Initialization: We initialize an empty object numIndices to store the indices of the elements.
\nIterate through the array: We loop through each element in the array nums.
\nCalculate complement: For each element, we calculate the complement (i.e., target - nums[i]).
\nCheck for complement: We check if the complement exists in the hash map numIndices. If it does, we return the indices of the complement and the current element.
\nStore the current element index: If the complement does not exist, we store the index of the current element in the hash map.
\nReturn result: If we find a pair that sums up to the target, we return their indices. If no such pair is found, we return an empty array.
\nThis approach ensures that we find the indices efficiently with a time complexity of
\n𝑂
\n(
\n𝑛
\n)
\nO(n), where
\n𝑛
\nn is the length of the input array nums.
-
|
Can anyone explain me this Two Sum DSA question |
Beta Was this translation helpful? Give feedback.
-
|
I also have same question @bibhuti9 |
Beta Was this translation helpful? Give feedback.
The given problem is attempting to solve the classic "Two Sum" problem using JavaScript. In the "Two Sum" problem, you are given an array of integers (nums) and a target integer (target). Your goal is to find two different indices in the array such that the numbers at those indices add up to the target. You should return these indices in an array.
Here’s the provided code and an explanation of why it doesn’t work as intended: