Intrigued by an interesting problem as following, I’d like to share some of my thoughts toward it. Hope it can be helpful.

Given an integer array arr to be divided into $m$ non-empty continuous subarrays, minimize the largest sum among these subarrays.

## Introduction

As you can see, the original problem assumes a strong constraint toward the range of each value in the array. Many have put forward greedy solutions performing binary search between the minimum and maximum of possible largest sums, which scan the array in one pass every time we check a possible value and thus result in a time complexity $O(n\log(sum))$ ($n$ is the length of array). This is efficient enough since $f(x) = \log(x)$ has a much smaller derivative compared to $f(x) = x$, providing a good performance in most cases even when sum is a somewhat large number.

Some others think of a more ‘general’ DP solution with time complexity $O(mn^2)$. This is not so tricky as the greedy solution above, and applies to a broader range of problems. However, $O(mn^2)$ is by no means satisfactory.

Is there another approach independent of the sum of array while achieving a relatively better time complexity?

## Intuition

### Beyond Bottom-up DP

Let’s start from the DP solutions. Most of them are implemented bottom-up. One significant defect of bottom-up DP is that redundant subproblems are computed. In many cases, we only need part of them to solve the original problem. This usually only causes a difference in constant factor though.

### Interesting Observations

Define the subroutine solving the problem above as splitArray(arr, m).

#### First Observation

Observe that if arr_1 is a subarray of arr_2, then splitArray(arr_1, m) $<$ splitArray(arr_2, m) holds for any $m$. This can be simply proved because the minimum of largest subarray sum cannot be larger when any subset of elements in arr_2 is removed.

##### Lemma 1

If arr_1 $\subseteq$ arr_2, then splitArray(arr_1, m) $\le$ splitArray(arr_2, m).

#### Second Observation

Another interesting fact: suppose we divide the array arr into two parts, the left part arr_left comprising $k$ subarrays and the right part arr_right comprising $m - k$ subarrays, then res is the result of splitArray(arr, m) iff res is the result of $\min \{ \max ($splitArray(arr_left, k), splitArray(arr_right, m - k)$\}$ in all possible divisions.

As each divison of $m$ subarrays can be mapped to a combination of $k$ subarrays on the left and $m - k$ subarrays on the right (called bijection), it can be easily shown that once either of splitArray(arr, m) and $\min \{\max($splitArray(arr_left, k), splitArray(arr_right, m - k)$)\}$ gives a smaller value, it will break the optimality of the other.

##### Lemma 2

Suppose that a division $d \in D_{k, m - k}$ split the array into arr_left and arr_right (that is, arr $=$ arr_left $\cup$ arr_right), then splitArray(arr, m) $\iff \min_{d \in D} \{ \max($splitArray(arr_left, k), splitArray(arr_right, m - k)$)\}$

### Combine the Lemmas

From lemma 2, it is enticing to adopt a Divide & Conquer approach:

Choose any legit k, then search $\min_{d \in D} \{ \max($splitArray(arr_left, k), splitArray(arr_right, m - k)$)\}$ in a total of $n - m + 1$ divisions.

But wait! Recall and apply lemma 1, we have

For a fixed $k$, as the division point moves rightward, the left part splitArray(arr_left, k) is monotonically increasing, and the right part splitArray(arr_right, m - k) is monotonically decreasing.

So there is NO need to search all of the $n - m + 1$ divisions! If we aim to find the division $d$ that minimizes $\max($splitArray(arr_left, k), splitArray(arr_right, m - k)$)$, just check the divison point at which the left and right parts have a minimized absolute value of difference.

To find the division point, employ binary search in logarithmic time. (And the search space shrinks to a logarithmic one too!)

## Solution

We have broken the problem into logarithmic subproblems now. A confusing point is how to choose the $k$. In fact, it’s all up to you. Any legit $k$ ($1 \le k < m$) is acceptable. For convenience, I’ll pick $k = 1$ for the left and hence $m - 1$ for the right part.

