Hungarian Algorithm for Assignment Problem (Introduction and Implementation) (original) (raw)

Last Updated : 26 Apr, 2025

You are the head of a company with n agents and n tasks. Every agent incurs a different cost to complete each task, as given by the cost[][] matrix, where cost[i][j] represents the cost for the ith agent to perform the jth task. Your objective is to assign exactly one agent to each task—and one task to each agent—in such a way that the total cost of all assignments is minimized.

**Example:

**Input: _n = 3
_cost[][] = [ [2500, 4000, 3500],
[4000, 6000, 3500],
[2000, 4000, 2500] ]
**Output: _9500
**Explanation: _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.

_This example can also be understood in the following way:
You work as a manager for a chip manufacturer, and you currently have 3 people on the road meeting clients. Your salespeople are in Jaipur, Pune and Bangalore, and you want them to fly to three other cities: Delhi, Mumbai and Kerala. The table below shows the cost of airline tickets in INR between the cities: The question: where would you send each of your salespeople in order to minimize fair? Possible assignment: Cost = 11000 INR Other Possible assignment: Cost = **9500 INR and this is the best of the **3! possible assignments.

**Approach:

The Hungarian algorithm (also known as the Munkres assignment algorithm) is designed to find an optimal assignment between n agents and n tasks with a worst-case time complexity of O(n³). Its key insight is that if you add or subtract a constant from all elements in any row or column of the cost matrix, the set of optimal assignments does not change. By leveraging this property, the algorithm reduces the original cost matrix to one containing zeros, thereby simplifying the assignment process to one where each agent can be assigned a task at zero penalty.

Follow the below given step-by-step approach:

1. **Row Reduction: For each row in the matrix, identify the smallest element and subtract it from every element in that row.
2. **Column Reduction: For each column in the matrix, identify the smallest element and subtract it from every element in that column.
3. **Zero Coverage: Cover all zeros in the resulting matrix using the minimum number of horizontal and vertical lines.
4. **Optimality Check:
   • If the number of covering lines equals n, then an optimal assignment is possible, and the algorithm terminates.
   • If the number of covering lines is less than n, proceed to the next step.
5. **Adjust the Matrix:
   • Determine the smallest entry that is not covered by any line.
   • Subtract this entry from every uncovered row.
   • Add the same entry to every column that is covered by a line.
   • Return to step 3 with the modified matrix.
6. This process is repeated until an optimal assignment is identified.

Consider few examples to understand the approach:

Let the 2D array be:
cost[][] = [ [2500, 4000, 3500],
[4000, 6000, 3500],
[2000, 4000, 2500] ]

**Step 1: Subtract minimum of every row. Thus, 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively.

cost[][] = [ [0, 1500, 1000],
[500, 2500, 0],
[0, 2000, 500] ]

**Step 2: Subtract minimum of every column. Thus 0, 1500 and 0 are subtracted from columns 1, 2 and 3 respectively.

cost[][] = [ [0, 0, 1000],
[500, 1000, 0],
[0, 500, 500] ]

**Step 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.

cost[][] = [ [2500, 4000, 3500],
[4000, 6000, 3500],
[**2000, 4000, 2500] ]

So the optimal cost is 4000 + 3500 + 2000 = 9500

In the above example, the first check for optimality did give us solution, but what if we the number covering lines is less than n. The below given example consider this case:

Let the 2D array be:
cost[][] = [ [1500, 4000, 4500],
[2000, 6000, 3500],
[2000, 4000, 2500] ]

**Step 1: Subtract minimum of every row. Thus, 1500, 2000 and 2000 are subtracted from rows 1, 2 and 3 respectively.

cost[][] = [ [0, 2500, 3000],
[0, 4000, 1500],
[0, 2000, 500] ]

**Step 2: Subtract minimum of every column. Thus 0, 2000 and 500 are subtracted from columns 1, 2 and 3 respectively.

cost[][] = [ [0, 500, 2500],
[0, 2000, 1000],
[0, 0, 0] ]

