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

Author: trekhleb

src/algorithms/sets/maximum-subarray/__test__/bfMaximumSubarray.test.js

@@ -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]);

src/algorithms/sets/maximum-subarray/__test__/dpMaximumSubarray.test.js

@@ -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]);

src/algorithms/sets/maximum-subarray/dpMaximumSubarray.js

@@ -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;