Split Array Largest Sum

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Problem

Imagine you have a long list of daily expenses and want to split them into a fixed number of months so that the maximum monthly expense is as small as possible.

Given an array of non-negative integers nums and an integer m, split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Return this minimized largest sum.

1 ≤ nums.length ≤ 10^50 ≤ nums[i] ≤ 10^91 ≤ m ≤ min(50, nums.length)

Edge cases: m = 1 → output is sum of entire arraym = length of nums → output is max element in numsnums contains zeros → zeros do not affect max sum but affect splits

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IDE

def splitArray(nums: List[int], m: int) -> int:public int splitArray(int[] nums, int m)int splitArray(vector<int>& nums, int m)function splitArray(nums, m)

def splitArray(nums, m):
    # Write your solution here
    pass

class Solution {
    public int splitArray(int[] nums, int m) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int splitArray(vector<int>& nums, int m) {
    // Write your solution here
    return 0;
}

function splitArray(nums, m) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: sum(nums)Always returning sum of entire array without splitting (ignoring m).✅ Implement binary search on answer space and greedy feasibility check to split into m subarrays.

Wrong: max(nums)Always returning max element without considering splits when m < length(nums).✅ Use binary search to find minimal largest sum, not just max element.

Wrong: Incorrect sums due to integer overflowUsing 32-bit int for sums causing overflow on large values.✅ Use 64-bit integer types (long/long long) for sums and binary search bounds.

Wrong: Too many or too few splits countedGreedy feasibility check incorrectly counts number of subarrays.✅ Ensure greedy check increments split count only when current sum exceeds mid and resets sum properly.

Wrong: Timeout or no outputUsing brute force or exponential approach instead of binary search.✅ Implement binary search on answer space with O(n) greedy check to meet time constraints.

✓

Test Cases

Focus on handling edge cases such as empty arrays, single elements, and boundary values for m.

Watch out for common pitfalls like greedy traps and integer overflow in sums.

Optimize your solution to use binary search on answer space with an O(n) feasibility check.

t1_01basic

Input{"nums":[7,2,5,10,8],"m":2}

Expected18

⏱ Performance - must finish in 2000ms

Split into [7,2,5] and [10,8], largest sums are 14 and 18, so answer is 18 which is minimal possible.

💡 Consider binary searching the answer space between max(nums) and sum(nums).

💡 Use a greedy check to verify if a candidate max sum can split nums into at most m subarrays.

💡 Adjust binary search bounds based on feasibility until minimal largest sum is found.

Why it failed: Incorrect output means binary search or greedy check logic is flawed; ensure you correctly count subarrays and update bounds.

✓ Correct minimal largest sum found using binary search and greedy feasibility check.

t1_02basic

Input{"nums":[1,4,4],"m":3}

Expected4

⏱ Performance - must finish in 2000ms

Split into [1], [4], [4], largest sums are 1,4,4; minimal largest sum is 4.

💡 When m equals nums.length, the answer is max element in nums.

💡 Check if your code handles the case where each subarray has exactly one element.

💡 Return max(nums) directly if m == len(nums) to optimize.

Why it failed: Failing this test usually means not handling the case m == nums.length correctly; fix by returning max(nums) when m equals length.

✓ Correctly handles splitting into single-element subarrays.

t2_01edge

Input{"nums":[0,0,0,0],"m":2}

Expected0

⏱ Performance - must finish in 2000ms

All zeros split into two subarrays; largest sum is 0.

💡 Zeros do not increase sums but affect how splits are counted.

💡 Ensure your greedy check counts subarrays correctly even if sums are zero.

💡 Check that sum resets properly when starting new subarray in greedy check.

Why it failed: Incorrect output means greedy check mishandles zero sums; fix by resetting current sum correctly and counting splits properly.

✓ Correctly handles zeros and counts splits accurately.

t2_02edge

Input{"nums":[10],"m":1}

Expected10

⏱ Performance - must finish in 2000ms

Single element array with one subarray; answer is the element itself.

💡 Single element with m=1 should return that element.

💡 Check base cases where nums length is 1.

💡 Avoid unnecessary binary search when array length is 1.

