#+title: Subarrays with Sum Equal to K
* Subarray Sum Equals K - Brute Force [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sum-equals-k/description/][LC 560]]. Find the total number of continuous subarrays whose sum equals k. Solve using the brute force approach.
Example: nums = [1,2,3], k = 3 → Output: 2 ([1,2] and [3])

** Back
#+begin_src c++
int subarraySum(vector<int>& nums, int k) {
    int count = 0;
    for (int i = 0; i < nums.size(); i++) {
        int sum = 0;
        for (int j = i; j < nums.size(); j++) {
            sum += nums[j];
            if (sum == k) count++;
        }
    }
    return count;
}
#+end_src
Time: O(n^2), Space: O(1)

* Subarray Sum Equals K - Prefix Sum + Hash Map [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sum-equals-k/description/][LC 560]]. Find the total number of continuous subarrays whose sum equals k. Solve optimally using prefix sum with a hash map.
Example: nums = [1,2,3], k = 3 → Output: 2

** Back
#+begin_src c++
int subarraySum(vector<int>& nums, int k) {
    unordered_map<int,int> prefixCount;
    prefixCount[0] = 1;  // base case: sum of 0 appears once
    int sum = 0, count = 0;
    for (int num : nums) {
        sum += num;
        // If (sum - k) exists, those prefixes form valid subarrays
        count += prefixCount[sum - k];
        prefixCount[sum]++;
    }
    return count;
}
#+end_src
Time: O(n), Space: O(n)

* Subarray Sum Equals K - Why prefixCount[0] = 1 [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sum-equals-k/description/][LC 560]]. In the optimal subarray sum solution (prefix sum + hash map), why is `prefixCount[0] = 1` initialized?

** Back
It handles the case where a subarray starts from index 0 and its sum equals k.

When sum == k at some index, we look up prefixCount[sum - k] = prefixCount[0].
Without the initialization, this lookup would return 0, missing valid subarrays like [3] where k = 3.

The base case means: "a prefix sum of 0 exists once (before any element)" — conceptually the empty prefix.

* Subarray Sum Divisible by K [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sums-divisible-by-k/description/][LC 974]]. Find the total number of continuous subarrays whose sum is divisible by k.
Example: nums = [23,2,4,6,7], k = 6 → Output: 2 ([2,4] and [7])

** Back
#+begin_src c++
int subarraysDivByK(vector<int>& nums, int k) {
    unordered_map<int,int> prefixCount;
    prefixCount[0] = 1;
    int sum = 0, count = 0;
    for (int num : nums) {
        sum += num;
        // Modulo can be negative in C++, fix with ((sum % k) + k) % k
        int remainder = ((sum % k) + k) % k;
        count += prefixCount[remainder];
        prefixCount[remainder]++;
    }
    return count;
}
#+end_src
Key insight: If two prefix sums have the same remainder mod k, their subarray sum is divisible by k.

Time: O(n), Space: O(min(n, k))

* Longest Subarray Sum Equals K [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/maximum-size-subarray-sum-equals-k/description/][LC 325]]. Find the maximum length of a contiguous subarray that sums to k.
Example: nums = [1,-1,5,-2,3], k = 3 → Output: 4 ([1,-1,5,-2])

** Back
#+begin_src c++
int maxSubArrayLen(vector<int>& nums, int k) {
    unordered_map<int,int> firstOccurrence;
    firstOccurrence[0] = 0;  // sum 0 at index 0 (1-based)
    int sum = 0, maxLen = 0;
    for (int i = 0; i < nums.size(); i++) {
        sum += nums[i];
        // Check if subarray ending here sums to k
        if (firstOccurrence.count(sum - k)) {
            maxLen = max(maxLen, i + 1 - firstOccurrence[sum - k]);
        }
        // Store first occurrence only (for longest)
        if (!firstOccurrence.count(sum)) {
            firstOccurrence[sum] = i + 1;
        }
    }
    return maxLen;
}
#+end_src
Key difference from counting: store only the *first* occurrence of each prefix sum to maximize length.