**Step 3: Cover all zeroes with minimum number of horizontal and vertical lines.

**Step 4: Since we need only 2 lines to cover all zeroes, the optimal assignment is not found.

**Step 5: We subtract the smallest uncovered entry from all uncovered rows. Smallest entry is 500.

cost[][] = [ [-500, 0, 2000],
[-500, 1500, 500],
[0, 0, 0] ]

**Step 6: Then we add the smallest entry to all covered columns, we get

cost[][] = [ [0, 0, 2000],
[0, 1500, 500],
[500, 0, 0] ]

Now we return to **Step 3: Here we cover again using lines. and go to **Step 4. Since we need 3 lines to cover, we found the optimal solution.

cost[][] = [ [1500, 4000, 4500],
[2000, 6000, 3500],
[2000, 4000, **2500] ]

So the optimal cost is 4000 + 2000 + 2500 = 8500

Below is given the **implementation:

C++ `

#include <bits/stdc++.h> using namespace std;

void labelIt(vector<vector> &cost, vector &lx) { int n = cost.size(); for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) lx[i] = max(lx[i], cost[i][j]); }

void addTree(int x, int prevX, vector &inTreeX, vector &prev, vector &slack, vector &slackX, vector &lx, vector &ly, vector<vector> &cost) { inTreeX[x] = true; prev[x] = prevX; for(int y = 0; y < slack.size(); y++) { if (lx[x] + ly[y] - cost[x][y] < slack[y]) { slack[y] = lx[x] + ly[y] - cost[x][y]; slackX[y] = x; } } }

void updateLabels(vector &inTreeX, vector &inTreeY, vector &slack, vector &lx, vector &ly) { int n = slack.size();

int delta = INT_MAX;

for(int y = 0; y < n; y++)
    if(!inTreeY[y])
        delta = min(delta, slack[y]);

for(int x = 0; x < n; x++)
    if(inTreeX[x])
        lx[x] -= delta;

for(int y = 0; y < n; y++)
    if(inTreeY[y])
        ly[y] += delta;

for(int y = 0; y < n; y++)
    if(!inTreeY[y])
        slack[y] -= delta;

}

void augment(vector<vector> &cost, int &match, vector &inTreeX, vector &inTreeY, vector &prev, vector &xy, vector &yx, vector &slack, vector &slackX, vector &lx, vector &ly) {

// augmenting path algorithm
int n = cost.size();

// check if we have found a perfect matching
if(match == n)
    return;

int x, y, root;

queue<int> q;

// find root of tree
for(int i = 0; i < n; i++) {

    if(xy[i] == -1) {
        q.push(root = i);
        prev[i] = -2;
        inTreeX[i] = true;
        break;
    }
}

// initialize slack
for(int i = 0; i < n; i++) {
    slack[i] = lx[root] + ly[i] - cost[root][i];
    slackX[i] = root;
}

// BFS to find augmenting path
while(true) {

    // building tree with BFS cycle
    while(!q.empty()) {

        // current vertex
        x = q.front();
        q.pop();

        //iterate through all edges in equality graph
        for(y = 0; y < n; y++) {
            if(lx[x] + ly[y] - cost[x][y] == 0 && !inTreeY[y]) {

                // if y is an exposed vertex in Y
                // found, so augmenting path exists
                if(yx[y] == -1) {
                    x = slackX[y];
                    break;
                } 
                
                // else just add y to inTreeY
                else {
                    inTreeY[y] = true;

                    // add vertex yx[y], which is 
                    // matched with y, to the queue
                    q.push(yx[y]);

                    // add edges (x, y) and (y, yx[y]) to the tree
                    addTree(yx[y], x, inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                }
            }
        }

        // augmenting path found
        if(y < n)
            break;
    }

    // augmenting path found
    if(y < n)
        break; 

