Two Sum
Given an array of integers nums
and an integer target
, return indices of the two numbers such that they add up to target
.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9 Output: [0,1] Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6 Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6 Output: [0,1]
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
hashmap = {}
for i in range(len(nums)):
if target-nums[i] in hashmap:
return [i,hashmap[target-nums[i]]]
else:
hashmap[nums[i]] = i
return []
TC : O(n) // Iterating over each element . SC: O(n) Extra space for Hashmap.
Add Two Numbers
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
tailA = l1
tailB = l2
carry = 0
dummyHead = ListNode(-1)
resNode = dummyHead
while tailA or tailB:
x = tailA.val if tailA else 0
y = tailB.val if tailB else 0
val = x + y + carry
resNode.next = ListNode(val % 10)
carry = val // 10
if tailA: tailA = tailA.next
if tailB: tailB = tailB.next
resNode = resNode.next
if carry == 1:
resNode.next = ListNode(carry)
resNode = resNode.next
return dummyHead.next
TC : O(max(m,n)) SC: O(max(m,n)+1)
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