Why it failed: Failing this test means base case handling is missing; fix by returning nums[0] if length is 1 and m=1.

✓ Correctly returns single element for single-element input.

t2_03edge

Input{"nums":[1,2,3,4,5],"m":5}

Expected5

⏱ Performance - must finish in 2000ms

Split into single elements; largest sum is max element 5.

💡 When m equals nums.length, answer is max element.

💡 Check if your code optimizes or handles this boundary condition.

💡 Return max(nums) directly if m == len(nums).

Why it failed: Failing this test means boundary condition m == nums.length not handled; fix by returning max(nums) when m equals length.

✓ Correctly handles exact boundary condition for m.

t3_01corner

Input{"nums":[1,2,3,4,5],"m":2}

Expected9

⏱ Performance - must finish in 2000ms

Optimal split is [1,2,3,4] and [5], largest sums 10 and 5; minimal largest sum is 9 by splitting [1,2,3] and [4,5].

💡 Greedy approach that always splits when sum exceeds mid can fail if not careful.

💡 Check if your greedy check tries to minimize number of splits correctly.

💡 Adjust binary search bounds carefully and verify greedy feasibility.

Why it failed: Failing this test indicates greedy check incorrectly counts splits or binary search bounds update is wrong; fix by ensuring greedy check counts splits properly and binary search mid updates correctly.

✓ Correctly handles greedy trap and finds minimal largest sum.

t3_02corner

Input{"nums":[2,3,1,2,4,3],"m":5}

Expected4

⏱ Performance - must finish in 2000ms

Splitting into 5 subarrays, minimal largest sum is 4.

💡 Check if your code confuses 0/1 splitting with unbounded splits.

💡 Ensure greedy check counts splits exactly and does not allow more than m.

💡 Verify binary search mid and feasibility check logic.

Why it failed: Failing this test means confusion between 0/1 and unbounded splits; fix by enforcing exactly m splits or fewer in greedy check.

✓ Correctly distinguishes 0/1 splitting constraints.

t3_03corner

Input{"nums":[1,2147483647,1,2147483647],"m":2}

Expected2147483648

⏱ Performance - must finish in 2000ms

Large values test handling of sums without overflow; split as [1,2147483647] and [1,2147483647].

💡 Check for integer overflow in sum calculations.

💡 Use 64-bit integers or long types to store sums.

💡 Ensure binary search bounds and sums use appropriate data types.

Why it failed: Failing this test means integer overflow in sums; fix by using 64-bit integer types for sums and binary search bounds.

✓ Correctly handles large sums without overflow.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100000, O(n log(sum(nums))) binary search solution must complete within 2 seconds.

💡 Use binary search on answer space with O(n) greedy check.

💡 Avoid brute force or exponential solutions.

💡 Optimize sum calculations and avoid unnecessary overhead.

Why it failed: TLE indicates brute force or inefficient approach; fix by implementing binary search on answer space with O(n) greedy feasibility check.

✓ Solution runs within time limits using O(n log(sum(nums))) complexity.

Practice

1. Consider the following Python code implementing the min-heap approach to find the kth smallest element in a sorted matrix. Given the matrix [[1,3],[2,4]] and k = 3, what value does the function return?

easy

A. 2

B. 4

C. 3

D. 1

Solution

1. Step 1: Initialize min-heap with first elements

   Push (1,0,0) and (2,1,0) into heap.

2. Step 2: Extract elements until count == k

   Pop 1 (count=1), push 3 from row 0; pop 2 (count=2), push 4 from row 1; pop 3 (count=3), return 3.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Third smallest element is 3 [OK]

Hint: Trace heap pops and pushes carefully [OK]

Common Mistakes:

• Off-by-one counting of popped elements
• Returning wrong popped value
• Confusing row and column indices

2. You are given two integers m and n representing the dimensions of a multiplication table where the value at row i and column j is i * j. Given an integer k, you need to find the k-th smallest number in this table. Which approach guarantees an optimal solution for large m and n?

easy

A. Use binary search on the value range with a counting function to find how many numbers are less than or equal to a candidate number.

