feat: add Kadane's Algorithm and Deque data structure with comprehensive tests and documentation

Algorithms.Tests/Other/KadanesAlgorithmTests.cs

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+using Algorithms.Other;
+using NUnit.Framework;
+using System;
+
+namespace Algorithms.Tests.Other;
+
+/// <summary>
+///     Comprehensive test suite for Kadane's Algorithm implementation.
+///     Tests cover various scenarios including:
+///     - Arrays with all positive numbers
+///     - Arrays with mixed positive and negative numbers
+///     - Arrays with all negative numbers
+///     - Edge cases (single element, empty array, null array)
+///     - Index tracking functionality
+///     - Long integer support for large numbers
+/// </summary>
+public static class KadanesAlgorithmTests
+{
+    [Test]
+    public static void FindMaximumSubarraySum_WithPositiveNumbers_ReturnsCorrectSum()
+    {
+        // Arrange: When all numbers are positive, the entire array is the maximum subarray
+        int[] array = { 1, 2, 3, 4, 5 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Sum of all elements = 1 + 2 + 3 + 4 + 5 = 15
+        Assert.That(result, Is.EqualTo(15));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithMixedNumbers_ReturnsCorrectSum()
+    {
+        // Arrange: Classic example with mixed positive and negative numbers
+        // The maximum subarray is [4, -1, 2, 1] starting at index 3
+        int[] array = { -2, 1, -3, 4, -1, 2, 1, -5, 4 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Maximum sum is 4 + (-1) + 2 + 1 = 6
+        Assert.That(result, Is.EqualTo(6)); // Subarray [4, -1, 2, 1]
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithAllNegativeNumbers_ReturnsLargestNegative()
+    {
+        // Arrange: When all numbers are negative, the algorithm returns the least negative number
+        // This represents a subarray of length 1 containing the largest (least negative) element
+        int[] array = { -5, -2, -8, -1, -4 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: -1 is the largest (least negative) number in the array
+        Assert.That(result, Is.EqualTo(-1));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithSingleElement_ReturnsThatElement()
+    {
+        // Arrange: Edge case with only one element
+        // The only possible subarray is the element itself
+        int[] array = { 42 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: The single element is both the subarray and its sum
+        Assert.That(result, Is.EqualTo(42));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithNullArray_ThrowsArgumentException()
+    {
+        // Arrange: Test defensive programming - null input validation
+        int[]? array = null;
+
+        // Act & Assert: Should throw ArgumentException for null input
+        Assert.Throws<ArgumentException>(() => KadanesAlgorithm.FindMaximumSubarraySum(array!));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithEmptyArray_ThrowsArgumentException()
+    {
+        // Arrange
+        int[] array = Array.Empty<int>();
+
+        // Act & Assert
+        Assert.Throws<ArgumentException>(() => KadanesAlgorithm.FindMaximumSubarraySum(array));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithAlternatingNumbers_ReturnsCorrectSum()
+    {
+        // Arrange: Alternating positive and negative numbers
+        // Despite negative values, the entire array gives the maximum sum
+        int[] array = { 5, -3, 5, -3, 5 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Sum of entire array = 5 - 3 + 5 - 3 + 5 = 9
+        Assert.That(result, Is.EqualTo(9)); // Entire array
+    }
+
+    [Test]
+    public static void FindMaximumSubarrayWithIndices_ReturnsCorrectIndices()
+    {
+        // Arrange: Test the variant that returns indices of the maximum subarray
+        // Array: [-2, 1, -3, 4, -1, 2, 1, -5, 4]
+        // Index:   0  1   2  3   4  5  6   7  8
+        int[] array = { -2, 1, -3, 4, -1, 2, 1, -5, 4 };
+
+        // Act
+        var (maxSum, startIndex, endIndex) = KadanesAlgorithm.FindMaximumSubarrayWithIndices(array);
+
+        // Assert: Maximum subarray is [4, -1, 2, 1] from index 3 to 6
+        Assert.That(maxSum, Is.EqualTo(6));
+        Assert.That(startIndex, Is.EqualTo(3));
+        Assert.That(endIndex, Is.EqualTo(6));
+    }
