update 0974 english readme

Author: Maddieee

solution/0900-0999/0974.Subarray Sums Divisible by K/README_EN.md

@@ -54,7 +54,29 @@ tags:
 
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-### Solution 1
+### Solution 1: Hash Table + Prefix Sum
+
+1. **Key Insight**:
+   - If there exist indices $i$ and $j$ such that $i \leq j$, and the sum of the subarray $nums[i, ..., j]$ is divisible by $k$, then $(s_j - s_i) \bmod k = 0$, this implies: $s_j \bmod k = s_i \bmod k$
+   - We can use a hash table to count the occurrences of prefix sums modulo $k$ to efficiently check for subarrays satisfying the condition.
+
+2. **Prefix Sum Modulo**:
+   - Use a hash table $cnt$ to count occurrences of each prefix sum modulo $k$.
+   - $cnt[i]$ represents the number of prefix sums with modulo $k$ equal to $i$.
+   - Initialize $cnt[0] = 1$ to account for subarrays directly divisible by $k$.
+
+3. **Algorithm**:
+   - Let a variable $s$ represent the running prefix sum, starting with $s = 0$.
+   - Traverse the array $nums$ from left to right.
+     - For each element $x$:
+       - Compute $s = (s + x) \bmod k$.
+       - Update the result: $ans += cnt[s]$.
+       - Increment $cnt[s]$ by $1$.
+   - Return the result $ans$.
+
+> Note: if $s$ is negative, adjust it to be non-negative by adding $k$ and taking modulo $k$ again.
+
+The time complexity is $O(n)$ and space complexity is $O(n)$ where $n$ is the length of the array $nums$.
 
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