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Popular posts from this blog

Leetcode 371. Sum of Two Integers. C++ / Java

371 .  Sum of Two Integers   Given two integers  a  and  b , return  the sum of the two integers without using the operators   +   and   - .   Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5   Constraints: -1000 <= a, b <= 1000 Solution :  C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java :  class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :
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Leetcode 217. Contains Duplicate. Python (Easiest Approach ✅)

217 .  Contains Duplicate   Given an integer array  nums , return  true  if any value appears  at least twice  in the array, and return  false  if every element is distinct.   Example 1: Input: nums = [1,2,3,1] Output: true Example 2: Input: nums = [1,2,3,4] Output: false Example 3: Input: nums = [1,1,1,3,3,4,3,2,4,2] Output: true   Constraints: 1 <= nums.length <= 10 5 -10 9  <= nums[i] <= 10 9 class Solution: def containsDuplicate(self, nums: List[int]) -> bool: hs = set() for n in nums: if n in hs: return True hs.add(n) return False Explaination :
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Leetcode 322. Coin Change. Python (Greedy? vs DP?)

322 .  Coin Change You are given an integer array  coins  representing coins of different denominations and an integer  amount  representing a total amount of money. Return  the fewest number of coins that you need to make up that amount . If that amount of money cannot be made up by any combination of the coins, return  -1 . You may assume that you have an infinite number of each kind of coin.   Example 1: Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2: Input: coins = [2], amount = 3 Output: -1 Example 3: Input: coins = [1], amount = 0 Output: 0   Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 2 31  - 1 0 <= amount <= 10 4   Solution : class Solution:     def coinChange(self, coins: List[int], amount: int) -> int:         dp = [amount + 1] * (amount + 1) #[0...7]         dp[0] = 0        ...
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