Advanced D&C Patterns

Multiplying two n-digit numbers “obviously” costs O(n²) — it’s how you learned to multiply in grade school, and there’s no way around it. Except a 23-year-old named Karatsuba found a way around it in 1960, by noticing he could replace four multiplications with three. That one swap dropped the exponent from 2 to about 1.585 and launched the entire field of fast algorithms. This page is a tour of combine steps clever enough to bend the exponent itself.

The classical D&C algorithms (merge sort, quicksort, binary search, fast pow) carry most interview problems. The five techniques on this page show up in the harder ones — competitive-programming flavored questions, the “now beat the obvious cubic” follow-ups, and the algorithms that built modern numerical computing.

1. Karatsuba multiplication — three sub-multiplies instead of four

Multiply two n-digit numbers x · y.

The grade-school algorithm is O(n²). Karatsuba (1960) does it in O(n^log₂ 3) ≈ O(n^1.585) with a clever combine step.

Split each number in half:

x = x1 · B + x0          (x1 = high digits, x0 = low; B = 10^(n/2))
y = y1 · B + y0

Then:

x · y = (x1 · B + x0)(y1 · B + y0)
       = x1·y1 · B²  +  (x1·y0 + x0·y1) · B  +  x0·y0

The middle term needs x1·y0 + x0·y1. Naïvely that’s two multiplications, total four (x1·y1, x0·y0, x1·y0, x0·y1).

Karatsuba’s trick:

(x1 + x0)(y1 + y0) = x1·y1 + x1·y0 + x0·y1 + x0·y0

So x1·y0 + x0·y1 = (x1 + x0)(y1 + y0) − x1·y1 − x0·y0. We’ve replaced two multiplies with one multiply plus a subtraction. Three sub-multiplies total, not four.

Recurrence: T(n) = 3 T(n / 2) + O(n) → O(n^log₂ 3) ≈ O(n^1.585).

def karatsuba(x, y):
    if x < 10 or y < 10: return x * y
    n = max(len(str(x)), len(str(y)))
    half = n // 2
    B = 10 ** half
    x1, x0 = divmod(x, B)
    y1, y0 = divmod(y, B)
    z2 = karatsuba(x1, y1)
    z0 = karatsuba(x0, y0)
    z1 = karatsuba(x1 + x0, y1 + y0) - z2 - z0
    return z2 * B * B + z1 * B + z0

Real big-number libraries use Karatsuba for moderate sizes and Schönhage-Strassen (or, since 2019, the Harvey-van der Hoeven O(n log n) algorithm) for truly enormous numbers.

2. Strassen’s matrix multiplication — 7 multiplies instead of 8

Multiply two n × n matrices A · B.

Naïve: O(n³) — three nested loops. Recursive block multiplication (“treat as 2×2 of (n/2) × (n/2) blocks”) gives T(n) = 8 T(n/2) + O(n²) = O(n³) — same asymptotic.

Strassen (1969) showed how to do the 2 × 2 block multiplication with 7 sub-multiplies instead of 8. The result:

Recurrence: T(n) = 7 T(n / 2) + O(n²) → O(n^log₂ 7) ≈ O(n^2.807).

That’s the first algorithm to break the cubic barrier — every subsequent improvement (Coppersmith-Winograd at O(n^2.376), the current best at around O(n^2.371552)) refined the same idea of reducing the number of sub-multiplies via clever linear combinations.

In practice, Strassen has large hidden constants and worse cache behavior than the naïve algorithm. It only wins for very large dense matrices (typically n >= ~1000).

3. Closest Pair of Points

Given n points in the plane, find the pair with the smallest Euclidean distance.

Brute force: check every pair. O(n²). D&C: O(n log n).

The algorithm:

1. Sort by x-coordinate.
2. Divide: split into left and right halves by x.
3. Conquer: recursively find the closest pair in each half.
4. Combine: let δ be the smaller of the two half-answers. Check all pairs straddling the dividing line that are within δ of it. Crucially, you only need to check the next few points (a constant number, at most 7 in any geometric configuration) sorted by y-coordinate within the strip.