    // else improve labeling
    updateLabels(inTreeX, inTreeY, slack, lx, ly);

    for(y = 0; y < n; y++) {

        if(!inTreeY[y] && slack[y] == 0) {

            // if y is an exposed vertex in Y
            // found, so augmenting path exists
            if(yx[y] == -1) {
                x = slackX[y];
                break;
            } 
            
            // else just add y to inTreeY
            else {
                inTreeY[y] = true;

                // add vertex yx[y], which is matched with y, to the queue
                if(!inTreeX[yx[y]]) {
                    q.push(yx[y]);

                    // add edges (x, y) and (y, yx[y]) to the tree
                    addTree(yx[y], slackX[y], inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                }
            }
        }
    }

    // augmenting path found
    if(y < n)
        break;
}

if(y < n) {

    // augmenting path found
    match++;

    // update xy and yx
    for(int cx = x, cy = y, ty; 
        cx != -2; cx = prev[cx], cy = ty) {
        ty = xy[cx];
        xy[cx] = cy;
        yx[cy] = cx;
    }

    // reset inTreeX and inTreeY
    fill(inTreeX.begin(), inTreeX.end(), false);
    fill(inTreeY.begin(), inTreeY.end(), false);

    // recall function, go to step 1 of the algorithm
    augment(cost, match, inTreeX, inTreeY, 
        prev, xy, yx, slack, slackX, lx, ly);
}

}

int findMinCost(vector<vector> &cost) { int n = cost.size();

// convert cost matrix to profit matrix
// by multiplying each element by -1
for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
        cost[i][j] = -1 * cost[i][j]; 

// to store the results
int result = 0;

// number of vertices in current matching
int match = 0;

vector<int> xy(n, -1), yx(n, -1), lx(n, 0);
vector<int> ly(n, 0), slack(n), slackX(n), prev(n);

vector<bool> inTreeX(n, false), inTreeY(n, false);

labelIt(cost, lx);

augment(cost, match, inTreeX, inTreeY, prev, 
    xy, yx, slack, slackX, lx, ly);

for(int i = 0; i < n; i++) {
    result += cost[i][xy[i]];
}
return -1 * result;

}

int main() { vector<vector> cost = { {2500, 4000, 3500}, {4000, 6000, 3500}, {2000, 4000, 2500} }; cout << findMinCost(cost); return 0; }

Java

import java.util.*;

public class GfG {

public static void labelIt(int[][] cost, int[] lx) {
    int n = cost.length;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            lx[i] = Math.max(lx[i], cost[i][j]);
}

public static void addTree(int x, int prevX, boolean[] inTreeX, 
    int[] prev, int[] slack, int[] slackX,
     int[] lx, int[] ly, int[][] cost) {
    inTreeX[x] = true;
    prev[x] = prevX;
    for(int y = 0; y < slack.length; y++) {
        if (lx[x] + ly[y] - cost[x][y] < slack[y]) {
            slack[y] = lx[x] + ly[y] - cost[x][y];
            slackX[y] = x;
        }
    }
}

public static void updateLabels(boolean[] inTreeX, boolean[] inTreeY,
     int[] slack, int[] lx, int[] ly) {
    int n = slack.length;

    int delta = Integer.MAX_VALUE;

    for(int y = 0; y < n; y++)
        if(!inTreeY[y])
            delta = Math.min(delta, slack[y]);

    for(int x = 0; x < n; x++)
        if(inTreeX[x])
            lx[x] -= delta;

    for(int y = 0; y < n; y++)
        if(inTreeY[y])
            ly[y] += delta;

    for(int y = 0; y < n; y++)
        if(!inTreeY[y])
            slack[y] -= delta;

}

public static void augment(int[][] cost, int[] match, boolean[] inTreeX, 
    boolean[] inTreeY, int[] prev, int[] xy, int[] yx, 
    int[] slack, int[] slackX, int[] lx, int[] ly) {

    // augmenting path algorithm
    int n = cost.length;

    // check if we have found a perfect matching
    if(match[0] == n)
        return;

    int x = 0, y = 0, root = 0;

