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

• Leetcode

• ProgramCreek

Examples

Example 1

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

Solutions

Solution 1

/**
 * 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");
}

Solution 2

/**
 * 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