Simplify dpMaximumSubarray (#189)

* 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

View File

@ -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);
}