K Stack Pops - Google Top Interview Questions


Problem Statement :


You are given two-dimensional list of integers stacks and an integer k. 

Assuming each list in stacks represents a stack, return the maximum possible sum that can be achieved from popping off exactly k elements from any combination of the stacks.

Constraints

n ≤ 500 where n is the number of rows in stacks.

m ≤ 200 where m is the maximum number of elements in a stack.

k ≤ 100

Example 1

Input

stacks = [

    [100, -3, -10],

    [1],

    [4, 5, 6]

]

k = 4

Output

93

Explanation

We can pop off all 3 elements from the first stack and pop the last element of the last stack to get -10 
+ -3 + 100 + 6.


Example 2
Input

stacks = [

    [1, 2, 4],

    [2, 3, 5],

    [3, 0, 6]

]

k = 3

Output

15

Explanation

Pop off the 4, 5 and 6 from each of the stacks.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<vector<int>>& rows, int k) {
    vector<int> dp(k + 1, -1e9);
    dp[0] = 0;

    for (const auto& nums : rows) {
        int n = nums.size();
        auto dpc = dp;

        for (int i = 1; i <= k; ++i) {
            int sum = 0;
            for (int j = 1; j <= min(i, n); ++j)
                dpc[i] = max(dpc[i], (sum += nums[n - j]) + dp[i - j]);
        }

        dp = dpc;
    }

    return dp[k];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    Integer[][] dp;
    List<List<Integer>> total;
    public int solve(int[][] stacks, int k) {
        dp = new Integer[stacks.length][k + 1];
        total = new ArrayList();
        for (int i = 0; i < stacks.length; i++) {
            List<Integer> list = new ArrayList();
            list.add(0);
            for (int j = stacks[i].length - 1; j > -1 && j >= stacks[i].length - k; j--) {
                list.add(stacks[i][j] + list.get(list.size() - 1));
            }
            total.add(list);
        }
        return recur(0, k);
    }
    public int recur(int ind, int k) {
        if (k == 0)
            return 0;
        if (ind == dp.length || k < 0)
            return -1 * (int) (1e7);
        if (dp[ind][k] != null)
            return dp[ind][k];
        int a = -1 * (int) (1e7);
        for (int i = 0; i < total.get(ind).size(); i++) {
            a = Math.max(a, total.get(ind).get(i) + recur(ind + 1, k - i));
        }
        dp[ind][k] = a;
        return a;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, stacks, k):
        NINF = float("-inf")
        dp = [NINF] * (k + 1)
        dp[0] = 0

        for stack in stacks:
            P = [0]
            for x in reversed(stack):
                P.append(P[-1] + x)

            for j in range(k, 0, -1):
                for i in range(1, min(j + 1, len(P))):
                    dp[j] = max(dp[j], dp[j - i] + P[i])

        return dp[k]


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