From 47dff8d93bd0f2eb9300900642c781acedc9ca3d Mon Sep 17 00:00:00 2001
From: Bruce Feldman <bfeldman@sunriselabs.com>
Date: Sat, 11 Aug 2018 18:05:54 -0400
Subject: [PATCH] Made maximum subarray solution more terse
---
.../maximum-subarray/dpMaximumSubarray.js | 61 +++----------------
1 file changed, 9 insertions(+), 52 deletions(-)
diff --git a/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js b/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
index 12f919d737..342e7f10e8 100644
--- a/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
+++ b/src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
@@ -1,62 +1,19 @@
/**
- * Dynamic Programming solution.
+ * Dynamic Programming solution using Kadane's algorithm.
* 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;
- }
+ // Use Kadane's algorithm to generate list of maximum sums ending/beginning at index.
+ const endMax = inputArray.reduce((a, x, i) => a.concat([x + Math.max(0, a[i - 1] || 0)]), []);
+ const begMax = inputArray.reduceRight((a, x) => [x + Math.max(0, a[0] || 0)].concat(a), []);
- if (highestElementValue === null || highestElementValue < inputArray[i]) {
- highestElementValue = inputArray[i];
- }
- }
+ // Obtain the maximum value for the sum of subarrays.
+ const maxVal = begMax.reduce((acc, val) => Math.max(acc, val), -Infinity);
- 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;
+ // Extract indices of maximum value from array & use them to slice input.
+ const maxCmp = val => (val === maxVal);
+ return inputArray.slice(begMax.findIndex(maxCmp), endMax.findIndex(maxCmp) + 1);
}