rbegin(),workers.rend());
        sort(tasks.rbegin(),tasks.rend());
        int j=0;
        int hi=m;
        int lo=0;
        int fin=0;
        int pills=p;
        while(lo<=hi){
            int mid= lo+hi;mid/=2;
            // check the mid smallest tasks;
            pills=p;
            if(mid>n){
                hi=mid-1;
                continue;
            }
            vector<bool> t(m+1,false);
            j=0;
            ans=0;
            for(int i=n-mid;i<n;i++){
                while(j<=m && t[j]) j++;
                if(j>=m) break;
                if(tasks[i]<=workers[j]){
                    t[j]=true;
                    ans++;
                }else{
                    if(strength+workers[j]<tasks[i]){
                        continue;
                    }
                    if(pills>0){
                        int req= tasks[i]-strength;
                        int llo=j,lhi=m-1;
                        int ind=j;
                        while(llo<=lhi){
                            int lmid= (llo+lhi)/2;
                            if(workers[lmid]<req){
                                lhi=lmid-1;
                            }else{
                                ind=lmid;
                                llo=lmid+1;
                            }
                        }
                        while(ind>0 && t[ind]) ind--;
                        t[ind]=true;
                        ans++;
                        pills--;
                    }
                }
            }
            if(ans>=mid){
                lo=mid+1;
                fin=mid;
            }else{
                hi=mid-1;
            }
        }
        return fin;
    }
};

this gives a right answer on all testcases but exceeds time limit. when i checked the correct solution it was with the same complexity using mulitsets

class Solution {
public:
    int maxTaskAssign(vector<int>& tasks, vector<int>& workers, int p, int strength) {
        int ans = 0;
        int n=tasks.size();
        int m=workers.size();
        sort(workers.rbegin(),workers.rend());
        sort(tasks.rbegin(),tasks.rend());
        int j=0;
        int hi=m;
        int lo=0;
        int fin=0;
        int pills=p;
        while(lo<=hi){
            int mid= lo+hi;mid/=2;
            // check the mid smallest tasks;
            pills=p;
            if(mid>n){
                hi=mid-1;
                continue;
            }
            vector<bool> t(m+1,false);
            j=0;
            ans=0;
            for(int i=n-mid;i<n;i++){
                while(j<=m && t[j]) j++;
                if(j>=m) break;
                if(tasks[i]<=workers[j]){
                    t[j]=true;
                    ans++;
                }else{
                    if(strength+workers[j]<tasks[i]){
                        continue;
                    }
                    if(pills>0){
                        int req= tasks[i]-strength;
                        int llo=j,lhi=m-1;
                        int ind=j;
                        while(llo<=lhi){
                            int lmid= (llo+lhi)/2;
                            if(workers[lmid]<req){
                                lhi=lmid-1;
                            }else{
                                ind=lmid;
                                llo=lmid+1;
                            }
                        }
                        while(ind>0 && t[ind]) ind--;
                        t[ind]=true;
                        ans++;
                        pills--;
                    }
                }
            }
            if(ans>=mid){
                lo=mid+1;
                fin=mid;
            }else{
                hi=mid-1;
            }
        }
        return fin;
    }
};

I am unable to figure out the reason for failure in my code if anyone could help me with it I'd really appreciate it.

Tags leetcode, time exceeded

History

 
 
 
 
Revisions
 
 
  Rev. Lang. By When Δ Comment
en2 English jobin491 2025-05-02 12:00:28 1243
en1 English jobin491 2025-05-02 11:57:12 5578 Initial revision (published)