leetcode/.notes/1365.md

# 1365. How Many Numbers Are Smaller Than the Current Number

**Difficulty:** Easy
**URL:** https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
**Topics:** Array, Hash Table, Sorting, Counting Sort

---

## The One Insight That Makes This Work

> If I know how many times each value appears, and I add those counts up from left to right, then `prefix[v]` tells me **how many elements are ≤ v** — instantly, for any v.

That's it. Everything below is just executing this idea.

---

## Step 1: Spot the Signal

Read the constraints:

```
0 <= nums[i] <= 100
```

Values are bounded to a tiny range (0–100). This is the **flashing neon sign** that says: don't sort, don't nest loops — build a frequency array indexed by value.

**Rule of thumb:** value range ≤ ~10⁶ and you need counting/ranking? Frequency array.

---

## Step 2: Count Every Value (the "bucket" pass)

Make an array with one slot per possible value. Walk the input once. Each number votes for its own slot.

```javascript
const freq = new Array(101).fill(0);  // slots for values 0..100
for (const x of nums) freq[x]++;
```

For `nums = [8, 1, 2, 2, 3]`:

```
value:  0  1  2  3  4  5  6  7  8  ...
freq:   0  1  2  1  0  0  0  0  1  ...
```

Read it as: "one 1, two 2s, one 3, one 8."

---

## Step 3: Prefix Sum (the magic pass)

Now transform `freq` in place: each slot becomes itself **plus everything before it**.

```javascript
for (let i = 1; i < 101; i++) freq[i] += freq[i - 1];
```

Same example after the pass:

```
value:  0  1  2  3  4  5  6  7  8  ...
freq:   0  1  3  4  4  4  4  4  5  ...
```

New meaning: `freq[v]` = **count of elements ≤ v**.

- `freq[3] = 4` → four numbers are ≤ 3 (they are 1, 2, 2, 3) ✓
- `freq[7] = 4` → still four numbers ≤ 7 ✓

---

## Step 4: Answer Queries in O(1)

"How many numbers are **strictly smaller** than x?" is the same question as "how many numbers are **≤ x − 1**?"

```javascript
return nums.map((x) => (x === 0 ? 0 : freq[x - 1]));
```

The `x === 0` guard exists because nothing can be smaller than the minimum possible value — and `freq[-1]` would be `undefined`.

Trace on `[8, 1, 2, 2, 3]`:

| x | lookup | answer |
|---|--------|--------|
| 8 | freq[7] | 4 |
| 1 | freq[0] | 0 |
| 2 | freq[1] | 1 |
| 2 | freq[1] | 1 |
| 3 | freq[2] | 3 |

→ `[4, 0, 1, 1, 3]` ✓

---

## Full Solution

```javascript
var smallerNumbersThanCurrent = function (nums) {
  // 1. Bucket counts
  const freq = new Array(101).fill(0);
  for (const x of nums) freq[x]++;

  // 2. Prefix sum: freq[v] = count of elements <= v
  for (let i = 1; i < 101; i++) freq[i] += freq[i - 1];

  // 3. Strictly smaller than x == count of elements <= x-1
  return nums.map((x) => (x === 0 ? 0 : freq[x - 1]));
};
```

**Complexity:** O(n + k) time, O(k) space, where k = value range (101 here). No sort, no log factor.

---

## The Reusable Pattern (memorize this shape)

```
1. BUCKET   — freq[value]++ for every element
2. PREFIX   — freq[i] += freq[i-1] left to right
3. QUERY    — freq[v] answers "how many ≤ v" in O(1)
              freq[v-1] answers "how many < v"
              n - freq[v] answers "how many > v"
```

### Where else this exact shape shows up

| Problem | Same pattern, different query |
|---|---|
| **Counting Sort** | Prefix sums become final sorted positions |
| **LC 315 / rank queries** | "How many smaller" is literally a rank |
| **LC 1122 Relative Sort Array** | Bucket + walk buckets in order |
| **Radix sort digit pass** | Bucket by digit, prefix for placement |
| **Histogram percentiles** | freq[v] / n = percentile of v |
| **"How many in range [a, b]?"** | freq[b] − freq[a−1] — the prefix subtraction trick |

### The generalization ladder

- Values bounded and small → **frequency array** (this pattern)
- Values huge but few distinct → **coordinate compression** first, then this pattern
- Need updates between queries → upgrade prefix array to a **Fenwick tree (BIT)** — same idea, log-time updates

---

## Common Mistakes

1. **Returning `freq[x]` instead of `freq[x-1]`** — that counts elements ≤ x (including x itself and its duplicates). Off-by-one between "≤" and "<" is where this pattern bites.
2. **Forgetting the `x === 0` edge** — smallest possible value has nothing below it.
3. **Sizing the array to `nums.length` instead of the value range** — the buckets are indexed by *value*, not position.