Number of subarrays with sum equals K

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        
        //For fast I/O in C++
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        int n = nums.size();
        if(n==0)
            return 0;
        
        unordered_map<int,int> mymap;   //Key = PrefixSUM, Value = Count of PrefixSUM.
        int currSUM = 0;
        int i = 0;
        int count = 0;
        
        while(i<n)
        {
            currSUM += nums[i];
            
            if(currSUM == k)    //We found a new subArray with SUM = k
                count += 1;
            
            if(mymap.find(currSUM-k)!=mymap.end())
                count += mymap[currSUM-k];
            
            mymap[currSUM] += 1;
            i += 1;
        }
        return count;
    }
};

Could you please explain how this thing works:
//For fast I/O in C++
ios_base::sync_with_stdio(false);
cin.tie(NULL);

@akshit0699
refer this https://www.google.com/amp/s/www.geeksforgeeks.org/fast-io-for-competitive-programming/amp/

the solution is not working

working

HashMap<Integer, Integer> map = new HashMap<>();
int count = 0;
int currentSum = 0;
int[] psa = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
currentSum += nums[i];
if (currentSum == k) {
count++;
}
if (map.containsKey(currentSum - k)) {
count += map.get(currentSum - k);
}
map.put(currentSum, map.getOrDefault(currentSum, 0) + 1);
}
return count;

please provide code for python

@praka07
the code to return the count/ number of subarrays with sum exactly equal to k is:

int i = 0, count = 0, currentSum = 0, n = arr.length;
HashMap<Integer, Integer> hm = new HashMap<>();
while (i < n) {
currentSum += arr[i];
i++;
if (currentSum == k || hm.containsKey(currentSum - k))
count ++;
hm.put(currentSum, hm.getOrDefault(currentSum, 0) + 1);
}
return count;

if (currentSum - k) is present in the hash map then it implies that there exists one subarray whose sum is equal to k from index 0 to i, therefore we increment count.