B. Generate all values in the multiplication table, sort them, and return the k-th element.

C. Use dynamic programming to build up the table and find the k-th smallest number by merging sorted rows.

D. Use a greedy approach by iteratively selecting the smallest unvisited product until the k-th is found.

Solution

1. Step 1: Understand the problem constraints

   The multiplication table is large (up to m*n), so generating and sorting all values is inefficient.

2. Step 2: Recognize the binary search pattern on answer space

   Binary searching the value range [1, m*n] and counting how many numbers are ≤ mid allows efficient narrowing down to the k-th smallest number.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Counting function with binary search is standard for this problem [OK]

Hint: Binary search on values, not indices [OK]

Common Mistakes:

• Trying to generate and sort entire table
• Using DP or greedy that don't scale

3. Consider the following buggy code snippet for searching a target in a rotated sorted array with duplicates. Which line contains the subtle bug that can cause infinite loops or incorrect results?

medium

A. Line 9: if nums[mid] == target: return True

B. Line 7: mid = (left + right) // 2

C. Line 13: if nums[left] <= target < nums[mid]: right = mid - 1

D. Line 4: while left < right and nums[left] == nums[left + 1]: left += 1

Solution

1. Step 1: Identify duplicate skipping logic

   The code only skips duplicates from the left side, ignoring duplicates on the right side.

2. Step 2: Understand impact of skipping duplicates only on one side

   If duplicates exist on the right side, pointers may not move correctly, causing infinite loops or incorrect search decisions.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Skipping duplicates only on one side breaks correctness [OK]

Hint: Must skip duplicates from both ends to avoid infinite loops [OK]

Common Mistakes:

• Skipping duplicates only from left or right
• Not checking target at mid first
• Incorrect pointer updates

4. Suppose the problem is modified so that the multiplication table allows repeated use of the same row and column indices (i.e., infinite repetitions of each product). Which of the following changes to the algorithm is correct to find the k-th smallest number under this new condition?

hard

A. Use the same binary search with the counting function unchanged, since repetition doesn't affect counts.

B. Switch to generating all products up to k and sorting them, since binary search no longer applies.

C. Use a min-heap to generate products in ascending order with duplicates allowed, extracting k elements.

D. Modify the counting function to remove the cap by n, counting all multiples less than or equal to x for each i.

Solution

1. Step 1: Understand the new problem condition

   Allowing infinite repetitions means the multiplication table is conceptually infinite, so counting by rows capped at n no longer applies.

2. Step 2: Analyze algorithmic implications

   Binary search with counting function fails because counts become infinite or undefined; generating all values is impossible.

3. Step 3: Use a min-heap approach

   A min-heap can generate products in ascending order with duplicates allowed, extracting the k-th smallest efficiently.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Min-heap approach handles infinite repetitions correctly [OK]

Hint: Infinite repetitions break counting; use min-heap [OK]

Common Mistakes:

• Assuming counting function unchanged
• Trying to generate all values

5. Suppose now you want to make bouquets but flowers can be reused multiple times (i.e., each flower can be counted multiple times for different bouquets). Which modification to the algorithm is correct?

hard

A. Keep the same binary search and feasibility check, but remove resetting flowers count to zero on encountering a flower that blooms after the day.

B. Binary search is no longer applicable; instead, use a greedy approach to pick flowers repeatedly until bouquets are formed.

C. Use binary search on days, but feasibility must count bouquets by grouping any flowers bloomed by day, allowing reuse by not resetting flowers count.

D. Modify feasibility to count total flowers bloomed by day and check if total flowers >= m * k, ignoring adjacency.

Solution

1. Step 1: Understand reuse changes adjacency constraint

   If flowers can be reused multiple times, adjacency no longer matters; only total count of bloomed flowers by day matters.

2. Step 2: Modify feasibility check

   Feasibility reduces to checking if total bloomed flowers by day >= m * k, ignoring adjacency and resetting.

3. Step 3: Binary search remains valid

   Binary search on days still applies, but feasibility is simplified to total count check.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Counting total bloomed flowers matches reuse scenario. [OK]

Hint: Reuse removes adjacency; feasibility counts total bloomed flowers [OK]

Common Mistakes:

• Ignoring adjacency but still resetting flowers count
• Removing binary search unnecessarily