    Queue<Integer> q = new LinkedList<>();

    // find root of tree
    for(int i = 0; i < n; i++) {

        if(xy[i] == -1) {
            q.add(root = i);
            prev[i] = -2;
            inTreeX[i] = true;
            break;
        }
    }

    // initialize slack
    for(int i = 0; i < n; i++) {
        slack[i] = lx[root] + ly[i] - cost[root][i];
        slackX[i] = root;
    }

    // BFS to find augmenting path
    while(true) {

        // building tree with BFS cycle
        while(!q.isEmpty()) {

            // current vertex
            x = q.poll();

            //iterate through all edges in equality graph
            for(y = 0; y < n; y++) {
                if(lx[x] + ly[y] - cost[x][y] == 0 && !inTreeY[y]) {

                    // if y is an exposed vertex in Y
                    // found, so augmenting path exists
                    if(yx[y] == -1) {
                        x = slackX[y];
                        break;
                    } 
                    
                    // else just add y to inTreeY
                    else {
                        inTreeY[y] = true;

                        // add vertex yx[y], which is 
                        // matched with y, to the queue
                        q.add(yx[y]);

                        // add edges (x, y) and (y, yx[y]) to the tree
                        addTree(yx[y], x, inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                    }
                }
            }

            // augmenting path found
            if(y < n)
                break;
        }

        // augmenting path found
        if(y < n)
            break; 

        // else improve labeling
        updateLabels(inTreeX, inTreeY, slack, lx, ly);

        for(y = 0; y < n; y++) {

            if(!inTreeY[y] && slack[y] == 0) {

                // if y is an exposed vertex in Y
                // found, so augmenting path exists
                if(yx[y] == -1) {
                    x = slackX[y];
                    break;
                } 
                
                // else just add y to inTreeY
                else {
                    inTreeY[y] = true;

                    // add vertex yx[y], which is matched with y, to the queue
                    if(!inTreeX[yx[y]]) {
                        q.add(yx[y]);

                        // add edges (x, y) and (y, yx[y]) to the tree
                        addTree(yx[y], slackX[y], inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                    }
                }
            }
        }

        // augmenting path found
        if(y < n)
            break;
    }

    if(y < n) {

        // augmenting path found
        match[0]++;

        // update xy and yx
        for(int cx = x, cy = y, ty; 
            cx != -2; cx = prev[cx], cy = ty) {
            ty = xy[cx];
            xy[cx] = cy;
            yx[cy] = cx;
        }

        // reset inTreeX and inTreeY
        Arrays.fill(inTreeX, false);
        Arrays.fill(inTreeY, false);

        // recall function, go to step 1 of the algorithm
        augment(cost, match, inTreeX, inTreeY, 
        prev, xy, yx, slack, slackX, lx, ly);
    }
}

public static int findMinCost(int[][] cost) {
    int n = cost.length;

    // convert cost matrix to profit matrix
    // by multiplying each element by -1
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            cost[i][j] = -1 * cost[i][j];

    // to store the results
    int result = 0;

    // number of vertices in current matching
    int[] match = new int[]{0};

    int[] xy = new int[n];
    int[] yx = new int[n];
    int[] lx = new int[n];
    int[] ly = new int[n];
    int[] slack = new int[n];
    int[] slackX = new int[n];
    int[] prev = new int[n];

    Arrays.fill(xy, -1);
    Arrays.fill(yx, -1);

    boolean[] inTreeX = new boolean[n];
    boolean[] inTreeY = new boolean[n];

    labelIt(cost, lx);

    augment(cost, match, inTreeX, inTreeY, prev, 
        xy, yx, slack, slackX, lx, ly);

    for(int i = 0; i < n; i++) {
        result += cost[i][xy[i]];
    }
    return -1 * result;
}

public static void main(String[] args) {
    int[][] cost = {
        {2500, 4000, 3500},
        {4000, 6000, 3500},
        {2000, 4000, 2500}
    };
    System.out.println(findMinCost(cost));
}

