LeetCode 560 Subarray Sum Equals K - HashMap Trick

Here's the observation. If $pref[j] - pref[i] = k$, then the subarray from $i+1$ to $j$ sums to $k$. Rearrange: $pref[j] - k = pref[i]$. So for each position $j$, you need to count how many earlier prefix sums equal $pref[j] - k$. Use a HashMap to store prefix sum frequencies. As you iterate:

$1.$ Check if $(pref[j] - k)$ exists in the map. Add that count to your answer.

$2.$ Add current $pref[j]$ to the map. One pass. $O(n)$ time, $O(n)$ space. This pattern appears in many problems.