### Back to DP

So far we’ve been exploring D&C and binary search. But how about the aforementioned bottom-up DP?

Think of splitArray(arr_left, 1) and splitArray(arr_right, m - 1). The left part is easy to evaluate by just calculating the sum of arr_left with a prefix sum array in $O(1)$ time. But the right part will invoke subroutine splitArray recursively. To avoid redundant computations, we can store the result every time we execute splitArray indexed by its two arguments (left bound of arr_right, m) (as right bound will always be the same as the original array).

This is the so-called memoization technique, aka top-down DP.

## Time Complexity Analysis

The complete solution has been constructed now. Last, but by no means the least, is the analysis toward its time complexity.

### Recurrence Relation

Again, let’s get back to the subproblem breaking step:

splitArray(arr, m) $\to \min_{d \in D} \{ \max($splitArray(arr_left, 1), splitArray(arr_right, m - 1)$)\}$, where splitArray(arr_left, 1) can be computed in $O(1)$ time and at most $O(\log n)$ subproblems (i.e. splitArray(arr_right, m - 1)) need to be searched when using binary search.

Define the time complexity of solving the original problem with the array length being $n$ and number of subarrays being $m$ as $T(n, m)$. Then we have the following recurrence relation equation:

$T(n, m) = \sum_{i=1}^{\lceil{\log n}\rceil} T(\frac{n}{2^i}, m - 1) + O(\log n)$

Unfortunately, it’s hard to derive an analytical solution from it. But we can get an upper bound instead of an asymtotically tight one (as big O notation defines).

### Loose Upper Bound

Note that $T(n, m - 1) \le T(n, m)$ as $T(n, m)$ can be broken into $T(n, m - 1)$ and $T(0, 1)$ and thus subproblem $Q(n, m - 1)$ can be reduced to $Q(n, m)$.

Therefore,

$T(n, m) \le \sum_{i=1}^{\lceil{\log n}\rceil} T(\frac{n}{2^i}, m) + O(\log n)$

$m$ can be regarded as a constant and removed now. Simplify $T(n, m)$ as $T(n)$, the equation can be rewritten in

$T(n) \le \sum_{i=1}^{\lceil{\log n}\rceil} T(\frac{n}{2^i}) + O(\log n)$

It’s very similar to the required form of master theorem.

$T(n) \le \lceil{\log n}\rceil \cdot T(\frac{n}{2}) + O(\log n)$

Well, it’s not deducible from master theorem for its non-constant coefficient. To remove the $\log n$ coefficient, transform it into another variable $k$ denoting $\lceil\log n\rceil$.

\begin{aligned} T(2^k) &\le kT(2^{k - 1}) + O(k) \\ &\le k \cdot ((k - 1) \cdot T(2^{k - 2}) + O(k - 1)) + O(k) \\ &\le k(k-1) \cdot T(2^{k - 2}) + O(k^2) \\ &\dotsb \\ &\le k! \cdot T(1) + O(k^{k}) \\ &= O(k!) + O(k^k) \\ \end{aligned}

Since $k!$ is asymtotically equal to $\sqrt{2\pi k}(\dfrac{k}{e})^k$ by Stirling’s approximation,

$T(2^k) \le O(k^k(\frac{\sqrt{k}}{e^k} + 1)) = O(k^k)$

Finally convert $k$ back to $n$, we get

$T(n) \le O({\log n}^{\log n})$

### Linear Complexity

Compare it with the approach using $sum$.

\begin{aligned} \log O({\log n}^{\log n}) &= O(\log n \log\log n) \\ \log O(n\log sum) &= O(\log n + \log\log sum) \end{aligned}

So it depends on the value of $n$ and $sum$. But it’s only a loose bound, and it can be tighter.