}

Python

from collections import deque import sys

def labelIt(cost, lx): n = len(cost) for i in range(n): for j in range(n): lx[i] = max(lx[i], cost[i][j])

def addTree(x, prevX, inTreeX, prev, slack, slackX, lx, ly, cost): inTreeX[x] = True prev[x] = prevX for y in range(len(slack)): if lx[x] + ly[y] - cost[x][y] < slack[y]: slack[y] = lx[x] + ly[y] - cost[x][y] slackX[y] = x

def updateLabels(inTreeX, inTreeY, slack, lx, ly): n = len(slack)

delta = sys.maxsize

for y in range(n):
    if not inTreeY[y]:
        delta = min(delta, slack[y])

for x in range(n):
    if inTreeX[x]:
        lx[x] -= delta

for y in range(n):
    if inTreeY[y]:
        ly[y] += delta

for y in range(n):
    if not inTreeY[y]:
        slack[y] -= delta

def augment(cost, match, inTreeX, inTreeY, prev, xy, yx, slack, slackX, lx, ly):

# augmenting path algorithm
n = len(cost)

# check if we have found a perfect matching
if match[0] == n:
    return

x = y = root = 0
q = deque()

# find root of tree
for i in range(n):
    if xy[i] == -1:
        root = i
        q.append(root)
        prev[i] = -2
        inTreeX[i] = True
        break

# initialize slack
for i in range(n):
    slack[i] = lx[root] + ly[i] - cost[root][i]
    slackX[i] = root

# BFS to find augmenting path
while True:
    
    # building tree with BFS cycle
    while q:
        x = q.popleft()
        
        #iterate through all edges in equality graph
        for y in range(n):
            if lx[x] + ly[y] - cost[x][y] == 0 and not inTreeY[y]:
                
                # if y is an exposed vertex in Y
                # found, so augmenting path exists
                if yx[y] == -1:
                    x = slackX[y]
                    break
                else:
                    # else just add y to inTreeY
                    inTreeY[y] = True
                    
                    # add vertex yx[y], which is 
                    # matched with y, to the queue
                    q.append(yx[y])
                    
                    # add edges (x, y) and (y, yx[y]) to the tree
                    addTree(yx[y], x, inTreeX, prev, slack, slackX, lx, ly, cost)
        if y < n:
            break
    
    # augmenting path found
    if y < n:
        break
    
    # else improve labeling
    updateLabels(inTreeX, inTreeY, slack, lx, ly)
    
    for y in range(n):
        if not inTreeY[y] and slack[y] == 0:
            if yx[y] == -1:
                x = slackX[y]
                break
            else:
                inTreeY[y] = True
                if not inTreeX[yx[y]]:
                    q.append(yx[y])
                    addTree(yx[y], slackX[y], inTreeX, prev, slack, slackX, lx, ly, cost)
    if y < n:
        break

if y < n:
    # augmenting path found
    match[0] += 1
    
    # update xy and yx
    cx = x
    cy = y
    while cx != -2:
        ty = xy[cx]
        xy[cx] = cy
        yx[cy] = cx
        cx = prev[cx]
        cy = ty
    
    # reset inTreeX and inTreeY
    for i in range(n):
        inTreeX[i] = False
        inTreeY[i] = False
    
    # recall function, go to step 1 of the algorithm
    augment(cost, match, inTreeX, inTreeY, prev, xy, yx, slack, slackX, lx, ly)

def findMinCost(cost): n = len(cost)

# convert cost matrix to profit matrix
# by multiplying each element by -1
for i in range(n):
    for j in range(n):
        cost[i][j] = -1 * cost[i][j]

# to store the results
result = 0

# number of vertices in current matching
match = [0]

xy = [-1] * n
yx = [-1] * n
lx = [0] * n
ly = [0] * n
slack = [0] * n
slackX = [0] * n
prev = [0] * n

inTreeX = [False] * n
inTreeY = [False] * n

labelIt(cost, lx)

augment(cost, match, inTreeX, inTreeY, prev, xy, yx, slack, slackX, lx, ly)

for i in range(n):
    result += cost[i][xy[i]]
return -1 * result

if name == "main": cost = [ [2500, 4000, 3500], [4000, 6000, 3500], [2000, 4000, 2500] ] print(findMinCost(cost))

