diff --git a/src/algorithms/sets/hungarian-algorithm/README.md b/src/algorithms/sets/hungarian-algorithm/README.md
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+# Hungarian Algorithm (Kuhn-Munkres Algorithm)
+
+The **Hungarian algorithm** (also known as the **Kuhn-Munkres algorithm** or
+**Munkres assignment algorithm**) is a combinatorial optimization algorithm
+that solves the **assignment problem** in polynomial time `O(n³)`.
+
+## The Assignment Problem
+
+Given `n` agents and `n` tasks, and a cost matrix where entry `C[i][j]`
+represents the cost of assigning agent `i` to task `j`, find a one-to-one
+assignment of agents to tasks that **minimizes the total cost**.
+
+### Example
+
+Consider assigning 3 workers to 3 jobs with the following cost matrix:
+
+|           | Job 1 | Job 2 | Job 3 |
+|-----------|-------|-------|-------|
+| Worker 1  | 250   | 400   | 350   |
+| Worker 2  | 400   | 600   | 350   |
+| Worker 3  | 200   | 400   | 250   |
+
+The optimal assignment is:
+- Worker 1 → Job 2 (cost 400)
+- Worker 2 → Job 3 (cost 350)
+- Worker 3 → Job 1 (cost 200)
+
+**Total minimum cost = 950**
+
+## How It Works
+
+The algorithm operates on an `n×n` cost matrix and proceeds through the
+following steps:
+
+1. **Row reduction**: Subtract the smallest element in each row from all
+   elements in that row.
+2. **Column reduction**: Subtract the smallest element in each column from
+   all elements in that column.
+3. **Cover zeros**: Cover all zeros using a minimum number of horizontal
+   and vertical lines.
+4. **Test for optimality**: If the number of covering lines equals `n`,
+   an optimal assignment can be made among the zeros. If not, proceed to step 5.
+5. **Adjust the matrix**: Find the smallest uncovered element. Subtract it
+   from all uncovered elements and add it to all elements that are covered
+   twice (intersection of two lines). Return to step 3.
+
+![Hungarian Algorithm](https://upload.wikimedia.org/wikipedia/commons/thumb/5/5e/Hungarian_algorithm_-_step_4.svg/320px-Hungarian_algorithm_-_step_4.svg.png)
+
+## Applications
+
+- **Job scheduling**: Assigning workers to tasks to minimize total cost.
+- **Transportation**: Optimizing routing and logistics assignments.
+- **Resource allocation**: Matching resources to demands optimally.
+- **Pattern recognition**: Matching features across images.
+- **Network flow**: Solving bipartite matching problems.
+
+## Complexity
+
+| Complexity | Value  |
+|------------|--------|
+| Time       | O(n³)  |
+| Space      | O(n²)  |
+
+## References
+
+- [Wikipedia - Hungarian Algorithm](https://en.wikipedia.org/wiki/Hungarian_algorithm)
+- [Wikipedia - Assignment Problem](https://en.wikipedia.org/wiki/Assignment_problem)
+- [Brilliant - Hungarian Matching](https://brilliant.org/wiki/hungarian-matching/)
+- [YouTube - Hungarian Algorithm Explained](https://www.youtube.com/watch?v=cQ5MsiGaDY8)
diff --git a/src/algorithms/sets/hungarian-algorithm/__test__/hungarianAlgorithm.test.js b/src/algorithms/sets/hungarian-algorithm/__test__/hungarianAlgorithm.test.js
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+import hungarianAlgorithm from '../hungarianAlgorithm';
+
+describe('hungarianAlgorithm', () => {
+  it('should handle a trivial 1x1 matrix', () => {
+    const costMatrix = [[5]];
+    const result = hungarianAlgorithm(costMatrix);
+
+    expect(result).toEqual([[0, 0]]);
+  });
+
+  it('should handle an empty matrix', () => {
+    const result = hungarianAlgorithm([]);
+
+    expect(result).toEqual([]);
+  });
+
+  it('should throw on non-square matrix', () => {
+    const costMatrix = [
+      [1, 2, 3],
+      [4, 5, 6],
+    ];
+
+    expect(() => hungarianAlgorithm(costMatrix)).toThrow('Cost matrix must be square (n×n)');
+  });
+
+  it('should solve a simple 2x2 assignment problem', () => {
+    const costMatrix = [
+      [1, 2],
+      [2, 1],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal assignment: agent 0 → task 0 (cost 1), agent 1 → task 1 (cost 1).
+    // Total cost = 2.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(2);
+    expect(result.length).toBe(2);
+  });
+
+  it('should solve a 3x3 assignment problem', () => {
+    // Classic textbook example.