def closest_pair(points):
    points.sort()                                             # by x
    def dist(p, q):
        return ((p[0] - q[0])**2 + (p[1] - q[1])**2) ** 0.5
    def solve(lo, hi):
        if hi - lo <= 3:                                       # base: brute force
            best = float('inf')
            for i in range(lo, hi):
                for j in range(i + 1, hi):
                    best = min(best, dist(points[i], points[j]))
            return best
        mid = (lo + hi) // 2
        mid_x = points[mid][0]
        d_left  = solve(lo, mid)
        d_right = solve(mid, hi)
        delta = min(d_left, d_right)
        strip = sorted([p for p in points[lo:hi] if abs(p[0] - mid_x) < delta], key=lambda p: p[1])
        for i in range(len(strip)):
            for j in range(i + 1, min(i + 8, len(strip))):     # at most 7 neighbors
                delta = min(delta, dist(strip[i], strip[j]))
        return delta
    return solve(0, len(points))

Recurrence: T(n) = 2 T(n/2) + O(n) (the strip step is O(n) with a clever proof). → O(n log n) total. (The sort itself adds O(n log n), which doesn’t change the bound.)

The geometric insight — “only the nearest 7 in the strip can possibly be closer than δ” — is the heart of the algorithm. Without it, the strip step is O(n²) and the whole thing collapses to O(n²).

4. The merge-with-extra-accounting pattern

Already glimpsed in inversion counting. Generalized: the merge step of merge sort lets you compute any quantity that depends on cross-pairs between left and right halves, while you sort.

Examples:

• Count Inversions: count pairs (i < j) with nums[i] > nums[j].
• Count of Smaller Numbers After Self: for each i, count j > i with nums[j] < nums[i].
• Reverse Pairs: count pairs (i < j) with nums[i] > 2 · nums[j].

All three solve in O(n log n) via merge sort, by adding a counting step during the merge.

For Reverse Pairs, the counting step is a separate two-pointer pass before the merge:

def reverse_pairs(nums):
    def merge_count(arr):
        if len(arr) <= 1: return arr, 0
        mid = len(arr) // 2
        left,  c1 = merge_count(arr[:mid])
        right, c2 = merge_count(arr[mid:])
        # count reverse pairs across the boundary
        c3 = 0
        j = 0
        for i in range(len(left)):
            while j < len(right) and left[i] > 2 * right[j]:
                j += 1
            c3 += j
        # standard merge
        out, i, j = [], 0, 0
        while i < len(left) and j < len(right):
            if left[i] <= right[j]:
                out.append(left[i]); i += 1
            else:
                out.append(right[j]); j += 1
        out.extend(left[i:]); out.extend(right[j:])
        return out, c1 + c2 + c3
    _, total = merge_count(nums)
    return total

O(n log n) — the count step is amortized O(n) (the j pointer is monotonic across the outer loop).

The whole Reverse Pairs trick lives in one line of the count step. Both halves are sorted, so as i advances, j only ever moves forward — and at each i, j is precisely the number of right-half elements that left[i] beats. Rebuild that line:

def count_step(left, right):
  c3 = 0
  j = 0
  for i in range(len(left)):
      # left and right are both sorted, so j never moves backward
      while j < len(right) and left[i] > 2 * right[j]:
          j += 1
      # at this point right[0..j-1] all satisfy the condition for left[i]
      # ??? j right-half elements satisfied left[i] > 2*right[j]
  return c3

Because j advances monotonically across all of i, the inner while runs at most len(right) times total — not per i. That’s why the count step is O(n), not O(n²), even though it looks like a nested loop.

5. FFT preview — multiplying polynomials in O(n log n)

Given two polynomials of degree n, compute their product (degree 2n).

Naïve: distribute, summing n² products. O(n²).

The Fast Fourier Transform uses D&C to do it in O(n log n):

1. Evaluate each polynomial at 2n cleverly-chosen complex roots of unity (this is the “DFT”).
2. Multiply pointwise — O(n) multiplications.
3. Interpolate back to coefficients (inverse DFT).

Steps 1 and 3 are each O(n log n) via D&C. The same algorithm underlies digital signal processing, JPEG / MP3 compression, big-integer multiplication (Schönhage-Strassen), and convolutional neural networks.

The recurrence: T(n) = 2 T(n / 2) + O(n) → O(n log n). Same shape as merge sort, applied to a completely different problem.

We won’t implement FFT here — it’s a chapter unto itself. But knowing it exists and it’s D&C is useful framing for the rest of CS.

Choosing the right tool

Quick check

These are the same classical shapes with the combine step turned into the star of the show — Karatsuba and Strassen bend the exponent, closest-pair and merge-accounting smuggle extra work into an O(n) merge. That exhausts the D&C toolkit. Next: basic warm-up questions — power function, find max/min, search in a sorted array, count nodes in a tree — to lock the template into muscle memory before the harder practice set.