Time: O(n), Space: O(n)

* Subarray Sum Equals K - Negative Numbers? [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sum-equals-k/description/][LC 560]]. Does the prefix sum + hash map approach for "subarray sum equals k" work when the array contains negative numbers? Why?

** Back
Yes, it works with negative numbers.

The prefix sum approach does NOT require positive-only elements. It works for any integer array because:
- prefix[j] - prefix[i] = sum[i+1..j] is always true regardless of sign
- The hash map tracks ALL prefix sums seen, positive or negative
- We only need (sum - k) to exist in the map

This makes it superior to the sliding window approach, which only works for non-negative arrays.

* Subarray Sum Equals K - Sliding Window Limitation [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/subarray-sum-equals-k/description/][LC 560]]. Can you use the sliding window technique to solve "subarray sum equals k"? When does it work?

** Back
Sliding window ONLY works when all elements are non-negative (positive or zero).

When all elements >= 0:
- Expanding the window increases the sum
- Shrinking the window decreases the sum
- This monotonicity lets us adjust the window

Sliding window fails with negative numbers because adding/removing elements doesn't monotonically change the sum.

Prefer prefix sum + hash map — it works for all cases (positive, negative, zero).

* Subarray Sum Variations — Master Table [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
What's the master table of subarray sum variations and their approaches?

** Back
| Question | Constraint | Approach | Key Data Structure | Time |
|----------|-----------|----------|-------------------|------|
| Count subarrays sum = K | Any integers | Prefix sum | unordered_map<sum, frequency> | O(n) |
| Longest subarray sum = K | Any integers | Prefix sum | unordered_map<sum, first_index> | O(n) |
| Shortest subarray sum ≥ K | Any integers | Prefix sum + monotonic deque | deque of indices | O(n) |
| Shortest subarray sum = K | Any integers | Prefix sum + ordered map | map<sum, last_index> | O(n log n) |
| Divisible by K | Any integers | Prefix sum + modulo | unordered_map<remainder, freq> | O(n) |
| Sum in range [lower, upper] | Any integers | Prefix sum + BST | multiset / Fenwick / segment tree | O(n log n) |
| Max circular subarray sum | Any integers | Kadane's + total - min_subarray | 2x Kadane | O(n) |
| Subarray sum with only positives | Positives only | Sliding window | Two pointers | O(n) |
| 2D matrix subarray sum = K | 2D grid | Fix rows, compress to 1D | Prefix sum on compressed row | O(n^3) |
| At most K distinct elements | Frequency constraint | Sliding window + freq map | Hash map of counts | O(n) |
| Subarray with exactly K ones | Binary array | Store indices of 1s | Vector of positions | O(n) |
| Two non-overlapping subarrays each = K | Any integers | Prefix sum + track both sides | 2 pass with prefix map | O(n) |

Core pattern: prefix[j] - prefix[i] = subarray(i+1..j). The variation changes what you store and query.

* Subarray Shortest Sum ≥ K — Monotonic Deque [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/description/][LC 862]]. Find the length of the shortest contiguous subarray with sum ≥ K. Works with negative numbers.
Example: nums = [84,-37,32,40,95], K = 167 → Output: 3

** Back
#+begin_src c++
int shortestSubarray(vector<int>& nums, int k) {
    int n = nums.size();
    vector<long long> prefix(n + 1, 0);
    for (int i = 0; i < n; i++) prefix[i + 1] = prefix[i] + nums[i];
    
    deque<int> dq;  // stores indices, prefix[dq[j]] is increasing
    int minLen = INT_MAX;
    
    for (int i = 0; i <= n; i++) {
        // If prefix[i] - prefix[dq.front()] >= K, pop front and record length
        while (!dq.empty() && prefix[i] - prefix[dq.front()] >= k) {
            minLen = min(minLen, i - dq.front());
            dq.pop_front();
        }
        // Maintain monotonic increasing order of prefix values in deque
        while (!dq.empty() && prefix[i] < prefix[dq.back()]) {
            dq.pop_back();
        }
        dq.push_back(i);
    }
    return minLen == INT_MAX ? -1 : minLen;
}
#+end_src
Time: O(n), Space: O(n)

Key insight: Maintain a deque of indices where prefix sums are increasing.
If prefix[i] - prefix[dq.front()] ≥ K, then any index after dq.front() gives a shorter valid subarray.