+
+    [Test]
+    public static void FindMaximumSubarrayWithIndices_WithSingleElement_ReturnsZeroIndices()
+    {
+        // Arrange
+        int[] array = { 10 };
+
+        // Act
+        var (maxSum, startIndex, endIndex) = KadanesAlgorithm.FindMaximumSubarrayWithIndices(array);
+
+        // Assert
+        Assert.That(maxSum, Is.EqualTo(10));
+        Assert.That(startIndex, Is.EqualTo(0));
+        Assert.That(endIndex, Is.EqualTo(0));
+    }
+
+    [Test]
+    public static void FindMaximumSubarrayWithIndices_WithNullArray_ThrowsArgumentException()
+    {
+        // Arrange
+        int[]? array = null;
+
+        // Act & Assert
+        Assert.Throws<ArgumentException>(() => KadanesAlgorithm.FindMaximumSubarrayWithIndices(array!));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithLongArray_ReturnsCorrectSum()
+    {
+        // Arrange: Test the long integer overload with same values as int test
+        // Verifies that the algorithm works correctly with long data type
+        long[] array = { -2L, 1L, -3L, 4L, -1L, 2L, 1L, -5L, 4L };
+
+        // Act
+        long result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Should produce same result as int version
+        Assert.That(result, Is.EqualTo(6L));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithLargeLongNumbers_ReturnsCorrectSum()
+    {
+        // Arrange: Test with large numbers that would overflow int type
+        // This demonstrates why the long overload is necessary
+        // Sum would be 1,500,000,000 which fits in long but is near int.MaxValue
+        long[] array = { 1000000000L, -500000000L, 1000000000L };
+
+        // Act
+        long result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Entire array sum = 1,000,000,000 - 500,000,000 + 1,000,000,000 = 1,500,000,000
+        Assert.That(result, Is.EqualTo(1500000000L));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithLongNullArray_ThrowsArgumentException()
+    {
+        // Arrange
+        long[]? array = null;
+
+        // Act & Assert
+        Assert.Throws<ArgumentException>(() => KadanesAlgorithm.FindMaximumSubarraySum(array!));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithZeros_ReturnsZero()
+    {
+        // Arrange: Edge case with all zeros
+        // Any subarray will have sum of 0
+        int[] array = { 0, 0, 0, 0 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Maximum sum is 0
+        Assert.That(result, Is.EqualTo(0));
+    }
+
+    [Test]
+    public static void FindMaximumSubarraySum_WithMixedZerosAndNegatives_ReturnsZero()
+    {
+        // Arrange: Mix of zeros and negative numbers
+        // The best subarray is any single zero (or multiple zeros)
+        int[] array = { -5, 0, -3, 0, -2 };
+
+        // Act
+        int result = KadanesAlgorithm.FindMaximumSubarraySum(array);
+
+        // Assert: Zero is better than any negative number
+        Assert.That(result, Is.EqualTo(0));
+    }
+}

Algorithms/Other/KadanesAlgorithm.cs

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+namespace Algorithms.Other;
+
+/// <summary>
+///     Kadane's Algorithm is used to find the maximum sum of a contiguous subarray
+///     within a one-dimensional array of numbers. It has a time complexity of O(n).
+///     This algorithm is a classic example of dynamic programming.
+///     Reference: "Introduction to Algorithms" by Cormen, Leiserson, Rivest, and Stein (CLRS).
+/// </summary>
+public static class KadanesAlgorithm
+{
+    /// <summary>
+    ///     Finds the maximum sum of a contiguous subarray using Kadane's Algorithm.
+    ///     The algorithm works by maintaining two variables:
+    ///     - maxSoFar: The maximum sum found so far (global maximum)
+    ///     - maxEndingHere: The maximum sum of subarray ending at current position (local maximum)
+    ///     At each position, we decide whether to extend the existing subarray or start a new one.
+    /// </summary>
+    /// <param name="array">The input array of integers.</param>
+    /// <returns>The maximum sum of a contiguous subarray.</returns>
+    /// <exception cref="ArgumentException">Thrown when the input array is null or empty.</exception>
+    /// <example>
+    ///     Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4].
+    ///     Output: 6 (subarray [4, -1, 2, 1]).
