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 :
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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