1. 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]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Example 3:
Solution:
Approach 1: Brute Force
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] result = new int[2];
for (int i = 0; i < nums.length; i++){
for(int j = i + 1; j < nums.length; j++){
if (nums[i] + nums[j] == target){
result[0] = i;
result[1] = j;
break;
}
}
}
return result;
}
}
// TC: O(n^2)
// SC: O(n)
Approach 2: Two-pass Hash Table
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] result = new int[2];
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
// value , index
for (int i = 0; i < nums.length; i++){
int cur = nums[i];
if (map.containsKey(target - cur)){
result[0] = map.get(target - cur);
result[1] = i;
return result;
}
map.put(cur, i);
}
return result;
}
}
// TC: O(n)
// SC: O(n)
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer,Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++){
int m = target - nums[i];
if (map.containsKey(m)){
return new int[]{map.get(m), i};
}
map.put(nums[i], i);
}
return new int[]{};
}
}
// TC: O(n)
// SC: O(n)
class Solution {
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++){
for (int j = i+1; j < nums.length; j++){
if( (nums[i] + nums[j]) == target){
return new int[]{i,j};
}
}
}
return null;
}
}
// TC:O(n^2)
// SC:O(1)