Algorithm (two-sum) time complexity comparison

Question

Relates to this leetcode question: https://leetcode.com/problems/two-sum/

This is my solution to 2 Sum, it takes ~45ms runtime:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res=new int[2];
        for (int i=0; i<nums.length; i++) {
            for (int j=i+1; j<nums.length; j++) {
                
                if(nums[i] + nums[j] == target) {
                    res[0] = i;
                    res[1] = j;
                    return res;
                }
            }
        }
        return res;
    }
}

and below is the best rated 0ms solution which beats 100% Java solutions:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        
        int[] arr = new int[2];
        
        for (int i = 1; i < nums.length; i++) {
            
            for (int j = i; j < nums.length; j++) {
                
                if (nums[j] + nums[j - i] == target) {
                    
                    arr[0] = j - i;
                    arr[1] = j;
                    
                    return arr;   
                }
            }
        }
        
        return arr;
    }
}

As far as I understand, both of the solutions make same number of array accesses to find the result. The only actual difference is the if-condition:

if(nums[i] + nums[j] == target) {

vs

if (nums[j] + nums[j - i] == target) {

I am really unable to get how does this make such a big difference!!

Please clarify this ambiguity to me. Thanks :)