* Subarray Sum in Range [lower, upper] [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/count-of-range-sum/description/][LC 327]]. Count the number of subarrays whose sum lies in range [lower, upper].
Example: nums = [-2,5,-1], lower = -2, upper = 2 → Output: 3 ([-2], [-1], [5,-1])

** Back
#+begin_src c++
int countRangeSum(vector<int>& nums, int lower, int upper) {
    int n = nums.size();
    vector<long long> prefix(n + 1, 0);
    for (int i = 0; i < n; i++) prefix[i + 1] = prefix[i] + nums[i];
    
    // Use merge sort to count valid pairs
    return mergeCount(prefix, 0, n, lower, upper);
}

int mergeCount(vector<long long>& P, int lo, int hi, int lower, int upper) {
    if (lo >= hi - 1) return 0;
    int mid = lo + (hi - lo) / 2;
    int count = mergeCount(P, lo, mid, lower, upper)
              + mergeCount(P, mid, hi, lower, upper);
    
    // Count valid (i, j) pairs where P[j] - P[i] in [lower, upper]
    int j = mid, k = mid, t = mid;
    for (int i = lo; i < mid; i++) {
        while (k <= hi && P[k] - P[i] < lower) k++;
        while (t <= hi && P[t] - P[i] <= upper) t++;
        count += (t - k);
    }
    
    // Merge step
    vector<long long> temp;
    int p1 = lo, p2 = mid;
    while (p1 < mid && p2 < hi) {
        if (P[p1] <= P[p2]) temp.push_back(P[p1++]);
        else temp.push_back(P[p2++]);
    }
    while (p1 < mid) temp.push_back(P[p1++]);
    while (p2 < hi) temp.push_back(P[p2++]);
    for (int i = 0; i < (int)temp.size(); i++) P[lo + i] = temp[i];
    
    return count;
}
#+end_src
Time: O(n log n), Space: O(n)

Why not hash map? Hash map can't count how many prefix sums fall in a range.
Merge sort counts (i, j) pairs where P[j] - P[i] ∈ [lower, upper] during the merge step.

* Max Circular Subarray Sum [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/maximum-sum-circular-subarray/description/][LC 918]]. Find the maximum sum of a non-empty subarray in a circular array.
Example: nums = [1,-2,3,-2] → Output: 3
Example: nums = [5,-3,5] → Output: 10 (wraps around: [5, 5])

** Back
#+begin_src c++
int maxSubarraySumCircular(vector<int>& nums) {
    int total = 0, maxSum = nums[0], curMax = 0;
    int minSum = nums[0], curMin = 0;
    
    for (int num : nums) {
        curMax = max(curMax + num, num);
        maxSum = max(maxSum, curMax);
        curMin = min(curMin + num, num);
        minSum = min(minSum, curMin);
        total += num;
    }
    
    // If all numbers are negative, maxSum is the answer (total - minSum = 0)
    return maxSum < 0 ? maxSum : max(maxSum, total - minSum);
}
#+end_src
Time: O(n), Space: O(1)

Two cases:
1. Maximum subarray does NOT wrap — standard Kadane's (maxSum)
2. Maximum subarray DOES wrap — total - minSubarray (remove the minimum subarray from the middle)

Edge case: all negative → total - minSum = 0, which is wrong. Return maxSum instead.