C#

using System; using System.Collections.Generic;

public class GfG {

public static void LabelIt(int[][] cost, int[] lx) {
    int n = cost.Length;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            lx[i] = Math.Max(lx[i], cost[i][j]);
}

public static void AddTree(int x, int prevX, bool[] inTreeX, 
    int[] prev, int[] slack, int[] slackX,
     int[] lx, int[] ly, int[][] cost) {
    inTreeX[x] = true;
    prev[x] = prevX;
    for (int y = 0; y < slack.Length; y++) {
        if (lx[x] + ly[y] - cost[x][y] < slack[y]) {
            slack[y] = lx[x] + ly[y] - cost[x][y];
            slackX[y] = x;
        }
    }
}

public static void UpdateLabels(bool[] inTreeX, bool[] inTreeY,
     int[] slack, int[] lx, int[] ly) {
    int n = slack.Length;

    int delta = int.MaxValue;

    for (int y = 0; y < n; y++)
        if (!inTreeY[y])
            delta = Math.Min(delta, slack[y]);

    for (int x = 0; x < n; x++)
        if (inTreeX[x])
            lx[x] -= delta;

    for (int y = 0; y < n; y++)
        if (inTreeY[y])
            ly[y] += delta;

    for (int y = 0; y < n; y++)
        if (!inTreeY[y])
            slack[y] -= delta;

}

public static void Augment(int[][] cost, int[] match, bool[] inTreeX, 
    bool[] inTreeY, int[] prev, int[] xy, int[] yx, 
    int[] slack, int[] slackX, int[] lx, int[] ly) {

    // augmenting path algorithm
    int n = cost.Length;

    // check if we have found a perfect matching
    if (match[0] == n)
        return;

    int x = 0, y = 0, root = 0;

    Queue<int> q = new Queue<int>();

    // find root of tree
    for (int i = 0; i < n; i++) {

        if (xy[i] == -1) {
            q.Enqueue(root = i);
            prev[i] = -2;
            inTreeX[i] = true;
            break;
        }
    }

    // initialize slack
    for (int i = 0; i < n; i++) {
        slack[i] = lx[root] + ly[i] - cost[root][i];
        slackX[i] = root;
    }

    // BFS to find augmenting path
    while (true) {

        // building tree with BFS cycle
        while (q.Count != 0) {

            // current vertex
            x = q.Dequeue();

            //iterate through all edges in equality graph
            for (y = 0; y < n; y++) {
                if (lx[x] + ly[y] - cost[x][y] == 0 && !inTreeY[y]) {

                    // if y is an exposed vertex in Y
                    // found, so augmenting path exists
                    if (yx[y] == -1) {
                        x = slackX[y];
                        break;
                    } 
                    
                    // else just add y to inTreeY
                    else {
                        inTreeY[y] = true;

                        // add vertex yx[y], which is 
                        // matched with y, to the queue
                        q.Enqueue(yx[y]);

                        // add edges (x, y) and (y, yx[y]) to the tree
                        AddTree(yx[y], x, inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                    }
                }
            }

            // augmenting path found
            if (y < n)
                break;
        }

        // augmenting path found
        if (y < n)
            break; 

        // else improve labeling
        UpdateLabels(inTreeX, inTreeY, slack, lx, ly);

        for (y = 0; y < n; y++) {

            if (!inTreeY[y] && slack[y] == 0) {

                // if y is an exposed vertex in Y
                // found, so augmenting path exists
                if (yx[y] == -1) {
                    x = slackX[y];
                    break;
                } 
                
                // else just add y to inTreeY
                else {
                    inTreeY[y] = true;

