Split List to Minimize Largest Sum - Amazon Top Interview Questions


Problem Statement :


Given a list of non-negative integers nums and an integer k, you can split the list into k non-empty sublists.

Return the minimum largest sum of the k sublists.

Constraints

k ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 3, 2, 4, 9]

k = 2

Output

10

Explanation

We can split the list into these 2 sublists: [1, 3, 2, 4] and [9].


Solution :



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                        Solution in C++ :

int sublistCount(vector<int>& nums, long long threshold) {
    int ret = 1;
    long long leftover = threshold;
    for (int val : nums) {
        if (val > leftover) {
            ret++;
            leftover = threshold;
        }
        leftover -= val;
    }
    return ret;
}

int solve(vector<int>& nums, int k) {
    long long lhs = *max_element(nums.begin(), nums.end());
    long long rhs = accumulate(nums.begin(), nums.end(), 0LL);
    while (lhs < rhs) {
        long long mid = (lhs + rhs) / 2;
        if (sublistCount(nums, mid) <= k)
            rhs = mid;
        else
            lhs = mid + 1;
    }
    return lhs;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int k) {
        int max = 0, sum = 0;
        for (int i : nums) {
            max = Math.max(i, max);
            sum += i;
        }
        int l = max, h = sum;
        while (l <= h) {
            int m = (l + h) >>> 1;
            if (can(nums, m, k))
                h = m - 1;
            else
                l = m + 1;
        }
        return l;
    }

    private boolean can(int[] nums, final int MAX, final int K) {
        int groups = 0, sum = -1;
        for (int i : nums) {
            if (sum + i > MAX)
                sum = -1;

            if (sum == -1) {
                groups++;
                sum = 0;
            }

            sum += i;
        }
        return groups <= K;
    }
}
                    


                        Solution in Python : 
                            
def parse(x):
    i = 0
    cnt = 0 # counts the number of sublists
    while i < n:
        tempX = x  
        while i < n and tempX - nums[i] >= 0:
            tempX -= nums[i]
            i += 1
        cnt +=1
    return cnt <= k


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