* 2D Matrix Subarray Sum = K [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/description/][Submatrices Sum to Target]]. Given a 2D matrix, find the number of submatrices whose sum equals K.
Example: matrix = [[0,1,0],[1,1,1],[0,1,0]], K = 0 → Output: 4

** Back
#+begin_src c++
int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
    int m = matrix.size(), n = matrix[0].size();
    int count = 0;
    
    // Fix top and bottom rows, compress to 1D array
    for (int r1 = 0; r1 < m; r1++) {
        vector<int> colSum(n, 0);
        for (int r2 = r1; r2 < m; r2++) {
            // Add row r2 to the compressed column sums
            for (int c = 0; c < n; c++) {
                colSum[c] += matrix[r2][c];
            }
            // Now solve 1D subarray sum = target
            unordered_map<int,int> prefixCount;
            prefixCount[0] = 1;
            int sum = 0;
            for (int val : colSum) {
                sum += val;
                count += prefixCount[sum - target];
                prefixCount[sum]++;
            }
        }
    }
    return count;
}
#+end_src
Time: O(m^2 * n), Space: O(n)

Key idea: Fix two rows (r1, r2), compress columns between them into a 1D array.
Then the problem reduces to 1D subarray sum = K.

* Exactly K Distinct Elements with Sum = K [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the length of the longest subarray with at most K distinct elements and sum = K.

** Back
#+begin_src c++
int longestSubarray(vector<int>& nums, int k) {
    unordered_map<int,int> freq;
    int sum = 0, distinct = 0, maxLen = 0;
    int left = 0;
    
    for (int right = 0; right < nums.size(); right++) {
        if (freq[nums[right]] == 0) distinct++;
        freq[nums[right]]++;
        sum += nums[right];
        
        // Shrink while distinct > K or sum > k
        while (distinct > k || sum > k) {
            freq[nums[left]]--;
            if (freq[nums[left]] == 0) distinct--;
            sum -= nums[left];
            left++;
        }
        
        if (sum == k) maxLen = max(maxLen, right - left + 1);
    }
    return maxLen;
}
#+end_src
Time: O(n), Space: O(min(n, K))

This combines sliding window (distinct elements constraint) with sum check.
The two constraints (distinct count + sum) make it trickier than simple sliding window.

* Binary Array — Exactly K Ones [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
See [[https://leetcode.com/problems/max-consecutive-ones-iii/description/][LC 1004]]. Given a binary array nums and an integer k, return the maximum number of consecutive 1's in the array if you can flip at most k 0's.
Example: nums = [1,1,1,0,0,0,1,1,1,1,0], k = 2 → Output: 6

** Back
#+begin_src c++
int longestOnes(vector<int>& nums, int k) {
    int left = 0, maxLen = 0;
    for (int right = 0; right < nums.size(); right++) {
        if (nums[right] == 0) k--;
        while (k < 0) {
            if (nums[left] == 0) k++;
            left++;
        }
        maxLen = max(maxLen, right - left + 1);
    }
    return maxLen;
}
#+end_src
Time: O(n), Space: O(1)

Sliding window: expand right, count zeros. When zeros exceed k, shrink from left.

* Two Non-Overlapping Subarrays Each Sum = K [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the maximum sum of two non-overlapping subarrays with lengths L and M.
Example: nums = [0,6,5,2,2,5,1,9,4], L = 1, M = 2 → Output: 20 ([9] and [6,5])

** Back
#+begin_src c++
int maxSumTwoNoOverlap(vector<int>& nums, int L, int M) {
    int n = nums.size();
    // Prefix sums for O(1) subarray sum queries
    vector<int> prefix(n + 1, 0);
    for (int i = 0; i < n; i++) prefix[i + 1] = prefix[i] + nums[i];
    
    auto sum = [&](int i, int j) { return prefix[j + 1] - prefix[i]; };
    
    int maxL = 0, maxM = 0, result = 0;
    
