up problem add binary

Author: sangtrandev00

src/leetcode/array/majority-element.ts

@@ -5,5 +5,24 @@ export function majorityElement(nums: number[]): number {
  return Math.floor(nums[Math.floor(nums.length / 2)])
 };
 
+export function majorityElementV2(nums: number[]): number {
+ if (nums.length == 0) return -1
+
+ const dict: Map<number, number> = new Map<number, number>()
+ for (let i = 0; i < nums.length; i++) {
+ if (dict.has(nums[i])) {
+ const occurs = dict.get(nums[i]) as number + 1
+ if (occurs > nums.length / 2) {
+ return nums[i]
+ }
+ dict.set(nums[i], occurs)
+
+ } else {
+ dict.set(nums[i], 1)
+ }
+ }
+ return nums[0] // Timecomplexity O(n), SpaceComplexity O(number)
+
+};
 
-console.log("major element: ", majorityElement([3, 3, 3, 4, 2]))
+console.log("major element: ", majorityElementV2([3, 3, 3, 4, 2]))

src/leetcode/string/add-binary.test.ts

@@ -0,0 +1,34 @@
+import { addBinary } from './add-binary';
+import { describe, test, expect } from 'bun:test';
+
+describe('addBinary', () => {
+ test('should return 0 for adding two zeros', () => {
+ expect(addBinary('0', '0')).toBe('0');
+ });
+
+ test('should return the same binary for adding zero', () => {
+ expect(addBinary('1', '0')).toBe('1');
+ expect(addBinary('0', '1')).toBe('1');
+ });
+
+ test('should return 10 for adding 1 and 1', () => {
+ expect(addBinary('1', '1')).toBe('10');
+ });
+
+ test('should correctly add two binary numbers of equal length', () => {
+ expect(addBinary('1010', '1011')).toBe('10101');
+ expect(addBinary('1101', '1011')).toBe('11000');
+ });
+
+ test('should correctly add binary numbers of different lengths', () => {
+ expect(addBinary('111', '1')).toBe('1000');
+ expect(addBinary('111111', '1')).toBe('1000000');
+ expect(addBinary('101010', '1100')).toBe('100110');
+ });
+
+ test('should handle empty strings', () => {
+ expect(addBinary('', '101')).toBe('101');
+ expect(addBinary('101', '')).toBe('101');
+ expect(addBinary('', '')).toBe('0');
+ });
+});

src/leetcode/string/add-binary.ts

@@ -0,0 +1,40 @@
+export function addBinary(a: string, b: string): string {
+ if (a.length == 0) return b
+ if (b.length == 0) return a
+ if (a.length && b.length == 0) return '0'
+ let sum = ''
+ let carry = 0
+ // 1010
+ // 101
+ while (a.length > 0 || b.length > 0 || carry != 0) {
+ const currentSum = Number(a[a.length - 1] ?? 0) + Number(b[b.length - 1] ?? 0)
+ if (currentSum == 2) {
+ if (carry > 0) {
+ sum = `${1}${sum}`
+ } else {
+ sum = `${0}${sum}`
+ carry += 1
+ }
+ } else {
+ if (carry > 0) {
+ if (currentSum + carry == 2) {
+ sum = `${0}${sum}`
+ } else {
+ sum = `${1}${sum}`
+ carry--
+ }
+ } else {
+ sum = `${currentSum}${sum}`
+ }
+ }
+ a = a.slice(0, -1)
+ b = b.slice(0, -1)
+ }
+ return sum
+};
+
+// console.log(addBinary('1010', '1011'));
+// console.log(addBinary('1010', '101'));
+// console.log(addBinary('11', '1'));
+// console.log(addBinary('1101', '1011')); // 11000
+// Time complexity: O(Max(n, m)), Có cách nào làm tốt hơn hay không ?