Partition to K Equal Sum Subsets

//TC. O(2^(N*K)) because we are either selecting or not selecting to 2^n elements because we come back to 0 after every //group with required sum
//space complexity : O(N)
    bool solve(int i,vector<int>&nums,int k,int sum,int req_sum,vector<bool>&vis){
        if(k == 1) return true;     //we are not checking for the last subset because sum is divisible by k and if k-1 subset is found so ofocurse kth
        if(i == nums.size()) return false;  //will also be present,same region for i==nums.size() return false;
        if(sum == req_sum) return solve(0,nums,k-1,0,req_sum,vis);

        bool op1 = false;

        if(!vis[i] and sum + nums[i] <= req_sum){
            vis[i] = true;
            if(solve(i+1,nums,k,sum + nums[i],req_sum,vis)) return true;
            vis[i] = false;
        }

        return solve(i+1,nums,k,sum,req_sum,vis);
    }
class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int n = nums.size();
        int sum = accumulate(nums.begin(),nums.end(),0);
        if(n < k or sum % k != 0) return false;
        int req_sum = sum/k;
        // sort(nums.begin(),nums.end(),greater<int>());
        vector<bool> vis(nums.size());
        return solve(0,nums,k,0,req_sum,vis);
    }

};