    // Case 1: L comes before M
    for (int i = L + M; i <= n; i++) {
        maxL = max(maxL, sum(i - L - M, i - M - 1));
        maxM = max(maxM, sum(i - M, i - 1));
        result = max(result, maxL + maxM);
    }
    
    // Case 2: M comes before L
    maxL = 0; maxM = 0; result = 0;
    for (int i = L + M; i <= n; i++) {
        maxM = max(maxM, sum(i - L - M, i - L - 1));
        maxL = max(maxL, sum(i - L, i - 1));
        result = max(result, maxL + maxM);
    }
    
    return result;
}
#+end_src
Time: O(n), Space: O(n)

Two cases: L before M, or M before L.
Track the running maximum of the first subarray while computing the second.

* Subarray Sum — All Relationships Diagram [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
What's the decision tree for choosing the right subarray sum approach?

** Back
1. **Are there negative numbers?**
   → YES: Prefix sum approach (hash map, BST, or merge sort)
   → NO: Sliding window is possible
   
2. **What are you counting?**
   → Count all: hash_map<sum, frequency>
   → Longest: hash_map<sum, first_index>
   → Shortest: hash_map<sum, last_index> or monotonic deque
   
3. **Is the target a range [lower, upper]?**
   → YES: Merge sort (count pairs in range) or Fenwick tree / BST
   
4. **Is it divisible by K?**
   → YES: hash_map<remainder, frequency> with modulo
   
5. **Is the array circular?**
   → YES: Kadane's twice — max(max_subarray, total - min_subarray)
   
6. **Is it 2D?**
   → YES: Fix 2 rows, compress to 1D, then prefix sum
   
7. **Are elements only 0/1?**
   → YES: Store indices of 1s, or sliding window counting zeros
   
8. **Are there constraints on distinct elements?**
   → YES: Sliding window + frequency map
   
9. **Are there multiple non-overlapping subarrays?**
   → YES: 2-pass — track running max of first while computing second

The fundamental identity: subarray(i,j) = prefix[j+1] - prefix[i].
Everything else is about what you do with the prefix array.

* Subarray Sum — Core Trigger Heuristic [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
What is the most reliable heuristic for recognizing when to use Prefix Sums?

** Back
When you see **"subarray"** and **"sum"** in the same problem description, **Prefix Sums** should be your immediate first thought.

The core formula:
  Sum(i..j) = PrefixSum[j] - PrefixSum[i-1]

This turns a **range query** into a **difference between two points**, eliminating the need to re-scan the array.

The mental trigger is simple: subarray + sum → prefix sums. Always.

* Subarray Sum — When to Use Sliding Window vs Prefix Sum [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
When should you use Sliding Window vs Prefix Sum + Hash Map for subarray sum problems?

** Back
| Condition | Use | Why |
|-----------|-----|-----|
| All numbers ≥ 0, need target sum or max length | **Sliding Window** | Monotonic: expanding always increases sum, shrinking always decreases. O(1) space vs O(n). |
| Numbers can be negative, need exact target sum | **Prefix Sum + Hash Map** | Monotonicity broken. Adding an element could make sum smaller. Must remember past states. |
| Frequent updates between queries | **Fenwick Tree / Segment Tree** | Prefix sum array takes O(n) to update. Trees give O(log n) update + query. |

Rule of thumb: check for negative numbers first. If none exist, sliding window wins.

* Subarray Product — Prefix Product Pattern [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
How does the prefix sum pattern translate to subarray **product** problems?

** Back
The prefix sum pattern translates directly into a **Prefix Product**:

  Product(i..j) = PrefixProduct[j] / PrefixProduct[i-1]

Instead of subtracting, you divide.

