Add maximum subarray.

Author: trekhleb

README.md

@@ -42,6 +42,7 @@
   * [Longest Increasing subsequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/longest-increasing-subsequence)
   * [Shortest Common Supersequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/shortest-common-supersequence) (SCS)
   * [Knapsack Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/knapsack-problem) - "0/1" and "Unbound" ones
+  * [Maximum Subarray](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/maximum-subarray) - "Brute Force" and "Dynamic Programming" versions
 * **String**
   * [Levenshtein Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
   * [Hamming Distance](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/string/hamming-distance) - number of positions at which the symbols are different
@@ -100,7 +101,7 @@
   * [Shortest Common Supersequence](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/shortest-common-supersequence)
   * [0/1 Knapsack Problem](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/knapsack-problem)
   * [Integer Partition](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/math/integer-partition)
-  * Maximum subarray
+  * [Maximum Subarray](https://github.com/trekhleb/javascript-algorithms/tree/master/src/algorithms/sets/maximum-subarray)
   * Maximum sum path
 * **Backtracking**
 * **Branch & Bound**

src/algorithms/sets/maximum-subarray/README.md

@@ -0,0 +1,19 @@
+# Maximum subarray problem
+
+The maximum subarray problem is the task of finding the contiguous 
+subarray within a one-dimensional array, `a[1...n]`, of numbers 
+which has the largest sum, where,
+
+![Maximum subarray](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8960f093107b71b21827e726e2bad8b023779b2)
+
+## Example
+
+The list usually contains both positive and negative numbers along 
+with `0`. For example, for the array of 
+values `−2, 1, −3, 4, −1, 2, 1, −5, 4` the contiguous subarray 
+with the largest sum is `4, −1, 2, 1`, with sum `6`.
+
+## References
+
+- [Wikipedia](https://en.wikipedia.org/wiki/Maximum_subarray_problem)
+- [YouTube](https://www.youtube.com/watch?v=ohHWQf1HDfU)

src/algorithms/sets/maximum-subarray/__test__/bfMaximumSubarray.test.js

@@ -0,0 +1,12 @@
+import bfMaximumSubarray from '../bfMaximumSubarray';
+
+describe('bfMaximumSubarray', () => {
+  it('should find maximum subarray using brute force algorithm', () => {
+    expect(bfMaximumSubarray([])).toEqual([]);
+    expect(bfMaximumSubarray([-1, -2, -3, -4, -5])).toEqual([-1]);
+    expect(bfMaximumSubarray([1, 2, 3, 2, 3, 4, 5])).toEqual([1, 2, 3, 2, 3, 4, 5]);
+    expect(bfMaximumSubarray([-2, 1, -3, 4, -1, 2, 1, -5, 4])).toEqual([4, -1, 2, 1]);
+    expect(bfMaximumSubarray([-2, -3, 4, -1, -2, 1, 5, -3])).toEqual([4, -1, -2, 1, 5]);
+    expect(bfMaximumSubarray([1, -3, 2, -5, 7, 6, -1, 4, 11, -23])).toEqual([7, 6, -1, 4, 11]);
+  });
+});

src/algorithms/sets/maximum-subarray/__test__/dpMaximumSubarray.test.js

@@ -0,0 +1,12 @@
+import dpMaximumSubarray from '../dpMaximumSubarray';
+
+describe('dpMaximumSubarray', () => {
+  it('should find maximum subarray using dynamic programming algorithm', () => {
+    expect(dpMaximumSubarray([])).toEqual([]);
+    expect(dpMaximumSubarray([-1, -2, -3, -4, -5])).toEqual([-1]);
+    expect(dpMaximumSubarray([1, 2, 3, 2, 3, 4, 5])).toEqual([1, 2, 3, 2, 3, 4, 5]);
+    expect(dpMaximumSubarray([-2, 1, -3, 4, -1, 2, 1, -5, 4])).toEqual([4, -1, 2, 1]);
+    expect(dpMaximumSubarray([-2, -3, 4, -1, -2, 1, 5, -3])).toEqual([4, -1, -2, 1, 5]);
+    expect(dpMaximumSubarray([1, -3, 2, -5, 7, 6, -1, 4, 11, -23])).toEqual([7, 6, -1, 4, 11]);
+  });
+});

src/algorithms/sets/maximum-subarray/bfMaximumSubarray.js

@@ -0,0 +1,26 @@
+/**
+ * Brute Force solution.
+ * Complexity: O(n^2)
+ *
+ * @param {Number[]} inputArray
+ * @return {Number[]}
+ */
+export default function bfMaximumSubarray(inputArray) {
+  let maxSubarrayStartIndex = 0;
+  let maxSubarrayLength = 0;
+  let maxSubarraySum = null;
+
+  for (let startIndex = 0; startIndex < inputArray.length; startIndex += 1) {
+    let subarraySum = 0;
+    for (let arrLength = 1; arrLength <= (inputArray.length - startIndex); arrLength += 1) {
+      subarraySum += inputArray[startIndex + (arrLength - 1)];
+      if (maxSubarraySum === null || subarraySum > maxSubarraySum) {
+        maxSubarraySum = subarraySum;
+        maxSubarrayStartIndex = startIndex;
+        maxSubarrayLength = arrLength;
+      }
+    }
+  }
+
+  return inputArray.slice(maxSubarrayStartIndex, maxSubarrayStartIndex + maxSubarrayLength);
+}

src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js

@@ -0,0 +1,62 @@
+/**
+ * Dynamic Programming solution.
+ * Complexity: O(n)
+ *
+ * @param {Number[]} inputArray
+ * @return {Number[]}
+ */
+export default function dpMaximumSubarray(inputArray) {
+  // Check if all elements of inputArray are negative ones and return the highest
+  // one in this case.
+  let allNegative = true;
+  let highestElementValue = null;
+  for (let i = 0; i < inputArray.length; i += 1) {
+    if (inputArray[i] >= 0) {
+      allNegative = false;
+    }
+
+    if (highestElementValue === null || highestElementValue < inputArray[i]) {
+      highestElementValue = inputArray[i];
+    }
+  }
+
+  if (allNegative && highestElementValue !== null) {
+    return [highestElementValue];
+  }
+
+  // Let's assume that there is at list one positive integer exists in array.
+  // And thus the maximum sum will for sure be grater then 0. Thus we're able
+  // to always reset max sum to zero.
+  let maxSum = 0;
+
+  // This array will keep a combination that gave the highest sum.
+  let maxSubArray = [];
+
+  // Current sum and subarray that will memoize all previous computations.
+  let currentSum = 0;
+  let currentSubArray = [];
+
+  for (let i = 0; i < inputArray.length; i += 1) {
+    // Let's add current element value to the current sum.
+    currentSum += inputArray[i];
+
+    if (currentSum < 0) {
+      // If the sum went below zero then reset it and don't add current element to max subarray.
+      currentSum = 0;
+      // Reset current subarray.
+      currentSubArray = [];
+    } else {
+      // If current sum stays positive then add current element to current sub array.
+      currentSubArray.push(inputArray[i]);
+
+      if (currentSum > maxSum) {
+        // If current sum became greater then max registered sum then update
+        // max sum and max subarray.
+        maxSum = currentSum;
+        maxSubArray = currentSubArray.slice();
+      }
+    }
+  }
+
+  return maxSubArray;
+}