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 Index Problem is we need to move all the zeros to the right and non-zeros to the left. Move Zeros to End Approaches we can follow are: 1st Approch :  1. Your approach of swapping each 0 element with the end ( j pointer) could work, but it requires some adjustments to handle the indices correctly. However, it’s less efficient than moving the non-zero elements to the front as it repeatedly swaps each 0 with elements toward the end, making it potentially  O ( n 2 ) O(n^2) O ( n 2 ) in the worst case due to unnecessary swaps. Here's a refined version of your approach to help clarify the issues and make it function correctly: Adjusted Code Using End-Pointer Swapping Initialize j to nums.size() - 1 (the last index). Traverse from the beginning, and for each 0 , swap it with nums[j] , then decrement j . However, keep in mind that this will cause the relative order of non-zero elements to change. class Solution { public:     void moveZeroes(vector<int>& nums) {         int