Approach Summary
Binary search on the answer (largest sum). Feasibility: count how many partitions are needed with given max sum — must be ≤ k.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Split Array Largest Sum optimally, we apply the binary search algorithmic strategy. Binary search on the answer (largest sum). Feasibility: count how many partitions are needed with given max sum — must be ≤ k. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n log(sum)) time, utilizing O(1) auxiliary memory.
Time Complexity: O(n log(sum)) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
class Solution {
// Time: O(n log(sum)), Space: O(1)
public int splitArrayLargestSum(int[] nums, int target) {
int low = 0, high = nums.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
}
}#include <vector>
using namespace std;
class Solution {
public:
// Time: O(n log(sum)), Space: O(1)
int splitArrayLargestSum(vector<int>& nums, int target) {
int low = 0, high = nums.size() - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
}
};class Solution:
def split_array_largest_sum(self, nums, target=None) -> int:
# Time: O(n log(sum)), Space: O(1)
low, high = 0, len(nums) - 1
ans = -1
while low <= high:
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
low = mid + 1
else:
high = mid - 1
return ansvar splitArrayLargestSum = function(nums, target) {
let low = 0, high = nums.length - 1;
while (low <= high) {
const mid = Math.floor(low + (high - low) / 2);
if (nums[mid] === target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
};Edge Cases to Watch
- Key recognition trigger: "Minimise the maximum subarray sum" with k splits.
- Constraint condition: Monotonic feasibility → binary search on answer.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n log(sum)) time budget without stack overflow or time limit exceeded.
Master the Binary Search Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Binary Search pattern.
How to Recognize This Pattern
- "Minimise the maximum subarray sum" with k splits
- Monotonic feasibility → binary search on answer
Complexity Analysis
Time Complexity
O(n log(sum))
Space Complexity
O(1)