Question
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Solution 1 : Brute Force
■ 1st Submission
class Solution {
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < nums.length && j != i; j++) { // [1]
if (nums[i] + nums[j] == target) {
return new int[]{i, j};
}
}
}
throw new IllegalArgumentException("No two sum solution");
}
}
Finished
Runtime: 0 ms
Your input
[2,7,11,15] 9
Output
[1,0]
Expected
[0,1]
- Oops, [1] shows that I am not fully understanding the procedure of the “for” loop
■ 2nd Submission
Brute Force with correct “for” loop
class Solution {
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < nums.length; j++) { // [2]
if (j == i) continue;
if (nums[i] + nums[j] == target) return new int[]{i, j};
}
}
throw new IllegalArgumentException("No two sum solution");
}
}
Status
Accepted
Runtime
75 ms
Memory
27.7 MB
- [2] makes a “drawback” that causes unnecessary calculation
- Time Complexity : O(n²), Space Complexity : O(1)
■ 3rd Submission
Time-optimized Brute Force
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};
}
}
throw new IllegalArgumentException("No two sum solution");
}
}
Status
Accepted
Runtime
35 ms
Memory
27.6 MB
- Is there any other better algorithm?
Solution 2 : ?
See you later