1. Two Sum
Description
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.
Constraints
Approach
Links
GeeksforGeeks
ProgramCreek
Examples
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
Solutions
/**
* Time complexity : O(n^2). For each element, we try to find its complement
* by looping through the rest of array which takes O(n) time.
* Therefore, the time complexity is O(n^2).
* Space complexity : O(1).
*/
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");
}/**
* Time complexity : O(n). We traverse the list containing n elements only once.
* Each look up in the table costs only O(1) time.
* Space complexity : O(n). The extra space required depends on the number of
* items stored in the hash table, which stores at most n elements.
*/
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap();
for(int i = 0; i < nums.length; i++) {
int d = target-nums[i];
if(map.containsKey(d)) {
return new int[] {map.get(d), i};
} else {
map.put(nums[i], i);
}
}
throw new IllegalArgumentException("Not found");
}
}Follow up
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