Let's take a look at solving Two Sum.
Problem
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.
Thought Process
The first thing that I thought of was to use two for loops so that I could iterate through all the integers inside nums and sum them both up to see if they equal target.
The second thing that I thought of was to ensure that the inner for loop is always +1 of the outer loop, so that we can ensure that the same element is not repeated.
Solution
public class Solution {
public int[] TwoSum(int[] nums, int target) {
for (int firstIndex = 0; firstIndex < nums.Length; firstIndex++) {
for (int secondIndex = firstIndex + 1; secondIndex < nums.Length; secondIndex++) {
if (nums[firstIndex] + nums[secondIndex] == target) {
return new int[] { firstIndex, secondIndex };
}
}
}
return new int[] { 0, 0 };
}
}public class Solution {
public int[] TwoSum(int[] nums, int target) {
for (int firstIndex = 0; firstIndex < nums.Length; firstIndex++) {
for (int secondIndex = firstIndex + 1; secondIndex < nums.Length; secondIndex++) {
if (nums[firstIndex] + nums[secondIndex] == target) {
return new int[] { firstIndex, secondIndex };
}
}
}
return new int[] { 0, 0 };
}
}This is probably not the best solution out there, but I think it is fast enough.