Edge case: zeros break division. Handle by:
1. Segmentation — split the array at zeros, solve each segment independently
2. Track position of last seen zero to reset boundaries

Example: count subarrays with product < K (all positive):
#+begin_src c++
int numSubarrayProductLessThanK(vector<int>& nums, int k) {
    if (k == 0) return 0;
    int count = 0, product = 1, left = 0;
    for (int right = 0; right < nums.size(); right++) {
        product *= nums[right];
        while (left <= right && product >= k) product /= nums[left++];
        count += right - left + 1;
    }
    return count;
}
#+end_src

* Subarray with Equal 0s and 1s — Value Mapping [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the maximum length of a contiguous subarray with equal number of 0s and 1s.
Example: nums = [1,0,1] → Output: 2 ([1,0] or [0,1])

** Back
Treat 0 as -1 and 1 as +1. The problem becomes: **"Find the longest subarray whose sum equals 0."**

#+begin_src c++
int findMaxLength(vector<int>& nums) {
    unordered_map<int,int> firstOccurrence;
    firstOccurrence[0] = 0;  // sum 0 at index 0 (1-based)
    int sum = 0, maxLen = 0;
    for (int i = 0; i < nums.size(); i++) {
        sum += (nums[i] == 1) ? 1 : -1;
        if (firstOccurrence.count(sum)) {
            maxLen = max(maxLen, i + 1 - firstOccurrence[sum]);
        } else {
            firstOccurrence[sum] = i + 1;
        }
    }
    return maxLen;
}
#+end_src
Time: O(n), Space: O(n)

This is the **"prefix state"** generalization: map values to +1/-1, then it's just prefix sum = 0.

* Subarray with Equal Odd and Even Numbers [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the longest subarray with equal number of odd and even numbers.

** Back
Map every even number to +1 and every odd number to -1 (or vice versa). The problem becomes:
**"Find the longest subarray whose sum equals 0."**

Same approach as equal 0s and 1s: prefix sum + hash map storing first occurrence.

This is the same "prefix state" generalization applied to a different domain.

* Multi-Category Balance — (A, B, C) Equal Counts [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the longest subarray with equal number of 'A's, 'B's, and 'C's.

** Back
You can't use a single scalar (+1/-1) for three categories. Instead, track **relative differences** between counts.

Maintain running counts: c_A, c_B, c_C.
At each step, compute the tuple of differences: (c_A - c_B, c_B - c_C).

If this tuple repeats later in the array, the elements between those indices have perfectly balanced A, B, C counts.

Data structure: hash map where the key is the state tuple:
  map<pair<int,int>, int> firstOccurrence;

The tuple (diff_AB, diff_BC) captures the full relative state. If two positions share the same tuple, the subarray between them has zero net change in all three relative differences → equal counts.

Time: O(n), Space: O(n)

* Subarray Product Is Positive / Negative [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Find the maximum length of a subarray with a positive (or negative) product.

** Back
Map: positive → +1, negative → -1, zero → resets the window.

A subarray has positive product if it contains an **even** number of negatives.
A subarray has negative product if it contains an **odd** number of negatives.

Track the parity (odd/even count) of negative numbers as you traverse:
- If parity is even at index i and was even at index j (j < i), the subarray (j+1..i) has positive product.
- If parity is odd at index i and was even at index j, the subarray (j+1..i) has negative product.

Data structure: two hash maps (or arrays) — first occurrence of even-parity index and first occurrence of odd-parity index.

Time: O(n), Space: O(n)

* Bitwise Subarray — OR / AND / XOR [algorithm:array]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Can you use prefix sums for subarray problems with Bitwise OR, AND, or XOR?

** Back
**XOR:** YES — XOR is invertible (XOR is its own inverse).
  XOR(i..j) = PrefixXOR[j] ^ PrefixXOR[i-1]
Same hash map pattern as prefix sum.

**OR / AND:** NO — not invertible. You cannot "undo" an OR or AND operation.

For OR/AND, exploit the key property: as you expand a subarray, the OR/AND result can only change at most **32 times** (for 32-bit integers) because bits only transition 0→1 (OR) or 1→0 (AND).

