# Split Array Largest Sum

This page explains Java solution to problem `Split Array Largest Sum` using `Binary Search` algorithm.

## Problem Statement

Given an array `nums` which consists of non-negative integers and an integer `m`, you can split the array into `m` non-empty continuous subarrays.

Write an algorithm to minimize the largest sum among these `m` subarrays.

Example 1:

Input: nums = [7,2,5,10,8], m = 2
Output: 18
Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

Example 2:

Input: nums = [1,2,3,4,5], m = 2
Output: 9

Example 3:

Input: nums = [1,4,4], m = 3
Output: 4

## Solution

If you have any suggestions in below code, please create a pull request by clicking here.
``````
package com.vc.hard;

class SplitArrayLargestSum {
public int splitArray(int[] nums, int m) {
if(nums == null || nums.length == 0) return 0;

int lo = Integer.MIN_VALUE;
int hi = 0;
for(int i = 0; i < nums.length; i++) {
lo = Math.max(lo, nums[i]);
hi += nums[i];
}

if(m == 1) return hi;

while(lo < hi) {
int mid = lo + (hi - lo) / 2;
if(isMorePartitions(nums, mid, m)) lo = mid + 1;
else hi = mid;
}
return lo;
}

private boolean isMorePartitions(int[] nums, int sum, int expectedPartitions) {
int total = 0;
int partitions = 1;
for(int i = 0; i < nums.length; i++) {
total += nums[i];
if(total > sum) {
total = nums[i];
partitions++;
if(partitions > expectedPartitions) return true;
}
}
return false;
}
}
``````

## Time Complexity

O(N * log(HI - LO)) Where
N is total number of elements in an input array
Lo is max element in an input array
HI is sum of elements in an input array

O(1)