hxx/javascript-algorithms
1
0
Fork 0
mirror of https://github.moeyy.xyz/https://github.com/trekhleb/javascript-algorithms.git synced 2024-12-26 23:21:18 +08:00
Code Issues Packages Projects Releases Wiki Activity

Simplify dpMaximumSubarray (#189)

Browse Source 
* Simplify dpMaximumSubarray

* change var name from currentMaxSum to currentSum

* fix comment with old variable name
This commit is contained in:
Kevin Brewer 2018-09-04 01:47:05 -05:00 committed by Oleksii Trekhleb
parent 6b0bacd993
commit 2a2b5daa7d
1 changed files with 31 additions and 47 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

76
src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
View File

@ -6,57 +6,41 @@
* @return {Number[]} * @return {Number[]}
*/ */
export default function dpMaximumSubarray(inputArray) { export default function dpMaximumSubarray(inputArray) {
// Check if all elements of inputArray are negative ones and return the highest // We iterate through the inputArray once, using a greedy approach
// one in this case. // to keep track of the maximum sum we've seen so far and the current sum
let allNegative = true; //
let highestElementValue = null; // currentSum gets reset to 0 everytime it drops below 0
for (let i = 0; i < inputArray.length; i += 1) { //
if (inputArray[i] >= 0) { // maxSum is set to -Infinity so that if all numbers
allNegative = false; // are negative, the highest negative number will constitute
} // the maximum subarray
let maxSum = -Infinity;
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 currentSum = 0;
let currentSubArray = [];
for (let i = 0; i < inputArray.length; i += 1) { // We need to keep track of the starting and ending indices that
// Let's add current element value to the current sum. // contributed to our maxSum so that we can return the actual subarray
currentSum += inputArray[i]; let maxStartIndex = 0;
let maxEndIndex = inputArray.length;
let currentStartIndex = 0;
if (currentSum < 0) { inputArray.forEach((num, currentIndex) => {
// If the sum went below zero then reset it and don't add current element to max subarray. currentSum += num;
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) { // Update maxSum and the corresponding indices
// If current sum became greater then max registered sum then update // if we have found a new max
// max sum and max subarray. if (maxSum < currentSum) {
maxSum = currentSum; maxSum = currentSum;
maxSubArray = currentSubArray.slice(); 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);
} }