# Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)

Given a 2D array, arr of size N*N where arr[i][j] denotes the cost to complete the jthjob by the ith worker. Any worker can be assigned to perform any job. The task is to assign the jobs such that exactly one worker can perform exactly one job in such a way that the total cost of the assignment is minimized.ExampleInput: arr[][] = {{3, 5}, {10, 1}}Output: 4Explanation: The optimal assignment is to assign job 1 to the 1st worker, job 2 to the 2nd worker.Hence, the optimal cost is 3 + 1 = 4.Input: arr[][] = {{2500, 4000, 3500}, {4000, 6000, 3500}, {2000, 4000, 2500}}Output: 4Explanation: The optimal assignment is to assign job 2 to the 1st worker, job 3 to the 2nd worker and job 1 to the 3rd worker.Hence, the optimal cost is 4000 + 3500 + 2000 = 9500.Different approaches to solve this problem are discussed in this article.Approach: The idea is to use the Hungarian Algorithm to solve this problem. The algorithm is as follows:For each row of the matrix, find the smallest element and subtract it from every element in its row.Repeat the step 1 for all columns.Cover all zeros in the matrix using the minimum number of horizontal and vertical lines.Test for Optimality: If the minimum number of covering lines is N, an optimal assignment is possible. Else if lines are lesser than N, an optimal assignment is not found and must proceed to step 5.Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.Consider an example to understand the approach:Let the 2D array be:2500 4000 35004000 6000 35002000 4000 2500Step 1: Subtract minimum of every row. 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively.0 1500 1000500 2500 00 2000 500Step 2: Subtract minimum of every column. 0, 1500 and 0 are subtracted from columns 1, 2 and 3 respectively.0 0 1000500 1000 00 500 500Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we need 3 lines to cover all zeroes, the optimal assignment is found. 2500 4000 3500 4000 6000 3500 2000 4000 2500So the optimal cost is 4000 + 3500 + 2000 = 9500For implementing the above algorithm, the idea is to use the max_cost_assignment() function defined in the dlib library. This function is an implementation of the Hungarian algorithm (also known as the Kuhn-Munkres algorithm) which runs in O(N3) time. It solves the optimal assignment problem. Below is the implementation of the above approach:Pythonimport dlib def minCost(arr): assignment = dlib.max_cost_assignment(arr) print(dlib.assignment_cost(arr, assignment)) arr = dlib.matrix([[3, 5], [10, 1]]) minCost(arr)Time Complexity: O(N3)Auxiliary Space: O(N2)Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.