Two Sum
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
Here is a solution that is using hash map.
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> temp = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int number = nums[i];
int difference = target - number;
if (temp.containsKey(difference)) {
return new int[] { temp.get(difference), i };
}
temp.put(number, i);
}
return null;
}
}
Two pass hash map.
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement) && map.get(complement) != i) {
return new int[] { i, map.get(complement) };
}
}
throw new IllegalArgumentException("No two sum solution");
}
Brute force approach.
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[j] == target - nums[i]) {
return new int[] { i, j };
}
}
}
throw new IllegalArgumentException("No two sum solution");
}
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