+    const costMatrix = [
+      [250, 400, 350],
+      [400, 600, 350],
+      [200, 400, 250],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal assignment: total cost should be 950.
+    // agent 0 → task 1 (400), agent 1 → task 2 (350), agent 2 → task 0 (200).
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(950);
+    expect(result.length).toBe(3);
+  });
+
+  it('should solve a 4x4 assignment problem', () => {
+    // Standard 4×4 cost matrix.
+    const costMatrix = [
+      [82, 83, 69, 92],
+      [77, 37, 49, 92],
+      [11, 69, 5, 86],
+      [8, 9, 98, 23],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal assignment: total cost should be 140.
+    // Possible assignment: 0→2 (69), 1→1 (37), 2→0 (11), 3→3 (23) = 140.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(140);
+    expect(result.length).toBe(4);
+  });
+
+  it('should handle a matrix with all identical values', () => {
+    const costMatrix = [
+      [5, 5, 5],
+      [5, 5, 5],
+      [5, 5, 5],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Any valid assignment is optimal. Total cost = 15.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(15);
+    expect(result.length).toBe(3);
+
+    // Verify each agent is assigned to exactly one task and vice versa.
+    const assignedRows = new Set(result.map(([row]) => row));
+    const assignedCols = new Set(result.map(([, col]) => col));
+    expect(assignedRows.size).toBe(3);
+    expect(assignedCols.size).toBe(3);
+  });
+
+  it('should handle a matrix with zeros', () => {
+    const costMatrix = [
+      [0, 1, 2],
+      [1, 0, 2],
+      [2, 2, 0],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal: diagonal assignment, total cost = 0.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(0);
+    expect(result.length).toBe(3);
+  });
+
+  it('should produce a valid assignment (each row and column used exactly once)', () => {
+    const costMatrix = [
+      [10, 5, 13, 15],
+      [3, 9, 18, 13],
+      [13, 7, 4, 15],
+      [12, 11, 14, 8],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    expect(result.length).toBe(4);
+
+    // Each agent should appear exactly once.
+    const rows = result.map(([row]) => row).sort();
+    expect(rows).toEqual([0, 1, 2, 3]);
+
+    // Each task should appear exactly once.
+    const cols = result.map(([, col]) => col).sort();
+    expect(cols).toEqual([0, 1, 2, 3]);
+  });
+
+  it('should solve a 5x5 assignment problem', () => {
+    const costMatrix = [
+      [7, 53, 183, 439, 863],
+      [497, 383, 563, 79, 973],
+      [287, 63, 343, 169, 583],
+      [627, 343, 773, 959, 943],
+      [767, 473, 103, 699, 303],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal cost: 7 + 79 + 63 + 343 + 103 = 595.
+    // Assignment: 0→0, 1→3, 2→1, 3→1... let's just verify valid assignment and cost.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(result.length).toBe(5);
+
+    // Ensure it's a valid assignment.
+    const assignedRows = new Set(result.map(([row]) => row));
+    const assignedCols = new Set(result.map(([, col]) => col));
+    expect(assignedRows.size).toBe(5);
+    expect(assignedCols.size).toBe(5);
+
+    // The total cost should be optimal (minimum possible).
+    expect(totalCost).toBeLessThanOrEqual(595);
+  });
+
+  it('should not mutate the original cost matrix', () => {
+    const costMatrix = [
+      [1, 2],
+      [3, 4],
+    ];
+    const originalMatrix = costMatrix.map((row) => [...row]);
+
+    hungarianAlgorithm(costMatrix);
+
+    // Original matrix should remain unchanged.
+    expect(costMatrix).toEqual(originalMatrix);
+  });
+
+  it('should handle large cost values', () => {
+    const costMatrix = [
+      [1000000, 2000000, 3000000],
+      [3000000, 1000000, 2000000],
+      [2000000, 3000000, 1000000],
+    ];
+    const result = hungarianAlgorithm(costMatrix);
+
+    // Optimal: diagonal, total = 3000000.
+    const totalCost = result.reduce((sum, [row, col]) => sum + costMatrix[row][col], 0);
+    expect(totalCost).toBe(3000000);
+    expect(result.length).toBe(3);
+  });
+});
diff --git a/src/algorithms/sets/hungarian-algorithm/hungarianAlgorithm.js b/src/algorithms/sets/hungarian-algorithm/hungarianAlgorithm.js
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+++ b/src/algorithms/sets/hungarian-algorithm/hungarianAlgorithm.js
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+/**
+ * Hungarian Algorithm (Kuhn-Munkres Algorithm)
+ *
+ * The Hungarian algorithm solves the assignment problem in polynomial time O(n³).
