LeetCode 410: Split Array Largest Sum Solution

Master LeetCode problem 410 (Split Array Largest Sum), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

410. Split Array Largest Sum

Hard

Array Binary Search Dynamic Programming Greedy Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use binary search on the answer. We define the search space as the range from the maximum value in the input array (max(nums)) to the sum of all elements in the input array (sum(nums)). We then perform binary search within this range to find the minimum largest sum of the split.

For each mid value in the binary search, we check if we can split the array into k subarrays such that none of the subarray sums exceed the mid value. If we can split the array into k or fewer subarrays, we move towards smaller values in the binary search. If we need more than k subarrays, we move towards larger values.

In each iteration of the binary search, we simulate the splitting process by iterating through the array and keeping track of the current sum. If the sum exceeds the mid value, we increment the number of subarrays and reset the sum to the current element's value.

Time complexity: O(n * log(sum(nums) - max(nums))) Space complexity: O(1)

Solution Code

class Solution {
    public int splitArray(int[] nums, int k) {
        int left = 0, right = 0;
        for (int num : nums) {
            left = Math.max(left, num);
            right += num;
        }
        
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (isValid(nums, k, mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        
        return left;
    }
    
    private boolean isValid(int[] nums, int k, int mid) {
        int count = 1;
        int sum = 0;
        
        for (int num : nums) {
            sum += num;
            if (sum > mid) {
                count++;
                sum = num;
            }
        }
        
        return count <= k;
    }
}

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