Strategy: maintain a **set** of all possible OR results ending at the current index. The set never exceeds size 32.

#+begin_src c++
int subarrayBitwiseORs(vector<int>& arr) {
    unordered_set<int> result, current;
    for (int x : arr) {
        unordered_set<int> next;
        next.insert(x);
        for (int val : current) {
            next.insert(val | x);
        }
        current = next;
        for (int val : current) result.insert(val);
    }
    return result.size();
}
#+end_src
Time: O(n * 32), Space: O(32) per step

* Subarray Sum — Keyword-to-Algorithm Mapping [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
What is the master keyword-to-algorithm mapping for subarray problems?

** Back
| Problem Phrase | Array Property | Algorithm |
|---------------|---------------|-----------|
| "Continuous subarray + Sum = K" | Only positive numbers | **Sliding Window** (O(1) space) |
| "Continuous subarray + Sum = K" | Positive & negative | **Prefix Sum + Hash Map** (O(n) space) |
| "Divisible by K" or "Multiple of X" | Any numbers | **Prefix Remainder + Hash Map** (sum % K) |
| "Equal number of X and Y" | Any numbers | **Value Mapping** (X→1, Y→-1) + Prefix Sum Map |
| "Maximum / Minimum Sum" | Any numbers | **Kadane's Algorithm** (DP) |
| "Subarray Sum + Frequent Updates" | Element mutations | **Fenwick Tree / Segment Tree** |
| "Subarray product = K" | No zeros | **Prefix Product** (division) |
| "Subarray product positive/negative" | Any numbers | **Parity tracking** of negative count |
| "Subarray XOR = K" | Any numbers | **Prefix XOR + Hash Map** |
| "Subarray OR / AND" | Any numbers | **Set of results** (bounded by 32 changes) |

* Subarray Sum — Modular Arithmetic Insight [algorithm:interview]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
Why does "subarray sum divisible by K" work with prefix remainders? Prove it.

** Back
If Sum(i..j) mod K = 0, then:
  (Prefix[j] - Prefix[i-1]) mod K = 0

By modular arithmetic:
  Prefix[j] mod K = Prefix[i-1] mod K

**Proof:**
  Prefix[j] - Prefix[i-1] = m * K  (for some integer m)
  Prefix[j] = Prefix[i-1] + m * K
  Prefix[j] mod K = (Prefix[i-1] + m * K) mod K
  Prefix[j] mod K = Prefix[i-1] mod K  (since m*K mod K = 0)

So we just need to find pairs of indices with the same prefix remainder.

**Caveat:** In C++/Java, % can return negative values for negative operands. Fix:
  remainder = ((prefix_sum % K) + K) % K

In Python, % always returns non-negative, so no fix needed.

* Subarray Sum — Prefix State Generalization [algorithm:concept]
:PROPERTIES:
:ANKI_NOTE_TYPE: Basic
:END:
** Front
What is the "prefix state" generalization, and how does it extend beyond addition?

** Back
The core philosophy of prefix sums is: **accumulate history as you traverse linearly, and use a hash map to track state.**

This generalizes far beyond addition:

| Problem | Mapping | Reduces To |
|---------|---------|-----------|
| Equal 0s and 1s | 0→-1, 1→+1 | Subarray sum = 0 |
| Equal odd/even | even→+1, odd→-1 | Subarray sum = 0 |
| Equal vowels/consonants | vowel→+1, consonant→-1 | Subarray sum = 0 |
| Equal A/B/C counts | Track (c_A-c_B, c_B-c_C) | Prefix state tuple repeats |
| Subarray product | Prefix product | Division (handle zeros) |
| Subarray XOR | Prefix XOR | XOR is invertible |
| Subarray sum | Prefix sum | Subtraction |

The pattern:
1. Define a state that accumulates as you traverse
2. Find a way to "undo" or compare states (subtract, XOR, divide, compare tuples)
3. Use a hash map to find when the same state (or target difference) appears