                    // add vertex yx[y], which is matched with y, to the queue
                    if (!inTreeX[yx[y]]) {
                        q.Enqueue(yx[y]);

                        // add edges (x, y) and (y, yx[y]) to the tree
                        AddTree(yx[y], slackX[y], inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                    }
                }
            }
        }

        // augmenting path found
        if (y < n)
            break;
    }

    if (y < n) {

        // augmenting path found
        match[0]++;

        // update xy and yx
        for (int cx = x, cy = y, ty; 
            cx != -2; cx = prev[cx], cy = ty) {
            ty = xy[cx];
            xy[cx] = cy;
            yx[cy] = cx;
        }

        // reset inTreeX and inTreeY
        for (int i = 0; i < n; i++) {
            inTreeX[i] = false;
            inTreeY[i] = false;
        }

        // recall function, go to step 1 of the algorithm
        Augment(cost, match, inTreeX, inTreeY, 
        prev, xy, yx, slack, slackX, lx, ly);
    }
}

public static int FindMinCost(int[][] cost) {
    int n = cost.Length;

    // convert cost matrix to profit matrix
    // by multiplying each element by -1
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cost[i][j] = -1 * cost[i][j];

    // to store the results
    int result = 0;

    // number of vertices in current matching
    int[] match = new int[]{0};

    int[] xy = new int[n];
    int[] yx = new int[n];
    int[] lx = new int[n];
    int[] ly = new int[n];
    int[] slack = new int[n];
    int[] slackX = new int[n];
    int[] prev = new int[n];

    for (int i = 0; i < n; i++) {
        xy[i] = -1;
        yx[i] = -1;
    }

    bool[] inTreeX = new bool[n];
    bool[] inTreeY = new bool[n];

    LabelIt(cost, lx);

    Augment(cost, match, inTreeX, inTreeY, prev, 
        xy, yx, slack, slackX, lx, ly);

    for (int i = 0; i < n; i++) {
        result += cost[i][xy[i]];
    }
    return -1 * result;
}

public static void Main(String[] args) {
    int[][] cost = new int[][] {
        new int[] {2500, 4000, 3500},
        new int[] {4000, 6000, 3500},
        new int[] {2000, 4000, 2500}
    };
    Console.WriteLine(FindMinCost(cost));
}

}

JavaScript

function labelIt(cost, lx) { let n = cost.length; for(let i = 0; i < n; i++) for(let j = 0; j < n; j++) lx[i] = Math.max(lx[i], cost[i][j]); }

function addTree(x, prevX, inTreeX, prev, slack, slackX, lx, ly, cost) { inTreeX[x] = true; prev[x] = prevX; for(let y = 0; y < slack.length; y++) { if (lx[x] + ly[y] - cost[x][y] < slack[y]) { slack[y] = lx[x] + ly[y] - cost[x][y]; slackX[y] = x; } } }

function updateLabels(inTreeX, inTreeY, slack, lx, ly) { let n = slack.length;

let delta = Number.MAX_SAFE_INTEGER;

for(let y = 0; y < n; y++)
    if(!inTreeY[y])
        delta = Math.min(delta, slack[y]);

for(let x = 0; x < n; x++)
    if(inTreeX[x])
        lx[x] -= delta;

for(let y = 0; y < n; y++)
    if(inTreeY[y])
        ly[y] += delta;

for(let y = 0; y < n; y++)
    if(!inTreeY[y])
        slack[y] -= delta;

}

function augment(cost, match, inTreeX, inTreeY, prev, xy, yx, slack, slackX, lx, ly) {

// augmenting path algorithm
let n = cost.length;

// check if we have found a perfect matching
if(match[0] == n)
    return;

let x, y, root;

let q = [];

// find root of tree
for(let i = 0; i < n; i++) {

    if(xy[i] == -1) {
        q.push(root = i);
        prev[i] = -2;
        inTreeX[i] = true;
        break;
    }
}

// initialize slack
for(let i = 0; i < n; i++) {
    slack[i] = lx[root] + ly[i] - cost[root][i];
    slackX[i] = root;
}

