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Add more test cases for finding max sub-array algorithm.

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This commit is contained in:
Oleksii Trekhleb 2018-09-04 11:35:13 +03:00
parent 2a2b5daa7d
commit 814fa773ca
3 changed files with 21 additions and 14 deletions
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4
src/algorithms/sets/maximum-subarray/__test__/bfMaximumSubarray.test.js
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@ -3,6 +3,10 @@ import bfMaximumSubarray from '../bfMaximumSubarray';
describe('bfMaximumSubarray', () => {
it('should find maximum subarray using brute force algorithm', () => {
expect(bfMaximumSubarray([])).toEqual([]);
expect(bfMaximumSubarray([0, 0])).toEqual([0]);
expect(bfMaximumSubarray([0, 0, 1])).toEqual([0, 0, 1]);
expect(bfMaximumSubarray([0, 0, 1, 2])).toEqual([0, 0, 1, 2]);
expect(bfMaximumSubarray([0, 0, -1, 2])).toEqual([2]);
expect(bfMaximumSubarray([-1, -2, -3, -4, -5])).toEqual([-1]);
expect(bfMaximumSubarray([1, 2, 3, 2, 3, 4, 5])).toEqual([1, 2, 3, 2, 3, 4, 5]);
expect(bfMaximumSubarray([-2, 1, -3, 4, -1, 2, 1, -5, 4])).toEqual([4, -1, 2, 1]);

4
src/algorithms/sets/maximum-subarray/__test__/dpMaximumSubarray.test.js
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@ -3,6 +3,10 @@ import dpMaximumSubarray from '../dpMaximumSubarray';
describe('dpMaximumSubarray', () => {
it('should find maximum subarray using dynamic programming algorithm', () => {
expect(dpMaximumSubarray([])).toEqual([]);
expect(dpMaximumSubarray([0, 0])).toEqual([0]);
expect(dpMaximumSubarray([0, 0, 1])).toEqual([0, 0, 1]);
expect(dpMaximumSubarray([0, 0, 1, 2])).toEqual([0, 0, 1, 2]);
expect(dpMaximumSubarray([0, 0, -1, 2])).toEqual([2]);
expect(dpMaximumSubarray([-1, -2, -3, -4, -5])).toEqual([-1]);
expect(dpMaximumSubarray([1, 2, 3, 2, 3, 4, 5])).toEqual([1, 2, 3, 2, 3, 4, 5]);
expect(dpMaximumSubarray([-2, 1, -3, 4, -1, 2, 1, -5, 4])).toEqual([4, -1, 2, 1]);

27
src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js
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@ -6,36 +6,35 @@
* @return {Number[]}
*/
export default function dpMaximumSubarray(inputArray) {
// 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
// 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
// The currentSum variable gets reset to 0 every time 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
// The maxSum variable 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;
// We need to keep track of the starting and ending indices that
// contributed to our maxSum so that we can return the actual subarray
// 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;
inputArray.forEach((num, currentIndex) => {
currentSum += num;
inputArray.forEach((currentNumber, currentIndex) => {
currentSum += currentNumber;
// Update maxSum and the corresponding indices
// if we have found a new max
// Update maxSum and the corresponding indices if we have found a new max.
if (maxSum < currentSum) {
maxSum = currentSum;
maxStartIndex = currentStartIndex;
maxEndIndex = currentIndex + 1;
}
// Reset currentSum and currentStartIndex
// if currentSum drops below 0
// Reset currentSum and currentStartIndex if currentSum drops below 0.
if (currentSum < 0) {
currentSum = 0;
currentStartIndex = currentIndex + 1;