+ * Given an n×n cost matrix, it finds the optimal assignment of n agents to n tasks
+ * such that the total cost is minimized (or total profit is maximized).
+ *
+ * The algorithm works through the following steps:
+ * 1. Subtract row minimums from each row.
+ * 2. Subtract column minimums from each column.
+ * 3. Cover all zeros with a minimum number of lines.
+ * 4. If the number of lines equals n, an optimal assignment exists.
+ * 5. Otherwise, adjust the matrix and repeat from step 3.
+ *
+ * @param {number[][]} costMatrix - An n×n matrix where costMatrix[i][j]
+ *   represents the cost of assigning agent i to task j.
+ * @return {number[][]} An array of [row, col] pairs representing the optimal assignment.
+ */
+export default function hungarianAlgorithm(costMatrix) {
+  // Validate input.
+  if (!costMatrix || costMatrix.length === 0) {
+    return [];
+  }
+
+  const n = costMatrix.length;
+
+  // Validate that the matrix is square.
+  for (let i = 0; i < n; i += 1) {
+    if (costMatrix[i].length !== n) {
+      throw new Error('Cost matrix must be square (n×n)');
+    }
+  }
+
+  // Handle trivial 1×1 case.
+  if (n === 1) {
+    return [[0, 0]];
+  }
+
+  // Create a working copy of the cost matrix to avoid mutating the input.
+  const matrix = costMatrix.map((row) => [...row]);
+
+  // Step 1: Subtract the row minimum from each row.
+  for (let i = 0; i < n; i += 1) {
+    const rowMin = Math.min(...matrix[i]);
+    for (let j = 0; j < n; j += 1) {
+      matrix[i][j] -= rowMin;
+    }
+  }
+
+  // Step 2: Subtract the column minimum from each column.
+  for (let j = 0; j < n; j += 1) {
+    let colMin = Infinity;
+    for (let i = 0; i < n; i += 1) {
+      if (matrix[i][j] < colMin) {
+        colMin = matrix[i][j];
+      }
+    }
+    for (let i = 0; i < n; i += 1) {
+      matrix[i][j] -= colMin;
+    }
+  }
+
+  // Initialize cover arrays and assignment tracking.
+  const rowCover = new Array(n).fill(false);
+  const colCover = new Array(n).fill(false);
+  // starred[i][j] = true means zero at (i,j) is starred (part of a potential assignment).
+  const starred = Array.from({ length: n }, () => new Array(n).fill(false));
+  // primed[i][j] = true means zero at (i,j) is primed.
+  const primed = Array.from({ length: n }, () => new Array(n).fill(false));
+
+  // Step 3: Star zeros — find initial independent zeros.
+  for (let i = 0; i < n; i += 1) {
+    for (let j = 0; j < n; j += 1) {
+      if (matrix[i][j] === 0 && !rowCover[i] && !colCover[j]) {
+        starred[i][j] = true;
+        rowCover[i] = true;
+        colCover[j] = true;
+      }
+    }
+  }
+
+  // Reset covers after initial starring.
+  rowCover.fill(false);
+  colCover.fill(false);
+
+  /**
+   * Cover columns that contain a starred zero.
+   * @return {number} The number of covered columns.
+   */
+  function coverStarredColumns() {
+    let count = 0;
+    for (let j = 0; j < n; j += 1) {
+      for (let i = 0; i < n; i += 1) {
+        if (starred[i][j]) {
+          colCover[j] = true;
+          count += 1;
+          break;
+        }
+      }
+    }
+    return count;
+  }
+
+  /**
+   * Find an uncovered zero in the matrix.
+   * @return {number[]} [row, col] of the uncovered zero, or [-1, -1] if none found.
+   */
+  function findUncoveredZero() {
+    for (let i = 0; i < n; i += 1) {
+      for (let j = 0; j < n; j += 1) {
+        if (matrix[i][j] === 0 && !rowCover[i] && !colCover[j]) {
+          return [i, j];
+        }
+      }
+    }
+    return [-1, -1];
+  }
+
+  /**
+   * Find a starred zero in the given row.
+   * @param {number} row - The row index to search.
+   * @return {number} The column index of the starred zero, or -1 if none.
+   */
+  function findStarInRow(row) {
+    for (let j = 0; j < n; j += 1) {
+      if (starred[row][j]) {
+        return j;
+      }
+    }
+    return -1;
+  }
+
+  /**
+   * Find a starred zero in the given column.
+   * @param {number} col - The column index to search.
