From 2a2b5daa7d7f265f2392ca5fa05a38dc08b26c41 Mon Sep 17 00:00:00 2001
From: Kevin Brewer <appleJax@users.noreply.github.com>
Date: Tue, 4 Sep 2018 01:47:05 -0500
Subject: [PATCH] Simplify dpMaximumSubarray (#189)

* Simplify dpMaximumSubarray

* change var name from currentMaxSum to currentSum

* fix comment with old variable name
---
 .../maximum-subarray/dpMaximumSubarray.js     | 78 ++++++++-----------
 1 file changed, 31 insertions(+), 47 deletions(-)

diff --git a/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js b/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
index 12f919d7..2abfd88d 100644
--- a/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
+++ b/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
@@ -6,57 +6,41 @@
  * @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.
+  // We iterate through the inputArray once, using a greedy approach
+  // to keep track of the maximum sum we've seen so far and the current sum
+  //
+  // currentSum gets reset to 0 everytime it drops below 0
+  //
+  // maxSum is set to -Infinity so that if all numbers
+  // are negative, the highest negative number will constitute
+  // the maximum subarray
+  let maxSum = -Infinity;
   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];
+  // We need to keep track of the starting and ending indices that
+  // contributed to our maxSum so that we can return the actual subarray
+  let maxStartIndex = 0;
+  let maxEndIndex = inputArray.length;
+  let currentStartIndex = 0;
 
-    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]);
+  inputArray.forEach((num, currentIndex) => {
+    currentSum += num;
 
-      if (currentSum > maxSum) {
-        // If current sum became greater then max registered sum then update
-        // max sum and max subarray.
-        maxSum = currentSum;
-        maxSubArray = currentSubArray.slice();
-      }
+    // Update maxSum and the corresponding indices
+    // if we have found a new max
+    if (maxSum < currentSum) {
+      maxSum = currentSum;
+      maxStartIndex = currentStartIndex;
+      maxEndIndex = currentIndex + 1;
     }
-  }
 
-  return maxSubArray;
+    // Reset currentSum and currentStartIndex
+    // if currentSum drops below 0
+    if (currentSum < 0) {
+      currentSum = 0;
+      currentStartIndex = currentIndex + 1;
+    }
+  });
+
+  return inputArray.slice(maxStartIndex, maxEndIndex);
 }