// BFS to find augmenting path
while(true) {

    // building tree with BFS cycle
    while(q.length > 0) {

        // current vertex
        x = q.shift();

        //iterate through all edges in equality graph
        for(y = 0; y < n; y++) {
            if(lx[x] + ly[y] - cost[x][y] == 0 && !inTreeY[y]) {

                // if y is an exposed vertex in Y
                // found, so augmenting path exists
                if(yx[y] == -1) {
                    x = slackX[y];
                    break;
                } 
                
                // else just add y to inTreeY
                else {
                    inTreeY[y] = true;

                    // add vertex yx[y], which is 
                    // matched with y, to the queue
                    q.push(yx[y]);

                    // add edges (x, y) and (y, yx[y]) to the tree
                    addTree(yx[y], x, inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                }
            }
        }

        // augmenting path found
        if(y < n)
            break;
    }

    // augmenting path found
    if(y < n)
        break; 

    // else improve labeling
    updateLabels(inTreeX, inTreeY, slack, lx, ly);

    for(y = 0; y < n; y++) {

        if(!inTreeY[y] && slack[y] == 0) {

            // if y is an exposed vertex in Y
            // found, so augmenting path exists
            if(yx[y] == -1) {
                x = slackX[y];
                break;
            } 
            
            // else just add y to inTreeY
            else {
                inTreeY[y] = true;

                // add vertex yx[y], which is matched with y, to the queue
                if(!inTreeX[yx[y]]) {
                    q.push(yx[y]);

                    // add edges (x, y) and (y, yx[y]) to the tree
                    addTree(yx[y], slackX[y], inTreeX, prev, 
                        slack, slackX, lx, ly, cost);
                }
            }
        }
    }

    // augmenting path found
    if(y < n)
        break;
}

if(y < n) {

    // augmenting path found
    match[0]++;

    // update xy and yx
    for(let cx = x, cy = y, ty; 
        cx != -2; cx = prev[cx], cy = ty) {
        ty = xy[cx];
        xy[cx] = cy;
        yx[cy] = cx;
    }

    // reset inTreeX and inTreeY
    for(let i = 0; i < n; i++) {
        inTreeX[i] = false;
        inTreeY[i] = false;
    }

    // recall function, go to step 1 of the algorithm
    augment(cost, match, inTreeX, inTreeY, 
        prev, xy, yx, slack, slackX, lx, ly);
}

}

function findMinCost(cost) { let n = cost.length;

// convert cost matrix to profit matrix
// by multiplying each element by -1
for(let i = 0; i < n; i++)
    for(let j = 0; j < n; j++)
        cost[i][j] = -1 * cost[i][j];

// to store the results
let result = 0;

// number of vertices in current matching
let match = [0];

let xy = new Array(n).fill(-1);
let yx = new Array(n).fill(-1);
let lx = new Array(n).fill(0);
let ly = new Array(n).fill(0);
let slack = new Array(n);
let slackX = new Array(n);
let prev = new Array(n);

let inTreeX = new Array(n).fill(false);
let inTreeY = new Array(n).fill(false);

labelIt(cost, lx);

augment(cost, match, inTreeX, inTreeY, prev, 
    xy, yx, slack, slackX, lx, ly);

for(let i = 0; i < n; i++) {
    result += cost[i][xy[i]];
}
return -1 * result;

}

let cost = [ [2500, 4000, 3500], [4000, 6000, 3500], [2000, 4000, 2500] ]; console.log(findMinCost(cost));

`

**Time complexity : O(n^3), where n is the number of workers and jobs. This is because the algorithm implements the Hungarian algorithm, which is known to have a time complexity of O(n^3).
**Space complexity : O(n^2), where n is the number of workers and jobs. This is because the algorithm uses a 2D cost matrix of size n x n to store the costs of assigning each worker to a job, and additional arrays of size n to store the labels, matches, and auxiliary information needed for the algorithm.