easiest approach
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nThe approach taken in t
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach taken in this code to solve the problem of assigning the maximum possible profit to workers based on their abilities and the available jobs can be summarized as follows:\n\n# **Initialization and Finding Maximum Difficulty:**\n\nInitialize n to the length of the difficulty array.\nIterate through the difficulty array to find the maximum difficulty level, n. This ensures n is the maximum of its initial value and the maximum difficulty in the array.\n\n# **Creating and Populating the Profit Array:**\n\nCreate an array arr of length n+1 initialized to zero. This array will store the maximum profit for each difficulty level up to n.\nIterate through the profit array and update arr at the index corresponding to each difficulty. This step ensures that arr[difficulty[i]] contains the highest profit available for that difficulty.\n\n# **Propagating Maximum Profits:**\n\nIterate through the arr array from index 1 to n and update each value to be the maximum of itself and the previous value. This step ensures that arr at any index j contains the maximum profit that can be obtained for any job difficulty less than or equal to j.\n\n# **Calculating Total Maximum Profit:**\n\nInitialize ans to 0 to store the total maximum profit.\nIterate through the worker array and for each worker, add the maximum profit they can achieve based on their ability. If the worker\'s ability is greater than n, add arr[n] to ans; otherwise, add arr[worker[i]].\n\n# **Return the Result:**\n\nReturn ans as the total maximum profit.\n\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {\n           int n = difficulty.length;\n           for(int i=0;i<difficulty.length;i++){\n              n = Math.max(n,difficulty[i]);\n           }\n           int[] arr = new int[n+1];\n           for(int i=0;i<profit.length;i++){\n            int index=difficulty[i];\n            arr[index] = Math.max(arr[index],profit[i]);\n           }\n           for(int j=1;j<arr.length;j++){\n            arr[j] = Math.max(arr[j],arr[j-1]);\n           }\n           int ans=0;\n           for(int i=0;i<worker.length;i++){\n             if(worker[i]>n){\n                ans += arr[n];\n             }\n             else{\n                ans += arr[worker[i]];\n             }\n           }\n           return ans;\n    }\n}\n```
['Java']