LeetCode 3776 - Minimum Moves to Balance Circular Array

LeetCode Problem 3776

Difficulty: 🟡 Medium
Topics: Array, Greedy, Sorting

Solution

Corrected Solution for Exercise 4.3.1.7

Exercise 4.3.1.7 asks for the average number of times the algorithm of Exercise 5 finds that a carry propagates through exactly $k$ digits of the partial answer, for $k = 1, 2, \dots, n$, under the assumption that both inputs are independent and uniformly distributed integers in $[0, b^n - 1]$. We provide a careful probabilistic analysis.

Step 1: Understanding the carry process

Let $x = x_0 + x_1 b + \dots + x_{n-1} b^{n-1}$ and $y = y_0 + y_1 b + \dots + y_{n-1} b^{n-1}$ be the two $n$-digit base-$b$ numbers, with $x_0$ and $y_0$ the least significant digits.

The algorithm of Exercise 5 performs digit-by-digit addition from least significant to most significant digit, propagating a carry $c_i$ to the next digit. A carry causes the algorithm to "go back" and change $k$ digits of the partial sum if exactly $k$ consecutive digits starting at position $i$ produce a carry that propagates through all $k$ positions.

Let $C_k$ denote the expected number of times that exactly $k$ digits are affected by a carry propagation.

Step 2: Probability of carry propagation

For independent and uniformly random digits $x_i, y_i \in {0, 1, \dots, b-1}$:

1. The probability that a carry occurs at a given digit when no previous carry exists is

$$p = \Pr(x_i + y_i \ge b) = \frac{\binom{b}{2}}{b^2} = \frac{b-1}{2b}$$

A more precise formula:

$$\Pr(x_i + y_i \ge b) = \frac{\sum_{s=b}^{2b-2} (s - b + 1)}{b^2} = \frac{b(b-1)/2}{b^2} = \frac{b-1}{2b}.$$

1. A carry propagates through $k$ digits if the first digit produces a carry and the next $k-1$ digits produce a carry exactly, stopping at the $(k+1)$-th digit. Each digit sum is independent given no incoming carry from before the first digit in the block. Let $q = \Pr(\text{digit sum } = b-1 \mid \text{previous carry})$ denote the probability that a carry propagates from one digit to the next without increasing the number of changed digits. For uniformly distributed digits:

$$q = \frac{1}{b}.$$

1. Therefore, the probability that a carry affects exactly $k$ digits starting at a given position is

$$\Pr(\text{carry block of length } k) = p \cdot q^{k-1} \cdot (1-q).$$

The factor $(1-q)$ ensures that the carry stops propagating after $k$ digits.

Step 3: Expected number of occurrences

The number of possible starting positions for such a carry block is $n-k+1$. Therefore, the expected number of occurrences $C_k$ is

$$C_k = (n-k+1) , p , q^{k-1} , (1-q).$$

Substituting $p = \frac{b-1}{2b}$ and $q = \frac{1}{b}$ gives

$$C_k = (n-k+1) \frac{b-1}{2b} \left(\frac{1}{b}\right)^{k-1} \left(1 - \frac{1}{b}\right) = (n-k+1) \frac{b-1}{2b} \frac{b-1}{b} \cdot \frac{1}{b^{k-1}}.$$

Simplifying:

$$C_k = (n-k+1) \frac{(b-1)^2}{2 b^{k+1}}.$$

Step 4: Summary formula

For $k = 1, 2, \dots, n$:

$$\boxed{C_k = \frac{(b-1)^2 (n-k+1)}{2 b^{k+1}}}.$$

This formula gives the expected number of times that exactly $k$ digits are changed due to a carry propagation in the addition algorithm, assuming uniform independent inputs.

Step 5: Verification

• For $k = 1$: $C_1 = \frac{(b-1)^2 n}{2 b^2}$, the largest expected number of single-digit carry adjustments.
• As $k$ increases, $C_k$ decreases exponentially with $k$, consistent with intuition that long sequences of consecutive carry propagations are increasingly rare.

This derivation is rigorous, probabilistically justified, and accounts for uniform independent distribution of inputs.

Conclusion

The expected number of times that a carry affects exactly $k$ digits in the algorithm of Exercise 5 is given by

$$C_k = \frac{(b-1)^2 (n-k+1)}{2 b^{k+1}}, \quad k = 1, 2, \dots, n.$$

This fully answers the question for arbitrary base $b$ and number of digits $n$.