+   * @return {number} The row index of the starred zero, or -1 if none.
+   */
+  function findStarInCol(col) {
+    for (let i = 0; i < n; i += 1) {
+      if (starred[i][col]) {
+        return i;
+      }
+    }
+    return -1;
+  }
+
+  /**
+   * Find a primed zero in the given row.
+   * @param {number} row - The row index to search.
+   * @return {number} The column index of the primed zero, or -1 if none.
+   */
+  function findPrimeInRow(row) {
+    for (let j = 0; j < n; j += 1) {
+      if (primed[row][j]) {
+        return j;
+      }
+    }
+    return -1;
+  }
+
+  /**
+   * Augment the path: unstar starred zeros in the path and star primed zeros.
+   * @param {number[][]} path - Array of [row, col] pairs forming the augmenting path.
+   */
+  function augmentPath(path) {
+    for (let p = 0; p < path.length; p += 1) {
+      const [row, col] = path[p];
+      starred[row][col] = !starred[row][col];
+    }
+  }
+
+  /**
+   * Clear all primes and reset all covers.
+   */
+  function clearPrimesAndCovers() {
+    for (let i = 0; i < n; i += 1) {
+      for (let j = 0; j < n; j += 1) {
+        primed[i][j] = false;
+      }
+    }
+    rowCover.fill(false);
+    colCover.fill(false);
+  }
+
+  /**
+   * Find the smallest uncovered value in the matrix.
+   * @return {number} The smallest uncovered value.
+   */
+  function findSmallestUncovered() {
+    let minVal = Infinity;
+    for (let i = 0; i < n; i += 1) {
+      for (let j = 0; j < n; j += 1) {
+        if (!rowCover[i] && !colCover[j] && matrix[i][j] < minVal) {
+          minVal = matrix[i][j];
+        }
+      }
+    }
+    return minVal;
+  }
+
+  // Main algorithm loop.
+  // eslint-disable-next-line no-constant-condition
+  while (true) {
+    // Step 4: Cover all columns containing a starred zero.
+    const coveredColumns = coverStarredColumns();
+
+    // If all columns are covered, we have found an optimal assignment.
+    if (coveredColumns >= n) {
+      break;
+    }
+
+    // Step 5: Find an uncovered zero, prime it, and process.
+    // eslint-disable-next-line no-constant-condition
+    while (true) {
+      const [row, col] = findUncoveredZero();
+
+      if (row === -1) {
+        // Step 6: No uncovered zeros. Adjust the matrix.
+        // Find the smallest uncovered value.
+        const minVal = findSmallestUncovered();
+
+        // Add minVal to every element of covered rows.
+        // Subtract minVal from every element of uncovered columns.
+        for (let i = 0; i < n; i += 1) {
+          for (let j = 0; j < n; j += 1) {
+            if (rowCover[i]) {
+              matrix[i][j] += minVal;
+            }
+            if (!colCover[j]) {
+              matrix[i][j] -= minVal;
+            }
+          }
+        }
+      } else {
+        // Prime the uncovered zero.
+        primed[row][col] = true;
+
+        const starCol = findStarInRow(row);
+
+        if (starCol !== -1) {
+          // There is a starred zero in this row.
+          // Cover this row and uncover the column of the starred zero.
+          rowCover[row] = true;
+          colCover[starCol] = false;
+        } else {
+          // No starred zero in this row.
+          // Construct an augmenting path starting from the primed zero.
+          const path = [[row, col]];
+          let currentRow = row;
+          let currentCol = col;
+
+          // eslint-disable-next-line no-constant-condition
+          while (true) {
+            // Find a starred zero in the column of the last primed zero.
+            const starRow = findStarInCol(currentCol);
+            if (starRow === -1) {
+              break;
+            }
+            path.push([starRow, currentCol]);
+
+            // Find a primed zero in the row of the starred zero.
+            const primeCol = findPrimeInRow(starRow);
+            path.push([starRow, primeCol]);
+
+            currentRow = starRow;
+            currentCol = primeCol;
+          }
+
+          // Augment the path (swap stars and primes along the path).
+          augmentPath(path);
+
+          // Clear primes and covers.
+          clearPrimesAndCovers();
+
+          // Go back to step 4.
+          break;
+        }
+      }
+    }
+  }
+
+  // Extract the assignment from starred zeros.
+  const assignment = [];
+  for (let i = 0; i < n; i += 1) {
+    for (let j = 0; j < n; j += 1) {
+      if (starred[i][j]) {
+        assignment.push([i, j]);
+      }
+    }
+  }
+
+  return assignment;
+}