Approach using binary search and Maximum Prefix array with proper explaination.
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem involves assigning jobs to workers to maximize total profit, given constrai
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves assigning jobs to workers to maximize total profit, given constraints on job difficulty and worker capability. Here\u2019s the intuition and step-by-step approach:\n\n1) Job-Worker Mapping:\n\nEach worker can do a job only if their ability is at least the difficulty of the job.\nTo maximize profit, we should assign each worker the highest paying job they can do.\n\n2) Sorting Jobs:\n\nSort jobs by difficulty. For jobs with the same difficulty, keep the one with the higher profit.\n\n3) Prefix Maximum Array:\n\nCreate a prefix array where each position holds the maximum profit achievable up to that job difficulty. This ensures that for any given difficulty, we know the best possible profit without rechecking all jobs.\n\n4) Binary Search for Efficiency:\n\nFor each worker, use binary search to find the highest difficulty job they can do, and fetch the corresponding profit using the prefix array.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nStep-by-Step Approach\n\n1) Combine Difficulty and Profit:\n\nCreate pairs of (difficulty, profit) for each job.\n\n\n2) Sort Jobs:\n\nSort the jobs first by difficulty. If two jobs have the same difficulty, sort by profit in descending order to maximize profit.\n\n3) Create Prefix Array:\n\nTraverse the sorted job list and update the profit to ensure it is the maximum profit up to that difficulty.\n\n4) Binary Search for Each Worker:\n\nFor each worker, use binary search on the sorted job list to find the maximum profit they can achieve given their ability.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nTime Complexity:\n\nSorting jobs: O(nlogn), where n is the number of jobs.\nBinary search for each worker: O(mlogn), where m is the number of workers.\nTotal: O(nlogn+mlogn).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nStoring job pairs and prefix array: O(n)\n\n# Code\n```\nclass Solution {\npublic:\n\n    int find(vector<pair<int,int>> &A,int d)\n    {\n        int l=0,h=A.size()-1,mid=(l+h)/2;\n        while(l<=h)\n        {   \n            mid=(l+h)/2;\n            if(A[mid].first==d)\n                return A[mid].second;\n            else if(A[mid].first<d)\n                l=mid+1;\n            else\n                h=mid-1;\n        }\n        if(h<0)\n            return 0;\n        return A[h].second;\n    }\n\n    int maxProfitAssignment(vector<int>& D, vector<int>& P, vector<int>& W) {\n        \n        int n=D.size();\n        vector<pair<int,int>> p;\n        for(int i=0;i<n;i++)\n        {\n            p.push_back({D[i],P[i]});\n        }\n        sort(p.begin(),p.end(),[](const pair<int,int> &a,const pair<int,int> &b){\n\n            if(a.first < b.first)\n                return true;\n            else if(a.first > b.first)\n                return false;\n            else\n                return (a.second > b.second);\n\n        });\n        \n        \n        for(int i=1;i<n;i++)\n        {\n            p[i].second=max(p[i].second,p[i-1].second);\n        }\n        \n        int ans=0;\n        for(int i=0;i<W.size();i++)\n        {   \n            ans+=find(p,W[i]);\n        }\n        return ans;\n    }\n};\n```
['Binary Search', 'C++']