Add maximum subarray.

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

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+/**
+ * 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;
+}