+    /// </example>
+    public static int FindMaximumSubarraySum(int[] array)
+    {
+        // Validate input to ensure array is not null or empty
+        if (array == null || array.Length == 0)
+        {
+            throw new ArgumentException("Array cannot be null or empty.", nameof(array));
+        }
+
+        // Initialize both variables with the first element
+        // maxSoFar tracks the best sum we've seen across all subarrays
+        int maxSoFar = array[0];
+
+        // maxEndingHere tracks the best sum ending at the current position
+        int maxEndingHere = array[0];
+
+        // Iterate through the array starting from the second element
+        for (int i = 1; i < array.Length; i++)
+        {
+            // Key decision: Either extend the current subarray or start fresh
+            // If adding current element to existing sum is worse than the element alone,
+            // it's better to start a new subarray from current element
+            maxEndingHere = Math.Max(array[i], maxEndingHere + array[i]);
+
+            // Update the global maximum if current subarray sum is better
+            maxSoFar = Math.Max(maxSoFar, maxEndingHere);
+        }
+
+        return maxSoFar;
+    }
+
+    /// <summary>
+    ///     Finds the maximum sum of a contiguous subarray and returns the start and end indices.
+    ///     This variant tracks the indices of the maximum subarray in addition to the sum.
+    ///     Useful when you need to know which elements form the maximum subarray.
+    /// </summary>
+    /// <param name="array">The input array of integers.</param>
+    /// <returns>A tuple containing the maximum sum, start index, and end index.</returns>
+    /// <exception cref="ArgumentException">Thrown when the input array is null or empty.</exception>
+    /// <example>
+    ///     Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4].
+    ///     Output: (MaxSum: 6, StartIndex: 3, EndIndex: 6).
+    ///     The subarray is [4, -1, 2, 1].
+    /// </example>
+    public static (int MaxSum, int StartIndex, int EndIndex) FindMaximumSubarrayWithIndices(int[] array)
+    {
+        // Validate input
+        if (array == null || array.Length == 0)
+        {
+            throw new ArgumentException("Array cannot be null or empty.", nameof(array));
+        }
+
+        // Initialize tracking variables
+        int maxSoFar = array[0];        // Global maximum sum
+        int maxEndingHere = array[0];   // Local maximum sum ending at current position
+        int start = 0;                   // Start index of the maximum subarray
+        int end = 0;                     // End index of the maximum subarray
+        int tempStart = 0;               // Temporary start index for current subarray
+
+        // Process each element starting from index 1
+        for (int i = 1; i < array.Length; i++)
+        {
+            // Decide whether to extend current subarray or start a new one
+            if (array[i] > maxEndingHere + array[i])
+            {
+                // Starting fresh from current element is better
+                maxEndingHere = array[i];
+                tempStart = i;  // Mark this as potential start of new subarray
+            }
+            else
+            {
+                // Extending the current subarray is better
+                maxEndingHere = maxEndingHere + array[i];
+            }
+
+            // Update global maximum and indices if we found a better subarray
+            if (maxEndingHere > maxSoFar)
+            {
+                maxSoFar = maxEndingHere;
+                start = tempStart;  // Commit the start index
+                end = i;            // Current position is the end
+            }
+        }
+
+        return (maxSoFar, start, end);
+    }
+
+    /// <summary>
+    ///     Finds the maximum sum of a contiguous subarray using Kadane's Algorithm for long integers.
+    ///     This overload handles larger numbers that exceed int range (up to 2^63 - 1).
+    ///     The algorithm logic is identical to the int version but uses long arithmetic.
+    /// </summary>
+    /// <param name="array">The input array of long integers.</param>
+    /// <returns>The maximum sum of a contiguous subarray.</returns>
+    /// <exception cref="ArgumentException">Thrown when the input array is null or empty.</exception>
+    /// <example>
+    ///     Input: [1000000000L, -500000000L, 1000000000L].
+    ///     Output: 1500000000L (entire array).
+    /// </example>
+    public static long FindMaximumSubarraySum(long[] array)
+    {
+        // Validate input
+        if (array == null || array.Length == 0)
+        {
+            throw new ArgumentException("Array cannot be null or empty.", nameof(array));
+        }
+
+        // Initialize with first element (using long arithmetic)
+        long maxSoFar = array[0];
+        long maxEndingHere = array[0];
+
+        // Apply Kadane's algorithm with long values
+        for (int i = 1; i < array.Length; i++)
+        {
+            // Decide: extend current subarray or start new one
+            maxEndingHere = Math.Max(array[i], maxEndingHere + array[i]);
+
+            // Update global maximum
+            maxSoFar = Math.Max(maxSoFar, maxEndingHere);
+        }
+
+        return maxSoFar;
+    }
+}