At first we have

$T(n) \le \sum_{i=1}^{\lceil{\log n}\rceil} T(\frac{n}{2^i}) + O(\log n)$

Note that $T(\dfrac{n}{2^i})$ was magnified. If we try expanding the expression,

\begin{aligned} T(n) &\le (T(\frac{n}{2}) + T(\frac{n}{2^2}) + \dotsb + T(1)) + O(\log n) \\ &\le (T(\frac{n}{2}) + O(1)) + (T(\frac{n}{2^2}) + O(1)) + \dotsb + (T(1) + O(1))) \\ & = T(\frac{n}{2}) + T(\frac{n}{2^2}) + \dotsb + T(1) \end{aligned}

As $T(n)$ is obviously non-trivial and is at least more complex than $O(1)$.

Observe that

\begin{aligned} T(1) &\le O(1) \\ T(2) &\le T(1) = O(1) \\ T(4) &\le T(2) + T(1) \le 2\cdot O(1) = O(2) \\ T(8) &\le T(4) + T(2) + T(1) \le O(4) \\ &\dotsm \end{aligned}

So we assume that $T(n) \le O(\dfrac{n}2) = O(n)$. It can be proved by mathematical induction technique.

Suppose $n = 2^k$ here for convenience. For $k = 0$, $T(1) = O(1)$ holds trivially. If $T(2^i) \le O(2^i)$ holds for all $0 \le i \le k, i \in \mathbb{N}$

\begin{aligned} T(2^{k + 1}) &\le \sum_{i=0}^k T(2^{i}) \\ &\le \sum_{i=0}^k O(2^i) \\ &= O(\sum_{i=0}^k 2^i) \\ &= O(2^{k + 1}) \end{aligned}

Therefore $T(n) \le O(n)$ is true.

## Appendix

A Java implementation.

// Prefix sum array
int[] sum;
// Memoization
int[][] memo;

public int splitArray(int[] arr, int m) {
// Initialize prefix sum array
sum = new int[arr.length + 1];
sum = 0;
for (int i = 0; i < arr.length; i++) {
sum[i + 1] = sum[i] + arr[i];
}

// Initialize values in memo as -1
memo = new int[arr.length][m + 1];
for (int i = 0; i < memo.length; i++) {
Arrays.fill(memo[i], -1);
}

return splitArray(arr, 0, m);
}

/*
* Param:
* - arr: original array
* - left: left boundary of the array to be split (i.e. arr[left..-1])
* - m: number of subarrays to be split
* Return:
*   the minimum of largest sum in all the subarrays
*/
private int splitArray(int[] arr, int left, int m) {
if (m == 1) return sum[arr.length] - sum[left];
if (memo[left][m] >= 0) return memo[left][m];

/*
* Left array should have at least 1 element, and right array
* should have at least m - 1 elements to be divided into m - 1 subarrays.
* Division point is defined as the index of the first element of right array,
* and can only be selected between leftVal and rightVal via binary search.
*
* leftVal - inclusive, rightVal - exclusive
*/
int leftVal = left + 1,
rightVal = arr.length - m + 2,
diff = 0;
while (leftVal < rightVal) {
int mid = (leftVal + rightVal) / 2;
// Note that the result of left array can be computed in O(1) immediately
diff = (sum[mid] - sum[left]) - splitArray(arr, mid, m - 1);

if (diff < 0) {
leftVal = mid + 1;
} else {
rightVal = mid;
}
}

int res = 0;
if (diff == 0) {
// Left and right arrays have equal results, so either is okay
res = sum[leftVal] - sum[left];
} else {
// There're no exact equal results of left and right arrays
if (leftVal == left + 1) {
// When binary search hits the left boundary,
// left array always has a larger value
res = sum[leftVal] - sum[left];
} else if (leftVal == arr.length - m + 2) {
// When binary search hits the right boundary,
// right array always has a larger value
res = splitArray(arr, leftVal - 1, m - 1);
} else {
// Compare when the dividing just results in a larger result
// of left or right array and select the smaller one
res = Math.min(sum[leftVal] - sum[left], splitArray(arr, leftVal - 1, m - 1));
}
}

memo[left][m] = res;

return res;
}